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Precalculus
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Learner’s Material
This learning resource was collaboratively developed and reviewed by educators from public and private schools, colleges, and/or universities. We encourage teachers and other education stakeholders to email their feedback, comments and recommendations to the Department of Education at
[email protected].
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We value your feedback and recommendations.
Department of Education Republic of the Philippines
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Precalculus Learner’s Material First Edition 2016 Republic Act 8293. Section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.
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Published by the Department of Education Secretary: Br. Armin A. Luistro FSC Undersecretary: Dina S. Ocampo, PhD
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Development Team of the Precalculus Learner’s Material Joy P. Ascano Jesus Lemuel L. Martin Jr. Arnel D. Olofernes Mark Anthony C. Tolentino, Ph.D Reviewers: Jerico B. Bacani, Ph.D Richard B. Eden, Ph.D Dr. Flordeliza F. Francisco Mark Anthony J. Vidallo Carly Mae Casteloy Angela Dianne Agustin Cover Art Illustrator: Quincy D. Gonzales Team Leader: Ian June L. Garces, Ph.D. Management Team of the Precalculus Learner’s Material Bureau of Curriculum Development Bureau of Learning Resources
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Table of Contents 1
DepEd Curriculum Guide for Precalculus
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Unit 1:
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To the Precalculus Learners
Analytic Geometry
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1.1.1: An Overview of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Lesson 1.1: Introduction to Conic Sections and Circles . . . . . . . . 1.1.2: Definition and Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . .
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1.1.3: More Properties of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.1.4: Situational Problems Involving Circles. . . . . . . . . . . . . . . . . . . . 12
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Lesson 1.2: Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.2.1: Definition and Equation of a Parabola . . . . . . . . . . . . . . . . . . . . 19
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1.2.2: More Properties of Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 26
Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 1.3.1: Definition and Equation of an Ellipse . . . . . . . . . . . . . . . . . . . . . 33 1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
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1.3.3: Situational Problems Involving Ellipses . . . . . . . . . . . . . . . . . . . 40
Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 1.4.1: Definition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 46 1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 54 Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 60 1.5.1: Identifying the Conic Section by Inspection . . . . . . . . . . . . . . . 60 1.5.2: Problems Involving Different Conic Sections . . . . . . . . . . . . . . 62
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Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 67 1.6.1: Review of Techniques in Solving Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 69 1.6.3: Solving Systems of Equations Using Elimination . . . . . . . . . . 70 1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 73 Unit 2:
Mathematical Induction
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Lesson 2.1: Review of Sequences and Series . . . . . . . . . . . . . . . . . . . 81 Lesson 2.2: Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 2.2.1: Writing and Evaluating Sums in Sigma Notation . . . . . . . . . 87
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2.2.2: Properties of Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Lesson 2.3: Principle of Mathematical Induction . . . . . . . . . . . . . . 96 2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.3.2: Proving Divisibility Statements. . . . . . . . . . . . . . . . . . . . . . . . . . . 101
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? 2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
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2.4.1: Pascal’s Triangle and the Concept of Combination . . . . . . . . 109 2.4.2: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 2.4.3: Terms of a Binomial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 114
? 2.4.4: Approximation and Combination Identities . . . . . . . . . . . . . . . 116 Trigonometry
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Unit 3:
Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 129 Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 136 3.2.2: Reference Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
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Lesson 3.3: Graphs of Circular Functions and Situational Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 3.3.1: Graphs of y = sin x and y = cos x . . . . . . . . . . . . . . . . . . . . . . . . 145 3.3.2: Graphs of y = a sin bx and y = a cos bx . . . . . . . . . . . . . . . . . . . 147 3.3.3: Graphs of y = a sin b(x − c) + d and y = a cos b(x − c) + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 154 3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 158
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3.3.6: Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Lesson 3.4: Fundamental Trigonometric Identities . . . . . . . . . . . . . 171 3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 171
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3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 173 3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 174 3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 176 Lesson 3.5: Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . . 181
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3.5.1: The Cosine Difference and Sum Identities . . . . . . . . . . . . . . . . 181 3.5.2: The Cofunction Identities and the Sine Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
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3.5.3: The Tangent Sum and Difference Identities . . . . . . . . . . . . . . . 186
Lesson 3.6: Double-Angle and Half-Angle Identities . . . . . . . . . . . 192 3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
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Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 201 3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 3.7.3: Inverse Tangent Function and the Remaining Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Lesson 3.8: Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 220 3.8.1: Solutions of a Trigonometric Equation . . . . . . . . . . . . . . . . . . . . 221 3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 227
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Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 236 3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 3.9.2: From Polar to Rectangular, and Vice Versa . . . . . . . . . . . . . . . 241 3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 244 Answers to Odd-Numbered Exercises in Supplementary Problems and All Exercises in Topic Tests 255 290
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References
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
To the Precalculus Learners
The Precalculus course bridges basic mathematics and calculus. This course completes your foundational knowledge on algebra, geometry, and trigonometry. It provides you with conceptual understanding and computational skills that are prerequisites for Basic Calculus and future STEM courses.
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Based on the Curriculum Guide for Precalculus of the Department of Education (see pages 2-5), the primary aim of this Learning Manual is to give you an adequate stand-alone material that can be used for the Grade 11 Precalculus course.
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The Manual is divided into three units: analytic geometry, summation notation and mathematical induction, and trigonometry. Each unit is composed of lessons that bring together related learning competencies in the unit. Each lesson is further divided into sub-lessons that focus on one or two competencies for effective learning.
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At the end of each lesson, more examples are given in Solved Examples to reinforce the ideas and skills being developed in the lesson. You have the opportunity to check your understanding of the lesson by solving the Supplementary Problems. Finally, two sets of Topic Test are included to prepare you for the exam.
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Answers, solutions, or hints to odd-numbered items in the Supplementary Problems and all items in the Topic Tests are provided at the end of the Manual to guide you while solving them. We hope that you will use this feature of the Manual responsibly. Some items are marked with a star. A starred sub-lesson means the discussion and accomplishment of the sub-lesson are optional. This will be decided by your teacher. On the other hand, a starred example or exercise means the use of calculator is required.
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We hope that you will find this Learning Manual helpful and convenient to use. We encourage you to carefully study this Manual and solve the exercises yourselves with the guidance of your teacher. Although great effort has been put into this Manual for technical correctness and precision, any mistake found and reported to the Team is a gain for other students. Thank you for your cooperation. The Precalculus LM Team
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016. Semester: First Semester No. of Hours/ Semester: 80 hours/ semester Pre-requisite (if needed):
key concepts of conic sections and systems of nonlinear equations
The learners demonstrate an understanding of...
CONTENT STANDARDS
model situations appropriately and solve problems accurately using conic sections and systems of nonlinear equations
The learners shall be able to...
PERFORMANCE STANDARDS
D LEARNING COMPETENCIES
5. 6. 7. 8. 9. 10. 11. 12.
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graph a circle in a rectangular coordinate system
4.
define a parabola determine the standard form of equation of a parabola graph a parabola in a rectangular coordinate system define an ellipse determine the standard form of equation of an ellipse graph an ellipse in a rectangular coordinate system define a hyperbola determine the standard form of equation of a hyperbola
determine the standard form of equation of a circle
3.
2.
illustrate the different types of conic sections: parabola, ellipse, circle, hyperbola, and degenerate cases.*** define a circle.
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The learners...
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Analytic Geometry
CONTENT
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STEM_PC11AG-Ia-5 STEM_PC11AG-Ib-1 STEM_PC11AG-Ib-2 STEM_PC11AG-Ic-1 STEM_PC11AG-Ic-2 STEM_PC11AG-Ic-3 STEM_PC11AG-Id-1 STEM_PC11AG-Id-2
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Subject Description: At the end of the course, the students must be able to apply concepts and solve problems involving conic sections, systems of nonlinear equations, series and mathematical induction, circular and trigonometric functions, trigonometric identities, and polar coordinate system.
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Grade: 11 Core Subject Title: Pre-Calculus
K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
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All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
key concepts of series and mathematical induction and the Binomial Theorem.
keenly observe and investigate patterns, and formulate appropriate mathematical statements and prove them using mathematical induction and/or Binomial Theorem.
PERFORMANCE STANDARDS
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CONTENT STANDARDS
recognize the equation and important characteristics of the different types of conic sections solves situational problems involving conic sections
14. 15.
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differentiate a series from a sequence use the sigma notation to represent a series illustrate the Principle of Mathematical Induction apply mathematical induction in proving identities illustrate Pascal’s Triangle in the expansion of 𝑥 + 𝑦 𝑛 for small positive integral values of 𝑛 prove the Binomial Theorem determine any term of 𝑥 + 𝑦 𝑛 , where 𝑛 is a positive integer, without expanding solve problems using mathematical induction and the Binomial Theorem
2. 3. 4. 5. 6.
9.
7. 8.
illustrate a series
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determine the solutions of systems of nonlinear equations using techniques such as substitution, elimination, and graphing*** solve situational problems involving systems of nonlinear equations
17.
18.
illustrate systems of nonlinear equations
16.
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graph a hyperbola in a rectangular coordinate system
LEARNING COMPETENCIES 13.
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Series and Mathematical Induction
CONTENT
STEM_PC11SMI-Ih-1
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STEM_PC11SMI-Ii-2
STEM_PC11SMI-Ih-2 STEM_PC11SMI-Ih-3 STEM_PC11SMI-Ih-4 STEM_PC11SMI-Ih-i-1
K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
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All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016. 3. formulate and solve accurately situational problems involving appropriate trigonometric functions 4. formulate and solve accurately situational problems involving the polar coordinate system
uses reference angles to find exact values of circular functions determine the domain and range of the different circular functions graph the six circular functions (a) amplitude, (b) period, and (c) phase shift solve problems involving circular functions determine whether an equation is an identity or a conditional equation derive the fundamental trigonometric identities derive trigonometric identities involving sum and difference of angles derive the double and half-angle formulas simplify trigonometric expressions prove other trigonometric identities solve situational problems involving trigonometric identities illustrate the domain and range of the inverse trigonometric functions. evaluate an inverse trigonometric expression. solve trigonometric equations. solve situational problems involving inverse trigonometric functions and trigonometric equations locate points in polar coordinate system convert the coordinates of a point from rectangular to polar systems and vice versa solve situational problems involving polar coordinate system
4. 5. 6. 7.
22.
20. 21.
17. 18. 19.
12. 13. 14. 15. 16.
10. 11.
8. 9.
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
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illustrate the different circular functions
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illustrate angles in standard position and coterminal angles
2.
LEARNING COMPETENCIES illustrate the unit circle and the relationship between the linear and angular measures of a central angle in a unit circle convert degree measure to radian measure and vice versa
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2. apply appropriate trigonometric identities in solving situational problems
PERFORMANCE STANDARDS 1. formulate and solve accurately situational problems involving circular functions
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CONTENT STANDARDS key concepts of circular functions, trigonometric identities, inverse trigonometric functions, and the polar coordinate system
***Suggestion for ICT-enhanced lesson when available and where appropriate
Trigonometry
CONTENT
K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – SCIENCE, TECHNOLOGY, ENGINEERING AND MATHEMATICS (STEM) SPECIALIZED SUBJECT
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STEM_PC11T-IIj-3
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STEM_PC11T-IIh-2 STEM_PC11T-IIh-i-1
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STEM_PC11T-IIf-1 STEM_PC11T-IIf-2 STEM_PC11T-IIf-g-1 STEM_PC11T-IIg-2
STEM_PC11T-IIe-3
STEM_PC11T-IIe-2
STEM_PC11T-IIe-1
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STEM_PC11T-IIb-2 STEM_PC11T-IIc-1
STEM_PC11T-IIb-1
STEM_PC11T-IIa-3
STEM_PC11T-IIa-1 STEM_PC11T-IIa-2
CODE
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All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016. Competency
Week
Quarter
Domain/Content/ Component/ Topic
Grade Level
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SAMPLE
Science, Technology, Engineering and Mathematics Pre-Calculus
illustrate the different types of conic sections: parabola, ellipse, circle, hyperbola, and degenerate cases
Week one
First Quarter
Analytic Geometry
Grade 11
Code Book Legend
1
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a
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STEM_PC11AG
DOMAIN/ COMPONENT
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Trigonometry
Series and Mathematical Induction
Analytic Geometry
Sample: STEM_PC11AG-Ia-1
K to 12 Senior High School STEM Specialized Subject – Pre-Calculus December 2013
Arabic Number
*Put a hyphen (-) in between letters to indicate more than a specific week
Lowercase Letter/s
*Zero if no specific quarter
Roman Numeral
Uppercase Letter/s
First Entry
Learning Area and Strand/ Subject or Specialization
LEGEND
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Unit 1
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Analytic Geometry
San Juanico Bridge, by Morten Nærbøe, 21 June 2009,
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https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG. Public Domain.
Stretching from Samar to Leyte with a total length of more than two kilometers, the San Juanico Bridge has been serving as one of the main thoroughfares of economic and social development in the country since its completion in 1973. Adding picturesque effect on the whole architecture, geometric structures are subtly built to serve other purposes. The arch-shaped support on the main span of the bridge helps maximize its strength to withstand mechanical resonance and aeroelastic flutter brought about by heavy vehicles and passing winds.
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Lesson 1.1. Introduction to Conic Sections and Circles Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) illustrate the different types of conic sections: parabola, ellipse, circle, hyperbola, and degenerate cases; (2) define a circle;
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(3) determine the standard form of equation of a circle; (4) graph a circle in a rectangular coordinate system; and
(5) solve situational problems involving conic sections (circles). Lesson Outline (2) Definition of a circle
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(1) Introduction of the four conic sections, along with the degenerate conics (3) Derivation of the standard equation of a circle (4) Graphing circles
Introduction
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(5) Solving situational problems involving circles
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We present the conic sections, a particular class of curves which sometimes appear in nature and which have applications in other fields. In this lesson, we first illustrate how each of these curves is obtained from the intersection of a plane and a cone, and then discuss the first of their kind, circles. The other conic sections will be covered in the next lessons.
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1.1.1. An Overview of Conic Sections We introduce the conic sections (or conics), a particular class of curves which oftentimes appear in nature and which have applications in other fields. One of the first shapes we learned, a circle, is a conic. When you throw a ball, the trajectory it takes is a parabola. The orbit taken by each planet around the sun is an ellipse. Properties of hyperbolas have been used in the design of certain telescopes and navigation systems. We will discuss circles in this lesson, leaving parabolas, ellipses, and hyperbolas for subsequent lessons. • Circle (Figure 1.1) - when the plane is horizontal • Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to form a bounded curve 7 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
• Parabola (Figure 1.2) - when the plane intersects only one cone to form an unbounded curve
Figure 1.1
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• Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersects both cones to form two unbounded curves (each called a branch of the hyperbola)
Figure 1.2
Figure 1.3
We can draw these conic sections (also called conics) on a rectangular coordinate plane and find their equations. To be able to do this, we will present equivalent definitions of these conic sections in subsequent sections, and use these to find the equations.
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There are other ways for a plane and the cones to intersect, to form what are referred to as degenerate conics: a point, one line, and two lines. See Figures 1.4, 1.5 and 1.6.
Figure 1.4
Figure 1.5
Figure 1.6
1.1.2. Definition and Equation of a Circle A circle may also be considered a special kind of ellipse (for the special case when the tilted plane is horizontal). As we get to know more about a circle, we will also be able to distinguish more between these two conics.
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See Figure 1.7, with the point C(3, 1) shown. From the figure, the distance of A(−2, 1) from p C is AC = 5. By the distance formula, the distance of B(6, 5) from C is BC = (6 − 3)2 + (5 − 1)2 = 5. There are other points P such that P C = 5. The collection of all such points which are 5 units away from C, forms a circle.
Figure 1.7
Figure 1.8
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Let C be a given point. The set of all points P having the same distance from C is called a circle. The point C is called the center of the circle, and the common distance its radius. The term radius is both used to refer to a segment from the center C to a point P on the circle, and the length of this segment.
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See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0. A point P (x, y) is on the circle if and only if P C = r. For any such point then, its coordinates should satisfy the following.
p
PC = r
(x − h)2 + (y − k)2 = r (x − h)2 + (y − k)2 = r2
This is the standard equation of the circle with center C(h, k) and radius r. If the center is the origin, then h = 0 and k = 0. The standard equation is then x2 + y 2 = r 2 .
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Example 1.1.1. In each item, give the standard equation of the circle satisfying the given conditions. (1) center at the origin, radius 4 √ (2) center (−4, 3), radius 7 (3) circle in Figure 1.7 (4) circle A in Figure 1.9 (5) circle B in Figure 1.9 (6) center (5, −6), tangent to the yaxis (7) center (5, −6), tangent to the x-axis
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Figure 1.9
Solution. (1) x2 + y 2 = 16 (2) (x + 4)2 + (y − 3)2 = 7
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(8) It has a diameter with endpoints A(−1, 4) and B(4, 2).
(3) The center is (3, 1) and the radius is 5, so the equation is (x − 3)2 + (y − 1)2 = 25.
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(4) By inspection, the center is (−2, −1) and the radius is 4. The equation is (x + 2)2 + (y + 1)2 = 16. (5) Similarly by inspection, we have (x − 3)2 + (y − 2)2 = 9.
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(6) The center is 5 units away from the y-axis, so the radius is r = 5 (you can make a sketch to see why). The equation is (x − 5)2 + (y + 6)2 = 25.
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(7) Similarly, since the center is 6 units away from the x-axis, the equation is (x − 5)2 + (y + 6)2 = 36. , 4+2 (8) The center C is the midpoint of A and B: C = −1+4 = 32 , 3 . The 2 2 q q 2 29 radius is then r = AC = −1 − 32 + (4 − 3)2 = . The circle has 4 2 equation x − 23 + (y − 3)2 = 29 . 2 4 1.1.3. More Properties of Circles
After expanding, the standard equation 2 3 29 x− + (y − 3)2 = 2 4 can be rewritten as x2 + y 2 − 3x − 6y + 4 = 0, an equation of the circle in general form. 10 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
If the equation of a circle is given in the general form Ax2 + Ay 2 + Cx + Dy + E = 0,
A 6= 0,
or x2 + y 2 + Cx + Dy + E = 0, we can determine the standard form by completing the square in both variables.
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Completing the square in an expression like x2 + 14x means determining the term to be added that will produce a perfect polynomial square. Since the coefficient of x2 is already 1, we take half the coefficient of x and square it, and we get 49. Indeed, x2 + 14x + 49 = (x + 7)2 is a perfect square. To complete the square in, say, 3x2 + 18x, we factor the coefficient of x2 from the expression: 3(x2 + 6x), then add 9 inside. When completing a square in an equation, any extra term introduced on one side should also be added to the other side.
(1) x2 + y 2 − 6x = 7 (2) x2 + y 2 − 14x + 2y = −14
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Example 1.1.2. Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center.
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(3) 16x2 + 16y 2 + 96x − 40y = 315
(1)
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Solution. The first step is to rewrite each equation in standard form by completing the square in x and in y. From the standard equation, we can determine the center and radius.
x2 − 6x + y 2 = 7 x2 − 6x + 9 + y 2 = 7 + 9 (x − 3)2 + y 2 = 16
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Center (3, 0), r = 4, Figure 1.10
(2)
x2 − 14x + y 2 + 2y = −14 x2 − 14x + 49 + y 2 + 2y + 1 = −14 + 49 + 1 (x − 7)2 + (y + 1)2 = 36 Center (7, −1), r = 6, Figure 1.11 (3) 16x2 + 96x + 16y 2 − 40y = 315 11 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
5 16(x + 6x) + 16 y − y = 315 2 25 25 5 2 2 = 315 + 16(9) + 16 16(x + 6x + 9) + 16 y − y + 2 16 16 2 5 16(x + 3)2 + 16 y − = 484 4 2 2 5 484 121 11 2 (x + 3) + y − = = = 4 16 4 2 Center −3, 54 , r = 5.5, Figure 1.12. 2
2
Figure 1.10
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2
Figure 1.11
Figure 1.12
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In the standard equation (x − h)2 + (y − k)2 = r2 , both the two squared terms on the left side have coefficient 1. This is the reason why in the preceding example, we divided by 16 at the last equation. 1.1.4. Situational Problems Involving Circles Let us now take a look at some situational problems involving circles.
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? Example 1.1.3. A street with two lanes, each 10 ft wide, goes through a semicircular tunnel with radius 12 ft. How high is the tunnel at the edge of each lane? Round off to 2 decimal places.
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Solution. We draw a coordinate system with origin at the middle of the highway, as shown. Because of the given radius, the tunnel’s boundary is on the circle x2 + y 2 = 122 . Point P is the point on the arc just above the edge of a lane, so its x-coordinate is 10. We√need its y-coordinate. We then solve 102 + y 2 = 122 2 for y > 0, giving us y = 2 11 ≈ 6.63 ft.
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Example 1.1.4. A piece of a broken plate was dug up in an archaeological site. It was put on top of a grid, as shown in Figure 1.13, with the arc of the plate passing through A(−7, 0), B(1, 4) and C(7, 2). Find its center, and the standard equation of the circle describing the boundary of the plate.
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Figure 1.13
Figure 1.14 Solution. We first determine the center. It is the intersection of the perpendicular bisectors of AB and BC (see Figure 1.14). Recall that, in a circle, the perpendicular bisector of any the center. Since the midpoint M chord passes through4−0 0+4 of AB is −7+1 , = (−3, 2), and m = = 12 , the perpendicular bisector AB 2 2 1+7 of AB has equation y − 2 = −2(x + 3), or equivalently, y = −2x − 4. 13 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
4+2 Since the midpoint N of BC is 1+7 , = (4, 3), and mBC = 2−4 = − 13 , 2 2 7−1 the perpendicular bisector of BC has equation y − 3 = 3(x − 4), or equivalently, y = 3x − 9. The intersection of the two lines y = 2x − 4 and y = 3x − 9 is (1, −6) (by solving a system of linear equations). We can take the radius as the distance of this point from any of A, B or C (it’s most convenient to use B in this case). We then get r = 10. The standard equation is thus (x − 1)2 + (y + 6)2 = 100. 2
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More Solved Examples
(b) center (1, 5), diameter 8 (c) circle A in Figure 1.15 (d) circle B in Figure 1.15
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(e) circle C in Figure 1.15
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1. In each item, give the standard equation of the circle satisying the given conditions. (a) center at the origin, contains (0, 3)
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(f) center (−2, −3), tangent to the yaxis (g) center (−2, −3), tangent to the xaxis Figure 1.15
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(h) contains the points (−2, 0) and (8, 0), radius 5 Solution: (a) The radius is 3, so the equation is x2 + y 2 = 9.
(b) The radius is 8/2 = 4, so the equation is (x − 1)2 + (y − 5)2 = 16.
(c) The center is (−2, 2) and the radius is 2, so the equation is (x + 2)2 + (y − 2)2 = 4. (d) The center is (2, 3) and the radius is 1, so the equation is (x − 2)2 + (y − 3)2 = 1. (e) The center is (1, −1) and by the Pythagorean Theorem, the radius (see √ √ 2 2 Figure 1.16) is 2 + 2 = 8, so the equation is (x − 1)2 + (x + 1)2 = 8.
Figure 1.16
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(f) The radius is 3, so the equation is (x + 2)2 + (y + 3)2 = 9. (g) The radius is 2, so the equation is (x + 2)2 + (y + 3)2 = 4. (h) The distance between (−2, 0) and (8, 0) is 10; since the radius is 5, these two points are endpoints of a diameter. Then the circle has center at (3, 0) and radius 5, so its equation is (x − 3)2 + y 2 = 25. 2. Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center. (a) x2 + y 2 + 8y = 33 (c) 4x2 + 12x + 4y 2 + 16y − 11 = 0 Solution:
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(a)
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(b) 4x2 + 4y 2 − 16x + 40y + 67 = 0
x2 + y 2 + 8y = 33 x2 + y 2 + 8y + 16 = 33 + 16 x2 + (y + 4)2 = 49 Center (0, −4), radius 7, see Figure 1.17.
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(b)
4x2 + 4y 2 − 16x + 40y + 67 = 0 67 4 67 x2 − 4x + 4 + y 2 + 10y + 25 = − + 4 + 25 4 2 49 7 2 2 (x − 2) + (y + 5) = = 4 2
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x2 − 4x + y 2 + 10y = −
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Center (2, −5), radius 3.5, see Figure 1.18.
(c)
4x2 + 12x + 4y 2 + 16y − 11 = 0 11 x2 + 3x + y 2 + 4y = 4 9 11 9 x2 + 3x + + y 2 + 4y + 4 = + +4 4 4 4 2 3 x+ + (y + 2)2 = 9 2 3 Center − , −2 , radius 3, see Figure 1.19. 2 15
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Figure 1.19
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Figure 1.17 Figure 1.18 3. A circular play area with radius 3 m is to be partitioned into two sections using a straight fence as shown in Figure 1.20. How long should the fence be? Solution: To determine the length of the fence, we need to determine the coordinates of its endpoints. From Figure 1.20, the endpoints have x coordinate −1 and are on the circle x2 + y√2 = 9. Then 1 + y 2 = 9, or y = ±2 2. √ Therefore, the length of the fence is 4 2 ≈ 5.66 m.
Figure 1.20
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4. A Cartesian coordinate system was used to identify locations on a circular track. As shown in Figure 1.21, the circular track contains the points A(−2, −4), B(−2, 3), C(5, 2). Find the total length of the track.
Figure 1.21
Figure 1.22
Solution: The segment AB is vertical and has midpoint (−2, −0.5), so its perpendicular bisector has equation y = −0.5. On the other hand, the segment BC has slope −1/7 and midpoint (1.5, 2.5), so its perpendicular bisector has equation y − 2.5 = 7(x − 1.5), or 7x − y − 8 = 0. The center of the circle is the intersection of y = −0.5 and 7x − y − 8 = 0; 15 that is, the center is at 14 , − 12 . The radius of the circle is the distance from ther center to any of the points A, 2125 5√ B, or C; by the distance formula, the radius is = 170. Therefore, 98 14 16 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
the total length of the track (its circumference), is 2×π×
5√ 170 ≈ 29.26 units. 14
Supplementary Problems 1.1 Identify the center and radius of the circle with the given equation in each item. Sketch its graph, and indicate the center.
2. 5x2 + 5y 2 = 125 2 3 2 3. (x + 4) + y − =1 4 4. x2 − 4x + y 2 − 4y − 8 = 0 5. x2 + y 2 − 14x + 12y = 36
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6. x2 + 10x + y 2 − 16y − 11 = 0
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1. x2 + y 2 =
7. 9x2 + 36x + 9y 2 + 72y + 155 = 0
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8. 9x2 + 9y 2 − 6x + 24y = 19
9. 16x2 + 80x + 16y 2 − 112y + 247 = 0 Find the standard equation of the circle which satisfies the given conditions.
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√ 10. center at the origin, radius 5 3 11. center at (17, 5), radius 12
12. center at (−8, 4), contains (−4, 2) 13. center at (15, −7), tangent to the x-axis 14. center at (15, −7), tangent to the y-axis 15. center at (15, −7), tangent to the line y = −10 16. center at (15, −7), tangent to the line x = 8 17. has a diameter with endpoints (3, 1) and (−7, 6) 17
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18. has a diameter with endpoints
9 3 , 4 and − , −2 2 2
19. concentric with x2 + 20x + y 2 − 14y + 145 = 0, diameter 12 20. concentric with x2 − 2x + y 2 − 2y − 23 = 0 and has 1/5 the area 21. concentric with x2 + 4x + y 2 − 6y + 9 = 0 and has the same circumference as x2 + 14x + y 2 + 10y + 62 = 0 22. contains the points (3, 3), (7, 1), (0, 2)
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23. contains the points (1, 4), (−1, 2), (4, −3) 24. center at (−3, 2) and tangent to the line 2x − 3y = 1
25. center at (−5, −1) and tangent to the line x + y + 10 = 0
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26. has center with x-coordinate 4 and tangent to the line −x + 3y = 9 at (3, 4)
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27. A stadium is shaped as in Figure 1.23, where its left and right ends are circular arcs both with center at C. What is the length of the stadium 50 m from one of the straight sides?
Figure 1.23
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28. A waterway in a theme park has a semicircular cross section with diameter 11 ft. The boats that are going to be used in this waterway have rectangular cross sections and are found to submerge 1 ft into the water. If the waterway is to be filled with water 4.5 ft deep, what is the maximum possible width of the boats?
Figure 1.24
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Lesson 1.2. Parabolas Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) define a parabola; (2) determine the standard form of equation of a parabola; (3) graph a parabola in a rectangular coordinate system; and
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(4) solve situational problems involving conic sections (parabolas). Lesson Outline (1) Definition of a parabola (3) Graphing parabolas
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(2) Derivation of the standard equation of a parabola (4) Solving situational problems involving parabolas Introduction
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A parabola is one of the conic sections. We have already seen parabolas which open upward or downward, as graphs of quadratic functions. Here, we will see parabolas opening to the left or right. Applications of parabolas are presented at the end. 1.2.1. Definition and Equation of a Parabola
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Consider the point F (0, 2) and the line ` having equation y = −2, as shown in Figure 1.25. What are the distances of A(4, 2) from F and from `? (The latter is taken as the distance of A from A` , the point on ` closest to A). How about the distances of B(−8, 8) from F and from ` (from B` )? AF = 4 and AA` = 4 p and BF = (−8 − 0)2 + (8 − 2)2 = 10
BB` = 10
There are other points P such that P F = P P` (where P` is the closest point on line `). The collection of all such points forms a shape called a parabola. Let F be a given point, and ` a given line not containing F . The set of all points P such that its distances from F and from ` are the same, is called a parabola. The point F is its focus and the line ` its directrix. 19 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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Figure 1.25
Figure 1.26
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Consider a parabola with focus F (0, c) and directrix ` having equation y = −c. See Figure 1.26. The focus and directrix are c units above and below, respectively, the origin. Let P (x, y) be a point on the parabola so P F = P P` , where P` is the point on ` closest to P . The point P has to be on the same side of the directrix as the focus (if P was below, it would be closer to ` than it is from F ). P F = P P`
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p
x2 + (y − c)2 = y − (−c) = y + c x2 + y 2 − 2cy + c2 = y 2 + 2cy + c2 x2 = 4cy
The vertex V is the point midway between the focus and the directrix. This equation, x2 = 4cy, is then the standard equation of a parabola opening upward with vertex V (0, 0). Suppose the focus is F (0, −c) and the directrix is y = c. In this case, a point P on the resulting parabola would be below the directrix (just like the focus). Instead of opening upward, it will open downward. Consequently, P F = p x2 + (y + c)2 and P P` = c − y (you may draw a version of Figure 1.26 for this case). Computations similar to the one done above will lead to the equation 20 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
x2 = −4cy.
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We collect here the features of the graph of a parabola with standard equation x2 = 4cy or x2 = −4cy, where c > 0.
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(1) vertex : origin V (0, 0)
• If the parabola opens upward, the vertex is the lowest point. If the parabola opens downward, the vertex is the highest point.
(2) directrix : the line y = −c or y = c • The directrix is c units below or above the vertex.
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(3) focus: F (0, c) or F (0, −c)
• The focus is c units above or below the vertex. • Any point on the parabola has the same distance from the focus as it has from the directrix.
(4) axis of symmetry: x = 0 (the y-axis) • This line divides the parabola into two parts which are mirror images of each other. Example 1.2.1. Determine the focus and directrix of the parabola with the given equation. Sketch the graph, and indicate the focus, directrix, vertex, and axis of symmetry. 21 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(1) x2 = 12y
(2) x2 = −6y
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Solution. (1) The vertex is V (0, 0) and the parabola opens upward. From 4c = 12, c = 3. The focus, c = 3 units above the vertex, is F (0, 3). The directrix, 3 units below the vertex, is y = −3. The axis of symmetry is x = 0.
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(2) The vertex is V (0, 0) and the parabola opens downward. From 4c = 6, c = 32 . The focus, c = 32 units below the vertex, is F 0, − 23 . The directrix, 32 units above the vertex, is y = 32 . The axis of symmetry is x = 0.
Example 1.2.2. What is the standard equation of the parabola in Figure 1.25? Solution. From the figure, we deduce that c = 2. The equation is thus x2 = 8y. 2
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1.2.2. More Properties of Parabolas The parabolas we considered so far are “vertical” and have their vertices at the origin. Some parabolas open instead horizontally (to the left or right), and some have vertices not at the origin. Their standard equations and properties are given in the box. The corresponding computations are more involved, but are similar to the one above, and so are not shown anymore.
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In all four cases below, we assume that c > 0. The vertex is V (h, k), and it lies between the focus F and the directrix `. The focus F is c units away from the vertex V , and the directrix is c units away from the vertex. Recall that, for any point on the parabola, its distance from the focus is the same as its distance from the directrix.
(y − k)2 = 4c(x − h)
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(x − h)2 = 4c(y − k)
(x − h)2 = −4c(y − k)
(y − k)2 = −4c(x − h)
directrix `: horizontal
directrix `: vertical
axis of symmetry: x=h, vertical
axis of symmetry: y=k, horizontal
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Note the following observations: • The equations are in terms of x − h and y − k: the vertex coordinates are subtracted from the corresponding variable. Thus, replacing both h and k with 0 would yield the case where the vertex is the origin. For instance, this replacement applied to (x − h)2 = 4c(y − k) (parabola opening upward) would yield x2 = 4cy, the first standard equation we encountered (parabola opening upward, vertex at the origin).
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• If the x-part is squared, the parabola is “vertical”; if the y-part is squared, the parabola is “horizontal.” In a horizontal parabola, the focus is on the left or right of the vertex, and the directrix is vertical. • If the coefficient of the linear (non-squared) part is positive, the parabola opens upward or to the right; if negative, downward or to the left.
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Example 1.2.3. Figure 1.27 shows the graph of parabola, with only its focus and vertex indicated. Find its standard equation. What is its directrix and its axis of symmetry? Solution. The vertex is V (5, −4) and the focus is F (3, −4). From these, we deduce the following: h = 5, k = −4, c = 2 (the distance of the focus from the vertex). Since the parabola opens to the left, we use the template (y − k)2 = −4c(x − h). Our equation is
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(y + 4)2 = −8(x − 5).
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Its directrix is c = 2 units to the right of V , which is x = 7. Its axis is the horizontal line through V : y = −4.
Figure 1.27 24 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
The standard equation (y + 4)2 = −8(x − 5) from the preceding example can be rewritten as y 2 + 8x + 8y − 24 = 0, an equation of the parabola in general form. If the equation is given in the general form Ax2 + Cx + Dy + E = 0 (A and C are nonzero) or By 2 + Cx + Dy + E = 0 (B and C are nonzero), we can determine the standard form by completing the square in both variables. Example 1.2.4. Determine the vertex, focus, directrix, and axis of symmetry of the parabola with the given equation. Sketch the parabola, and include these points and lines.
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(1) y 2 − 5x + 12y = −16 (2) 5x2 + 30x + 24y = 51
Solution. (1) We complete the square on y, and move x to the other side.
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y 2 + 12y = 5x − 16 y 2 + 12y + 36 = 5x − 16 + 36 = 5x + 20 (y + 6)2 = 5(x + 4)
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The parabola opens to the right. It has vertex V (−4, −6). From 4c = 5, we get c = 45 = 1.25. The focus is c = 1.25 units to the right of V : F (−2.75, −6). The (vertical) directrix is c = 1.25 units to the left of V : x = −5.25. The (horizontal) axis is through V : y = −6.
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(2) We complete the square on x, and move y to the other side. 5x2 + 30x = −24y + 51 5(x2 + 6x + 9) = −24y + 51 + 5(9) 5(x + 3)2 = −24y + 96 = −24(y − 4) 24 (x + 3)2 = − (y − 4) 5 In the last line, we divided by 5 for the squared part not to have any coefficient. The parabola opens downward. It has vertex V (−3, 4).
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From 4c = 24 , we get c = 65 = 1.2. The focus is c = 1.2 units below V : 5 F (−3, 2.8). The (horizontal) directrix is c = 1.2 units above V : y = 5.2. The (vertical) axis is through V : x = −3.
Example 1.2.5. A parabola has focus F (7, 9) and directrix y = 3. Find its standard equation.
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Solution. The directrix is horizontal, and the focus is above it. The parabola then opens upward and its standard equation has the form (x − h)2 = 4c(y − k). Since the distance from the focus to the directrix is 2c = 9 − 3 = 6, then c = 3. Thus, the vertex is V (7, 6), the point 3 units below F . The standard equation is then (x − 7)2 = 12(y − 6). 2
1.2.3. Situational Problems Involving Parabolas Let us now solve some situational problems involving parabolas. Example 1.2.6. A satellite dish has a shape called a paraboloid, where each cross-section is a parabola. Since radio signals (parallel to the axis) will bounce 26 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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off the surface of the dish to the focus, the receiver should be placed at the focus. How far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5 ft deep at the vertex?
Solution. The second figure above shows a cross-section of the satellite dish drawn on a rectangular coordinate system, with the vertex at the origin. From the problem, we deduce that (6, 4.5) is a point on the parabola. We need the distance of the focus from the vertex, i.e., the value of c in x2 = 4cy.
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x2 = 4cy 62 = 4c(4.5) 62 =2 c= 4 · 4.5
Thus, the receiver should be 2 ft away from the vertex.
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Example 1.2.7. The cable of a suspension bridge hangs in the shape of a parabola. The towers supporting the cable are 400 ft apart and 150 ft high. If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high is the cable 50 ft away (horizontally) from either tower?
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Solution. Refer to the figure above, where the parabolic cable is drawn with its vertex on the y-axis 30 ft above the origin. We may write its equation as (x − 0)2 = a(y − 30); since we don’t need the focal distance, we use the simpler variable a in place of 4c. Since the towers are 150 ft high and 400 ft apart, we deduce from the figure that (200, 150) is a point on the parabola.
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x2 = a(y − 30) 2002 = a(150 − 30) 2002 1000 a= = 120 3 (y − 30), or equivalently, The parabola has equation x2 = 1000 3 y = 0.003x2 + 30. For the two points on the parabola 50 ft away from the towers, x = 150 or x = −150. If x = 150, then
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y = 0.003(1502 ) + 30 = 97.5.
Thus, the cable is 97.5 ft high 50 ft away from either tower. (As expected, we get the same answer from x = −150.) 2
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More Solved Examples
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For Examples 1 and 2, determine the focus and directrix of the parabola with the given equation. Sketch the graph, and indicate the focus, directrix, and vertex. 1. y 2 = 20x 2. 3x2 = −12y Solution: Solution: 3x2 = −12y ⇔ x2 = −4y Vertex: V (0, 0), opens to the right Vertex: V (0, 0), opens downward 4c = 20 ⇒ c = 5 4c = 4 ⇒ c = 1 Focus: F (5, 0), Directrix: x = −5 Focus: F (0, −1), Directrix: y = 1 See Figure 1.28. See Figure 1.29.
Figure 1.28
Figure 1.29 28
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3. Determine the standard equation of the parabola in Figure 1.30 given only its vertex and focus. Then determine its directix and axis of symmetry.
3 Equation: (y − 4)2 = −10 x + 2 Directrix: x = 1, Axis: y = 4
Figure 1.30
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4. Determine the standard equation of the parabola in Figure 1.31 given only its vertex and diretrix. Then determine its focus and axis of symmetry.
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Solution: 3 V − , 4 , F (−4, 4) 2 5 c= ⇒ 4c = 10 2 Parabola opens to the left
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Solution: 13 15 V 5, , directrix: y = 2 2 c = 1 ⇒ 4c = 4 Parabola opens downward 2 13 = −4 (x − 5) Equation: y − 2 11 Focus: 5, , Axis: x = 5 2
Figure 1.31
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For Examples 5 and 6, determine the vertex, focus, directrix, and axis of symmetry of the parabola with the given equation. Sketch the parabola, and include these points and lines. 5. x2 − 6x − 2y + 9 = 0 Solution: x2 − 6x = 2y − 9 x2 − 6x + 9 = 2y (x − 3)2 = 2y
V (3, 0), parabola opens upward 1 1 , 4c = 2 ⇒ c = , F 3, 2 2 1 directrix: y = − , axis: x = 3 2 See Figure 1.32.
Figure 1.32
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6. 3y 2 + 8x + 24y + 40 = 0 Solution: 3y 2 + 24y = −8x − 40 3(y 2 + 8y) = −8x − 40 3(y 2 + 8y + 16) = −8x − 40 + 48 3(y + 4)2 = −8x + 8 8 (y + 4)2 = − (x − 1) 3
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V (1, −4), parabola opens to the left 8 2 1 4c = ⇒ c = , F , −4 , 3 3 3 5 directrix: x = , axis: y = −4 3 See Figure 1.33.
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Figure 1.33
7. A parabola has focus F (−11, 8) and directrix x = −17. Find its standard equation.
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Solution: Since the focus is 6 units to the right of the directrix, the parabola opens to the right with 2c = 6. Then c = 3 and V (−14, 8). Hence, the equation is (y − 8)2 = 12(x + 14). 8. A flashlight is shaped like a paraboloid and the light source is placed at the focus so that the light bounces off parallel to the axis of symmetry; this is done to maximize illumination. A particular flashlight has its light source located 1 cm from the base and is 6 cm deep; see Figure 1.34. What is the width Figure 1.34 of the flashlight’s opening? Solution: Let the base (the vertex) of the flashlight be the point V (0, 0). Then the light source (the focus) is at F (0, 1); so c = 1. Hence, the parabola’s equation is x2 = 4y. To get the width of the opening, we need the x coordinates of the points on the parabola with y coordinate 6. √ x2 = 4(6) ⇒ x = ±2 6 √ √ Therefore, the width of the opening is 2 × 2 6 = 4 6 ≈ 9.8 cm.
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9. An object thrown from a height of 2 m above the ground follows a parabolic path until the object falls to the ground; see Figure 1.35. If the object reaches a maximum height (measured from the ground) of 7 m after travelling a horizontal distance of 4 m, determine the horizontal distance between the object’s initial and final positions.
Figure 1.35
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Solution: Let V (0, 7) be the parabola’s vertex, which corresponds to the highest point reached by the object. Then the parabola’s equation is of the form x2 = −4c(y − 7) and the object’s starting point is at (−4, 2). Then 16 4 = . 20 5
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(−4)2 = −4c(2 − 7) ⇒ c =
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16 Hence, the equation of the parabola is x2 = − (y − 7). When the object hits 5 the ground, the y coordinate is 0 and r 112 16 7 ⇒ x = ±4 . x2 = − (0 − 7) = 5 5 5 r 7 Since this point is to the right of the vertex, we choose x = +4 . Therefore, 5 r 7 the total distance travelled is 4 − (−4) ≈ 8.73 m. 5
Supplementary Problems 1.2
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Determine the vertex, focus, directrix, and axis of symmetry of the parabola with the given equation. Sketch the graph, and include these points and lines. 1. y 2 = −36x
2. 5x2 = 100y 3. y 2 + 4x − 14y = −53 4. y 2 − 2x + 2y − 1 = 0 5. 2x2 − 12x + 28y = 38 6. (3x − 2)2 = 84y − 112 31 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Find the standard equation of the parabola which satisfies the given conditions. 7. vertex (7, 11), focus (16, 11) 8. vertex (−10, −5), directrix y = −1 11 23 , directrix y = − 9. focus −10, 2 2 3 37 10. focus − , 3 , directrix x = − 2 2
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11. axis of symmetry y = 9, directrix x = 24, vertex on the line 3y − 5x = 7 12. vertex (0, 7), vertical axis of symmetry, through the point P (4, 5)
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13. vertex (−3, 8), horizontal axis of symmetry, through the point P (−5, 12) 14. A satellite dish shaped like a paraboloid has its receiver located at the focus. How far is the receiver from the vertex if the dish is 10 ft across and 3 ft deep at the center?
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15. A flashlight shaped like a paraboloid has its light source at the focus located 1.5 cm from the base and is 10 cm wide at its opening. How deep is the flashlight at its center?
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16. The ends of a rope are held in place at the top of two posts, 9 m apart and each one 8 m high. If the rope assumes a parabolic shape and touches the ground midway between the two posts, how high is the rope 2 m from one of the posts?
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17. Radiation is focused to an unhealthy area in a patient’s body using a parabolic reflector, positioned in such a way that the target area is at the focus. If the reflector is 30 cm wide and 15 cm deep at the center, how far should the base of the reflector be from the target area? 18. A rectangular object 25 m wide is to pass under a parabolic arch that has a width of 32 m at the base and a height of 24 m at the center. If the vertex of the parabola is at the top of the arch, what maximum height should the rectangular object have?
4
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Lesson 1.3. Ellipses Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) define an ellipse; (2) determine the standard form of equation of an ellipse; (3) graph an ellipse in a rectangular coordinate system; and
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(4) solve situational problems involving conic sections (ellipses). Lesson Outline (1) Definition of an ellipse (3) Graphing ellipses
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(2) Derivation of the standard equation of an ellipse (4) Solving situational problems involving ellipses Introduction
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Unlike circle and parabola, an ellipse is one of the conic sections that most students have not encountered formally before. Its shape is a bounded curve which looks like a flattened circle. The orbits of the planets in our solar system around the sun happen to be elliptical in shape. Also, just like parabolas, ellipses have reflective properties that have been used in the construction of certain structures. These applications and more will be encountered in this lesson. 1.3.1. Definition and Equation of an Ellipse
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Consider the points F1 (−3, 0) and F2 (3, 0), as shown in Figure 1.36. What is the sum of the distances of A(4, 2.4) from F1 and from F2 ? How about the sum of the distances of B (and C(0, −4)) from F1 and from F2 ? AF1 + AF2 = 7.4 + 2.6 = 10 BF1 + BF2 = 3.8 + 6.2 = 10 CF1 + CF2 = 5 + 5 = 10 There are other points P such that P F1 + P F2 = 10. The collection of all such points forms a shape called an ellipse.
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Figure 1.36
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Figure 1.37
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Let F1 and F2 be two distinct points. The set of all points P , whose distances from F1 and from F2 add up to a certain constant, is called an ellipse. The points F1 and F2 are called the foci of the ellipse.
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Given are two points on the x-axis, F1 (−c, 0) and F2 (c, 0), the foci, both c units away from their center (0, 0). See Figure 1.37. Let P (x, y) be a point on the ellipse. Let the common sum of the distances be 2a (the coefficient 2 will make computations simpler). Thus, we have P F1 + P F2 = 2a.
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P F1 = 2a − P F2 p p (x + c)2 + y 2 = 2a − (x − c)2 + y 2 p x2 + 2cx + c2 + y 2 = 4a2 − 4a (x − c)2 + y 2 + x2 − 2cx + c2 + y 2 p a (x − c)2 + y 2 = a2 − cx a2 x2 − 2cx + c2 + y 2 = a4 − 2a2 cx + c2 x2 (a2 − c2 )x2 + a2 y 2 = a4 − a2 c2 = a2 (a2 − c2 ) √ b2 x 2 + a2 y 2 = a2 b2 by letting b = a2 − c2 , so a > b x2 y 2 + 2 =1 a2 b
√ When we let b = a2 − c2 , we assumed a > c. To see why this is true, look at 4P F1 F2 in Figure 1.37. By the Triangle Inequality, P F1 + P F2 > F1 F2 , which implies 2a > 2c, so a > c. We collect here the features of the graph of an ellipse with standard equation √ x2 y 2 + 2 = 1, where a > b. Let c = a2 − b2 . 2 a b
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PY (2) foci : F1 (−c, 0) and F2 (c, 0)
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(1) center : origin (0, 0)
• Each focus is c units away from the center.
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• For any point on the ellipse, the sum of its distances from the foci is 2a. (3) vertices: V1 (−a, 0) and V2 (a, 0)
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• The vertices are points on the ellipse, collinear with the center and foci. • If y = 0, then x = ±a. Each vertex is a units away from the center. • The segment V1 V2 is called the major axis. Its length is 2a. It divides the ellipse into two congruent parts.
(4) covertices: W1 (0, −b) and W2 (0, b)
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• The segment through the center, perpendicular to the major axis, is the minor axis. It meets the ellipse at the covertices. It divides the ellipse into two congruent parts. • If x = 0, then y = ±b. Each covertex is b units away from the center. • The minor axis W1 W2 is 2b units long. Since a > b, the major axis is longer than the minor axis.
Example 1.3.1. Give the coordinates of the foci, vertices, and covertices of the ellipse with equation x2 y 2 + = 1. 25 9 Sketch the graph, and include these points. 35 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and c = foci: F1 (−4, 0), F2 (4, 0)
√
a2 − b2 = 4.
vertices: V1 (−5, 0), V2 (5, 0)
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covertices: W1 (0, −3), W2 (0, 3)
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Example 1.3.2. Find the (standard) equation of the ellipse whose foci are F1 (−3, 0) and F2 (3, 0), such that for any point on it, the sum of its distances from the foci is 10. See Figure 1.36. Solution. We have 2a = 10 and c = 3, so a = 5 and b = equation is x2 y 2 + = 1. 25 16
√
a2 − c2 = 4. The 2
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1.3.2. More Properties of Ellipses
The ellipses we have considered so far are “horizontal” and have the origin as their centers. Some ellipses have their foci aligned vertically, and some have centers not at the origin. Their standard equations and properties are given in the box. The derivations are more involved, but are similar to the one above, and so are not shown anymore. √ In all four cases below, a > b and c = a2 − b2 . The foci F1 and F2 are c units away from the center. The vertices V1 and V2 are a units away from the center, the major axis has length 2a, the covertices W1 and W2 are b units away from the center, and the minor axis has length 2b. Recall that, for any point on the ellipse, the sum of its distances from the foci is 2a. 36
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Center
Corresponding Graphs
x2 y 2 + 2 = 1, b > a b2 a
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(h, k)
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x2 y 2 + 2 = 1, a > b a2 b
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(0, 0)
(x − h)2 (y − k)2 + =1 a2 b2 a>b
(x − h)2 (y − k)2 + =1 b2 a2 b>a
major axis: horizontal
major axis: vertical
minor axis: vertical
minor axis: horizontal
In the standard equation, if the x-part has the bigger denominator, the ellipse is horizontal. If the y-part has the bigger denominator, the ellipse is vertical. Example 1.3.3. Give the coordinates of the center, foci, vertices, and covertices of the ellipse with the given equation. Sketch the graph, and include these points.
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(x + 3)2 (y − 5)2 + =1 24 49 (2) 9x2 + 16y 2 − 126x + 64y = 71 (1)
√ Solution. √(1) From a2 = 49 and b2 = 24, we have a = 7, b = 2 6 ≈ 4.9, and c = a2 − b2 = 5. The ellipse is vertical. center: foci:
(−3, 5) F1 (−3, 0), F2 (−3, 10)
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vertices: V1 (−3, −2), V2 (−3, 12) √ covertices: W1 (−3 − 2 6, 5) ≈ (−7.9, 5) √ W2 (−3 + 2 6, 5) ≈ (1.9, 5)
(2) We first change the given equation to standard form. 9(x2 − 14x) + 16(y 2 + 4y) = 71 9(x2 − 14x + 49) + 16(y 2 + 4y + 4) = 71 + 9(49) + 16(4) 9(x − 7)2 + 16(y + 2)2 = 576 (x − 7)2 (y + 2)2 + =1 64 36 38 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
We have a = 8 and b = 6. Thus, c = horizontal. center: foci:
√
√ a2 − b2 = 2 7 ≈ 5.3. The ellipse is
(7, −2) √ F1 (7 − 2 7, −2) ≈ (1.7, −2) √ F2 (7 + 2 7, −2) ≈ (12.3, −2)
vertices: V1 (−1, −2), V2 (15, −2) W1 (7, −8), W2 (7, 4)
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covertices:
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Example 1.3.4. The foci of an ellipse are (−3, −6) and (−3, 2). For any point on the ellipse, the sum of its distances from the foci is 14. Find the standard equation of the ellipse.
Solution. The midpoint (−3, −2) of the foci is the center of the ellipse. The ellipse is vertical (because the foci are vertically √ aligned)√and c = 4. From the given sum, 2a = 14 so a = 7. Also, b = a2 − c2 = 33. The equation is (x + 3)2 (y + 2)2 + = 1. 2 33 49 √ √ Example 1.3.5. An ellipse has vertices (2 − 61, −5) and (2 + 61, −5), and its minor axis is 12 units long. Find its standard equation and its foci.
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Solution. The midpoint (2, −5)√of the vertices is the center of the ellipse, which is horizontal. Each vertex is a = 61 units away from the center. From the length of (x − 2)2 (y + 5)2 the minor axis, 2b = 12 so b = 6. The standard equation is + = 61 36 √ 1. Each focus is c = a2 − b2 = 5 units away from (2, −5), so their coordinates are (−3, −5) and (7, −5). 2 1.3.3. Situational Problems Involving Ellipses Let us now apply the concept of ellipse to some situational problems.
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? Example 1.3.6. A tunnel has the shape of a semiellipse that is 15 ft high at the center, and 36 ft across at the base. At most how high should a passing truck be, if it is 12 ft wide, for it to be able to fit through the tunnel? Round off your answer to two decimal places.
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Solution. Refer to the figure above. If we draw the semiellipse on a rectangular coordinate system, with its center at the origin, an equation of the ellipse which contains it, is x2 y2 + = 1. 182 152 To maximize its height, the corners of the truck, as shown in the figure, would have to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n) be the corner of the truck in the first quadrant, where n > 0, is the (maximum) height of the truck. Since this point is on the ellipse, it should fit the equation. Thus, we have 62 n2 + =1 182 152 √ n = 10 2 ≈ 14.14 ft
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Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on one of the foci is the star around which it revolves. The planet is closest to the star when it is at one vertex. It is farthest from the star when it is at the other vertex. Suppose the closest and farthest distances of the planet from this star, are 420 million kilometers and 580 million kilometers, respectively. Find the equation of the ellipse, in standard form, with center at the origin and the star at the x-axis. Assume all units are in millions of kilometers.
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Solution. In the figure above, the orbit is drawn as a horizontal ellipse with center at the origin. From the planet’s distances from the star, at its closest and farthest points, it follows that the major axis is 2a = 420 + 580 = 1000 (million kilometers), so a = 500. If we place the star at the positive x-axis, then it is c = 500 − 420 = 80 units away from the center. Therefore, we get b2 = a2 − c2 = 5002 − 802 = 243600. The equation then is x2 y2 + = 1. 250000 243600 The star could have been placed on the negative x-axis, and the answer would still be the same. 2
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1. Give the coordinates of the foci, vertices, and covertices of the ellipse with equay2 x2 + = 1. Then sketch the tion 169 144 graph and include these points. Solution: The ellipse is horizontal. a2 =√169 ⇒ a = 13, b2 = 144 ⇒ b = 12, c = 169 − 144 = 5 Foci: F1 (−5, 0), F2 (5, 0) Vertices: V1 (−13, 0), V2 (13, 0) Covertices: W1 (0, −12), W2 (0, 12) See Figure 1.38.
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More Solved Examples
Figure 1.38
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2. Find the standard equation of the ellipse whose foci are F1 (0, −8) and F2 (0, 8), such that for any point on it, the sum of its distances from the foci is 34. Solution: The ellipse is vertical and has center at (0, 0). 2a = 34 ⇒ a = 17 √ c = 8 ⇒ b = 172 − 82 = 15 y2 x2 + = 1. 225 289
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The equation is
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For Examples 3 and 4, give the coordinates of the center, foci, vertices, and covertices of the ellipse with the given equation. Sketch the graph, and include these points. (x − 7)2 (y + 2)2 3. + =1 64 25 Solution: The ellipse is horizontal. a2 =√64 ⇒ a = 8,√b2 = 25 ⇒ b = 5 c = 64 − 25 = 39 ≈ 6.24 center:
(7, −2)
foci:
F1 (7 −
√ 39, −2) ≈ (0.76, −2) √ F2 (7 + 39, −2) ≈ (13.24, −2)
vertices: V1 (−1, −2), V2 (15, −2) Figure 1.39
covertices: W1 (7, −7), W2 (7, 3) See Figure 1.39.
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4. 16x2 + 96x + 7y 2 + 14y + 39 = 0 Solution:
The ellipse is vertical. √ a2 = 16 ⇒ a = 4, b2 = 7 ⇒ b = 7 ≈ 2.65√ c = 16 − 7 = 3 foci:
(−3, −1) F1 (−3, −4), F2 (−3, 2)
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center:
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16x2 + 96x + 7y 2 + 14y = −39 16(x2 + 6x + 9) + 7(y 2 + 2y + 1) = −39 + 151 16(x + 3)2 + 7(y + 1)2 = 112 (x + 3)2 (y + 1)2 + =1 7 16
Figure 1.40
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vertices: V1 (−3, −5), V2 (−3, 3) √ covertices: W1 (−3 − 7, −1) ≈ (−5.65, −1) √ W2 (−3 + 7, −1) ≈ (−0.35, −1) See Figure 1.40.
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5. The covertices of an ellipse are (5, 6) and (5, 8). For any point on the ellipse, the sum of its distances from the foci is 12. Find the standard equation of the ellipse.
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Solution: The ellipse is horizontal with center at the midpoint (5, 7) of the (x − 5)2 + covertices. Also, 2a = 12 so a = 6 while b = 1. The equation is 36 (y − 7)2 = 1. 1 √ √ 6. An ellipse has foci (−4 − 15, 3) and (−4 + 15, 3), and its major axis is 10 units long. Find its standard equation and its vertices. Solution: The √ ellipse is horizontal with center at the midpoint (−4, 3) of the foci; also c = 15. √ Since the √length of the major axis is 10, 2a = 10 and a = 5. Thus b = 52 − 15 = 10. Therefore, the equation of the ellipse is (x + 4)2 (y − 3)2 + = 1 and its vertices are (−9, 3) and (1, 3). 25 10
7. A whispering gallery is an enclosure or room where whispers can be clearly heard in some parts of the gallery. Such a gallery can be constructed by making its ceiling in the shape of a semi-ellipse; in this case, a whisper from one focus can be clearly heard at the other focus. If an elliptical whispering 43 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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gallery is 90 feet long and the foci are 50 feet apart, how high is the gallery at its center?
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Solution: We set up a Cartesian coordinate system by assigning the center of the semiellipse as the origin. The point on the ceiling right above the center is a covertex of the ellipse. Since√2a = 90 and 2c = 50; then b2 = 452 − 252 = 1400. The height is given by b = 1400 ≈ 37.4 ft.
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8. A spheroid (or oblate spheroid) is the surface obtained by rotating an ellipse around its minor axis. The bowl in Figure 1.41 is in the shape of the lower half of a spheroid; that is, its horizontal cross sections are circles while its vertical cross sections that pass through the center are semiellipses. If this bowl is 10 √ in wide at the opening and 10 in deep at the center, how deep does a circular cover with diameter 9 in go into the bowl?
Figure 1.41
Solution: We set up a Cartesian coordinate system √ by assigning the center of the semiellipse as the origin. Then a = 5, b = 10, and the equation of 2 2 the ellipse is x25 + y10 = 1. We want the y-coordinate of the points on the q 2 ellipse that has x = ±4.5. This coordinate is y = − 10 1 − x25 ≈ −1.38. Therefore, the cover will go 1.38 inches into the bowl.
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Supplementary Problems 1.3 Give the coordinates of the center, foci, vertices, and covertices of the ellipse with the given equation. Sketch the graph, and include these points. 1.
x2 y 2 + =1 8 4
2.
x2 (y − 2)2 + =1 16 25
4.
(x + 5)2 (y − 2)2 + =1 49 121
5. 16x2 − 224x + 25y 2 + 250y − 191 = 0
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6. 25x2 − 200x + 16y 2 − 160y = 800
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3. (x − 1)2 + (2y − 2)2 = 4
Find the standard equation of the ellipse which satisfies the given conditions.
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√ √ 7. foci (2 − 33, 8) and (2 + 33, 8), the sum of the distances of any point from the foci is 14 8. center (−3, −7), vertical major axis of length 20, minor axis of length 12
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9. foci (−21, 10) and (3, 10), contains the point (−9, 15) 10. a vertex at (−3, −18) and a covertex at (−12, −7), major axis is either horizontal or vertical 11. a focus at (−9, 15) and a covertex at (1, 10), with vertical major axis
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12. A 40-ft wide tunnel has the shape of a semiellipse that is 5 ft high a distance of 2 ft from either end. How high is the tunnel at its center? 13. The moon’s orbit is an ellipse with Earth as one focus. If the maximum distance from the moon to Earth is 405 500 km and the minimum distance is 363 300 km, find the equation of the ellipse in a Cartesian coordinate system where Earth is at the origin. Assume that the ellipse has horizontal major axis and that the minimum distance is achieved when the moon is to the right of Earth. Use 100 km as one unit. 14. Two friends visit a whispering gallery (in the shape of a semiellipsoid) where they stand 100 m apart to be at the foci. If one of them is 6 m from the nearest wall, how high is the gallery at its center? 45
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15. A jogging path is in the shape of an ellipse. If it is 120 ft long and 40 ft wide, what is the width of the track 15 ft from either vertex? 16. Radiation is focused to an unhealthy area in a patient’s body using a semielliptic reflector, positioned in such a way that the target area is at one focus while the source of radiation is at the other. If the reflector is 100 cm wide and 30 cm high at the center, how far should the radiation source and the target area be from the ends of the reflector?
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Lesson 1.4. Hyperbolas
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Learning Outcomes of the Lesson
At the end of the lesson, the student is able to: (1) define a hyperbola;
(2) determine the standard form of equation of a hyperbola; (3) graph a hyperbola in a rectangular coordinate system; and
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Lesson Outline
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(4) solve situational problems involving conic sections (hyperbolas).
(1) Definition of a hyperbola
(2) Derivation of the standard equation of a hyperbola (3) Graphing hyperbolas
(4) Solving situational problems involving hyperbolas
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Introduction
Just like ellipse, a hyperbola is one of the conic sections that most students have not encountered formally before. Its graph consists of two unbounded branches which extend in opposite directions. It is a misconception that each branch is a parabola. This is not true, as parabolas and hyperbolas have very different features. An application of hyperbolas in basic location and navigation schemes are presented in an example and some exercises. 1.4.1. Definition and Equation of a Hyperbola
Consider the points F1 (−5, 0) and F2 (5, 0) as shown in Figure 1.42. What is the absolute value of the difference of the distances of A(3.75, −3) from F1 and from 46 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
F2 ? How about the absolute value of the difference of the distances of B −5, 16 3 from F1 and from F2 ?
|AF1 − AF2 | = |9.25 − 3.25| = 6 16 34 |BF1 − BF2 | = − = 6 3 3
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There are other points P such that |P F1 − P F2 | = 6. The collection of all such points forms a shape called a hyperbola, which consists of two disjoint branches. For points P on the left branch, P F2 − P F1 = 6; for those on the right branch, P F1 − P F2 = 6.
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Figure 1.42
Figure 1.43
47 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Let F1 and F2 be two distinct points. The set of all points P , whose distances from F1 and from F2 differ by a certain constant, is called a hyperbola. The points F1 and F2 are called the foci of the hyperbola.
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In Figure 1.43, given are two points on the x-axis, F1 (−c, 0) and F2 (c, 0), the foci, both c units away from their midpoint (0, 0). This midpoint is the center of the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolute value of the difference of the distances of P from F1 and F2 , be 2a (the coefficient 2 will make computations simpler). Thus, |P F1 − P F2 | = 2a, and so p p 2 2 2 2 (x + c) + y − (x − c) + y = 2a. Algebraic manipulations allow us to rewrite this into the much simpler where b =
√
c2 − a2 .
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x2 y 2 − 2 = 1, a2 b
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√ When we let b = c2 − a2 , we assumed c > a. To see why this is true, suppose that P is closer to F2 , so P F1 − P F2 = 2a. Refer to Figure 1.43. Suppose also that P is not on the x-axis, so 4P F1 F2 is formed. From the triangle inequality, F1 F2 + P F2 > P F1 . Thus, 2c > P F1 − P F2 = 2a, so c > a.
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Now we present a derivation. For now, assume P is closer to F2 so P F1 > P F2 , and P F1 − P F2 = 2a. P F1 = 2a + P F2 p (x + c)2 + y 2 = 2a + (x − c)2 + y 2 2 2 p p 2 2 2 2 (x + c) + y = 2a + (x − c) + y p cx − a2 = a (x − c)2 + y 2 2 p (cx − a2 )2 = a (x − c)2 + y 2
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p
(c2 − a2 )x2 − a2 y 2 = a2 (c2 − a2 ) b 2 x 2 − a2 y 2 = a2 b2 x2 y 2 − 2 =1 a2 b
by letting b =
√
c2 − a2 > 0
We collect here the features of the graph of a hyperbola with standard equation x2 y 2 − 2 = 1. a2 b √ Let c = a2 + b2 . 48 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Figure 1.45
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Figure 1.44
(2) foci : F1 (−c, 0) and F2 (c, 0)
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(1) center : origin (0, 0)
• Each focus is c units away from the center.
• For any point on the hyperbola, the absolute value of the difference of its distances from the foci is 2a.
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(3) vertices: V1 (−a, 0) and V2 (a, 0)
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• The vertices are points on the hyperbola, collinear with the center and foci. • If y = 0, then x = ±a. Each vertex is a units away from the center. • The segment V1 V2 is called the transverse axis. Its length is 2a.
(4) asymptotes: y = ab x and y = − ab x, the lines `1 and `2 in Figure 1.45
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• The asymptotes of the hyperbola are two lines passing through the center which serve as a guide in graphing the hyperbola: each branch of the hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends. (We need the concept of limits from calculus to explain this.) • An aid in determining the equations of the asymptotes: in the standard 2 2 equation, replace 1 by 0, and in the resulting equation xa2 − yb2 = 0, solve for y. • To help us sketch the asymptotes, we point out that the asymptotes `1 and `2 are the extended diagonals of the auxiliary rectangle drawn in Figure 1.45. This rectangle has sides 2a and 2b with its diagonals intersecting at the center C. Two sides are congruent and parallel to the transverse axis V1 V2 . The other two sides are congruent and parallel 49
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to the conjugate axis, the segment shown which is perpendicular to the transverse axis at the center, and has length 2b.
2 2 Solution. With √ a = 9 and √ b = 7, we have a = 3, b = 7, and c = a2 + b2 = 4. foci: F1 (−4, 0) and F2 (4, 0)
vertices: V1 (−3, 0) and V2 (3, 0) √
√
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asymptotes: y = 37 x and y = − 37 x The graph is shown at the right. The conju√ gate axis drawn has its endpoints b = 7 ≈ 2.7 units above and below the center. 2
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Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbola with equation x2 y 2 − = 1. 9 7 Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle.
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Example 1.4.2. Find the (standard) equation of the hyperbola whose foci are F1 (−5, 0) and F2 (5, 0), such that for any point on it, the absolute value of the difference of its distances from the foci is 6. See Figure 1.42.
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Solution. We have 2a = 6 and c = 5, so a = 3 and b = x2 y 2 hyperbola then has equation − = 1. 9 16
√
c2 − a2 = 4. The 2
1.4.2. More Properties of Hyperbolas
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The hyperbolas we considered so far are “horizontal” and have the origin as their centers. Some hyperbolas have their foci aligned vertically, and some have centers not at the origin. Their standard equations and properties are given in the box. The derivations are more involved, but are similar to the one above, and so are not shown anymore. √ In all four cases below, we let c = a2 + b2 . The foci F1 and F2 are c units away from the center C. The vertices V1 and V2 are a units away from the center. The transverse axis V1 V2 has length 2a. The conjugate axis has length 2b and is perpendicular to the transverse axis. The transverse and conjugate axes bisect each other at their intersection point, C. Each branch of a hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends. The equations of the asymptotes can be determined by replacing 1 in the standard equation by 0. The asymptotes can be drawn as the extended diagonals of the auxiliary rectangle determined by the transverse and conjugate axes. Recall that, 50
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for any point on the hyperbola, the absolute value of the difference of its distances from the foci is 2a. Center
Corresponding Hyperbola
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(0, 0)
y 2 x2 − 2 =1 a2 b
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x2 y 2 − 2 =1 a2 b
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(h, k)
(x − h)2 (y − k)2 − =1 a2 b2
(y − k)2 (x − h)2 − =1 a2 b2
transverse axis: horizontal
transverse axis: vertical
conjugate axis: vertical
conjugate axis: horizontal
In the standard equation, aside from being positive, there are no other restrictions on a and b. In fact, a and b can even be equal. The orientation of the hyperbola is determined by the variable appearing in the first term (the positive term): the corresponding axis is where the two branches will open. For example, if the variable in the first term is x, the hyperbola is “horizontal”: the transverse 51 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
axis is horizontal, and the branches open to the left and right in the direction of the x-axis. Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle. (y + 2)2 (x − 7)2 − =1 25 9 (2) 4x2 − 5y 2 + 32x + 30y = 1
(1)
foci: F1 (7, −2 −
√
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center: C(7, −2)
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2 2 Solution. (1) From √ a = 25 and b = 9, we have a = 5, b = 3, and c = √ a2 + b2 = 34 ≈ 5.8. The hyperbola is vertical. To determine the asymp2 2 totes, we write (y+2) − (x−7) = 0, which is equivalent to y + 2 = ± 53 (x − 7). 25 9 We can then solve this for y.
34) ≈ (7, −7.8) and F2 (7, −2 +
√ 34) ≈ (7, 3.8)
vertices: V1 (7, −7) and V2 (7, 3) asymptotes: y = 35 x −
41 3
and y = − 35 x +
29 3
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The conjugate axis drawn has its endpoints b = 3 units to the left and right of the center.
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(2) We first change the given equation to standard form.
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4(x2 + 8x) − 5(y 2 − 6y) = 1 4(x2 + 8x + 16) − 5(y 2 − 6y + 9) = 1 + 4(16) − 5(9) 4(x + 4)2 − 5(y − 3)2 = 20 (x + 4)2 (y − 3)2 − =1 5 4 √ √ We have a = 5 ≈ 2.2 and b = 2. Thus, c = a2 + b2 = 3. The hyperbola 2 2 − (y−3) = 0 is horizontal. To determine the asymptotes, we write (x+4) 5 4 2 which is equivalent to y − 3 = ± √5 (x + 4), and solve for y. center: C(−4, 3)
foci: F1 (−7, 3) and F2 (−1, 3) √ √ vertices: V1 (−4 − 5, 3) ≈ (−6.2, 3) and V2 (−4 + 5, 3) ≈ (−1.8, 3) √2 x 5
+
√8 5
+ 3 and y = − √25 x −
√8 5
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asymptotes: y =
+3
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The conjugate axis drawn has its endpoints b = 2 units above and below the center.
Example 1.4.4. The foci of a hyperbola are (−5, −3) and (9, −3). For any point on the hyperbola, the absolute value of the difference of its of its distances from the foci is 10. Find the standard equation of the hyperbola. 53 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Solution. The midpoint (2, −3) of the foci is the center of the hyperbola. Each focus is c = 7 units away from the center. From the given difference, 2a = 10 so a = 5. Also, b2 = c2 − a2 = 24. The hyperbola is horizontal (because the foci are horizontally aligned), so the equation is (x − 2)2 (y + 3)2 − = 1. 25 24
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Example 1.4.5. √ A hyperbola has vertices (−4, −5) and (−4, 9), and one of its foci is (−4, 2 − 65). Find its standard equation.
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Solution. The midpoint (−4, 2) of the vertices is the center of the hyperbola, which is vertical (because the vertices are vertically aligned). Each vertex is √ a = 7 units away from the center. The given focus is c = 65 units away from the center. Thus, b2 = c2 − a2 = 16, and the standard equation is 2
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(y − 2)2 (x + 4)2 − = 1. 49 16
1.4.3. Situational Problems Involving Hyperbolas
Let us now give an example on an application of hyperbolas.
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Example 1.4.6. An explosion was heard by two stations 1200 m apart, located at F1 (−600, 0) and F2 (600, 0). If the explosion was heard in F1 two seconds before it was heard in F2 , identify the possible locations of the explosion. Use 340 m/s as the speed of sound.
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Solution. Using the given speed of sound, we can deduce that the sound traveled 340(2) = 680 m farther in reaching F2 than in reaching F1 . This is then the difference of the distances of the explosion from the two stations. Thus, the explosion is on a hyperbola with foci are F1 and F2 , on the branch closer to F1 .
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We have c = 600 and 2a = 680, so a = 340 and b2 = c2 − a2 = 244400. The explosion could therefore be anywhere on the left branch of the hyperbola y2 x2 − 244400 = 1. 2 115600
More Solved Examples
center: foci:
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Solution: The hyperbola is horizontal. a2 = 16 ⇒ a = √ 4, 2 b =√33 ⇒ b = 33, c = 16 + 33 = 7
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1. Determine the foci, vertices, and asymptotes of the hyperbola with equation y2 x2 − = 1. Sketch the graph, and include these points and lines, the 16 33 transverse and conjugate axes, and the auxiliary rectangle.
(0, 0)
F1 (−7, 0), F2 (7, 0)
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vertices: V1 (−4, 0), V2 (4, 0) √ √ 33 33 asymptotes: y = x, y = − x 4 4 The √conjugate axis √ has endpoints (0, − 33) and (0, 33). See Figure 1.46.
Figure 1.46
2. Find the standard equation of the hyperbola whose foci are F1 (0, −10) and F2 (0, 10), such that for any point on it, the absolute value of the difference of its distances from the foci is 12.
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Solution: The hyperbola is vertical and has center at (0, 0). We have 2a = 12, √ y 2 x2 2 2 so a = 6; also, c = 10. Then b = 10 − 6 = 8. The equation is − = 1. 36 64
For Examples 3 and 4, give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle. 3.
(y + 6)2 (x − 4)2 − =1 25 39 Solution: The hyperbola is vertical. √ √ a2 = 25 ⇒ a = 5, b2 = 39 ⇒ b = 39, c = 25 + 39 = 8 55
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center: foci:
(4, −6) F1 (4, −14), F2 (4, 2)
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vertices: V1 (4, −11), V2 (4, −1) (y + 6)2 (x − 4)2 − =0 asymptotes: 25 39 5 ⇔ y + 6 = ± √ (x − 4) 39 √ The conjugate axis has endpoints b = 39 units to the left and to the right of the center. See Figure 1.47.
Figure 1.47
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4. 9x2 + 126x − 16y 2 − 96y + 153 = 0 Solution:
9x2 + 126x − 16y 2 − 96y = −153 9(x2 + 14x + 49) − 16(y 2 + 6y + 9) = −153 + 9(49) − 16(9) 9(x + 7)2 − 16(y + 3)2 = 144 (x + 7)2 (y + 3)2 − =1 16 9
The hyperbola is horizontal. a2 = 16 ⇒ a = 4,
b2 = 9 ⇒ b = 3,
c=
√
16 + 9 = 5
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center: foci:
(−7, −3) F1 (−12, −3), F2 (−2, −3)
vertices: V1 (−11, −3), V2 (−3, −3) 3 (x + 7)2 (y + 3)2 − = 0 ⇔ y + 3 = ± (x + 7) asymptotes: 16 9 4
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The conjugate axis have endpoints (−7, −6) and (−7, 0). See Figure 1.48.
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Figure 1.48
5. The foci of a hyperbola are (−17, −3) and (3, −3). For any point on the hyperbola, the absolute value of the difference of its distances from the foci is 14. Find the standard equation of the hyperbola.
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Solution: The hyperbola is horizontal with center at the midpoint (−7, −3) of the foci. Also, 2a = 14 so a = 7 while c = 10. Then b2 = 102 − 72 = 51. The (x + 7)2 (y + 3)2 equation is − = 1. 49 51
6. The auxiliary rectangle of a hyperbola has vertices (−24, −15), (−24, 9), (10, 9), and (10, −15). Find the equation of the hyperbola if its conjugate axis is horizontal. Solution: The hyperbola is vertical. Using the auxiliary rectangle’s dimensions, we see that the length of the transverse axis is 2a = 24 while the length of the conjugate axis is 2b = 34. Thus, a = 12 and b = 17. The hyperbola’s vertices are the midpoints (−7, −15) and (−7, 9) of the bottom 57 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
and top sides, respectively, of the auxiliary rectangle. Then the hyperbola’s center is (−7, −3), which is the midpoint of the vertices. The equation is (y + 3)2 (x + 7)2 − = 1. 144 289 7. Two LORAN (long range navigation) stations A and B are situated along a straight shore, where A is 200 miles west of B. These stations transmit radio signals at a speed 186 miles per millisecond. The captain of a ship travelling on the open sea intends to enter a harbor that is located 40 miles east of station A.
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Due to the its location, the harbor experiences a time difference in receiving the signals from both stations. The captain navigates the ship into the harbor by following a path where the ship experiences the same time difference as the harbor.
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(a) What time difference between station signals should the captain be looking for in order the ship to make a successful entry into the harbor? (b) If the desired time difference is achieved, determine the location of the ship if it is 75 miles offshore. Solution:
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(a) Let H represent the harbor on the shoreline. Note that BH −AH = 160− 40 = 120. The time difference on the harbor is given by 120÷186 ≈ 0.645 milliseconds. This is the time difference needed to be maintained in order to for the ship to enter the harbor.
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(b) Situate the stations A and B on the Cartesian plane so that A (−100, 0) and B (100, 0). Let P represent the ship on the sea, which has coordinates (h, 75). Since P B − P A = 120, then it should follow that h < 0. Moreover, P should lie on the left branch of the hyperbola whose equation is given by x2 y 2 − 2 =1 a2 b √ √ where 2a = 120 ⇒ a = 60, and b = c2 − a2 = 1002 − 602 = 80. Therefore, h2 752 − = 1 602 802 s 752 h = − 1 + 2 602 ≈ −82.24 80 This means that the ship is around 17.76 miles to the east of station A.
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Supplementary Problems 1.4 Give the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph and the auxiliary rectangle, then include these points and lines. 1.
y2 x2 − =1 100 81
2. y − x =
3. 4x2 − 15(y − 5)2 = 60
1 y+x
(y − 6)2 (x − 8)2 − =1 64 36
4.
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5. 9y 2 + 54y − 6x2 − 36x − 27 = 0 6. 16x2 + 64x − 105y 2 + 840y − 3296 = 0
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Find the standard equation of the hyperbola which satisfies the given conditions. 7. foci (−7, −17) and (−7, 17), the absolute value of the difference of the distances of any point from the foci is 24 8. foci (−3, −2) and (15, −2), a vertex at (9, −2)
71 3
− 43 x and y = 34 x −
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10. asymptotes y =
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9. center (−10, −4), one corner of auxiliary rectangle at (−1, 12), with horizontal transverse axis
5 x+ 11. asymptotes y = − 12
19 3
and y =
17 3
5 x 12
+
and a vertex at (17, 9) 29 3
and a focus at (−4, −5)
12. horizontal conjugate axis, one corner of auxiliary rectangle at (3, 8), and an asymptote 4x + 3y = 12
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13. two corners of auxiliary rectangle at (2, 3) and (16, −1), and horizontal transverse axis 14. Two radio stations are located 150 miles apart, where station A is west of station B. Radio signals are being transmitted simultaneously by both stations, travelling at a rate of 0.2 miles/µsec. A plane travelling at 60 miles above ground level has just passed by station B and is headed towards the other station. If the signal from B arrives at the plane 480 µsec before the signal sent from A, determine the location of the plane.
4
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Lesson 1.5. More Problems on Conic Sections Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) recognize the equation and important characteristics of the different types of conic sections; and
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(2) solve situational problems involving conic sections. Lesson Outline
(1) Conic sections with associated equations in general form
(2) Problems involving characteristics of various conic sections
Introduction
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(3) Solving situational problems involving conic sections
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In this lesson, we will identify the conic section from a given equation. We will analyze the properties of the identified conic section. We will also look at problems that use the properties of the different conic sections. This will allow us to synthesize what has been covered so far.
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1.5.1. Identifying the Conic Section by Inspection The equation of a circle may be written in standard form Ax2 + Ay 2 + Cx + Dy + E = 0,
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that is, the coefficients of x2 and y 2 are the same. However, it does not follow that if the coefficients of x2 and y 2 are the same, the graph is a circle.
(A) (B)
General Equation Standard Equation graph 2 2 1 3 point 2x2 + 2y 2 − 2x + 6y + 5 = 0 x − 2 + y + 2 = 0 2 2 2 2 x + y − 6x − 8y + 50 = 0 (x − 3) + (y − 4) = −25 empty set
For a circle with equation (x − h)2 + (y − k)2 = r2 , we have r2 > 0. This is not the case for the standard equations of (A) and (B). In (A), because the sum of two squares can only be 0 if and only if each square is 0, it follows that x − 21 = 0 and y + 32 = 0. The graph is thus the single point 1 , − 32 . 2
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In (B), no real values of x and y can make the nonnegative left side equal to the negative right side. The graph is then the empty set. Let us recall the general form of the equations of the other conic sections. We may write the equations of conic sections we discussed in the general form Ax2 + By 2 + Cx + Dy + E = 0. Some terms may vanish, depending on the kind of conic section. (1) Circle: both x2 and y 2 appear, and their coefficients are the same
Example: 18x2 + 18y 2 − 24x + 48y − 5 = 0
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Ax2 + Ay 2 + Cx + Dy + E = 0
Degenerate cases: a point, and the empty set
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(2) Parabola: exactly one of x2 or y 2 appears
Ax2 + Cx + Dy + E = 0 (D 6= 0, opens upward or downward) By 2 + Cx + Dy + E = 0 (C 6= 0, opens to the right or left)
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Examples: 3x2 − 12x + 2y + 26 = 0 (opens downward) − 2y 2 + 3x + 12y − 15 = 0 (opens to the right)
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(3) Ellipse: both x2 and y 2 appear, and their coefficients A and B have the same sign and are unequal Examples: 2x2 + 5y 2 + 8x − 10y − 7 = 0 (horizontal major axis) 4x2 + y 2 − 16x − 6y + 21 = 0 (vertical major axis) If A = B, we will classify the conic as a circle, instead of an ellipse. Degenerate cases: a point, and the empty set
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(4) Hyperbola: both x2 and y 2 appear, and their coefficients A and B have different signs Examples: 5x2 − 3y 2 − 20x − 18y − 22 = 0 (horizontal transverse axis) − 4x2 + y 2 + 24x + 4y − 36 = 0 (vertical transverse axis) Degenerate case: two intersecting lines The following examples will show the possible degenerate conic (a point, two intersecting lines, or the empty set) as the graph of an equation following a similar pattern as the non-degenerate cases. 2
2
(1) 4x + 9y − 16x + 18y + 25 = 0
=⇒ =⇒
(x − 2)2 (y + 1)2 + =0 32 22 one point: (2, −1)
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(2) 4x2 + 9y 2 − 16x + 18y + 61 = 0
=⇒ =⇒
(3) 4x2 − 9y 2 − 16x − 18y + 7 = 0
=⇒ =⇒
(x − 2)2 (y + 1)2 + = −1 32 22 empty set (x − 2)2 (y + 1)2 − =0 32 22 2 two lines: y + 1 = ± (x − 2) 3
A Note on Identifying a Conic Section by Its General Equation
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It is only after transforming a given general equation to standard form that we can identify its graph either as one of the degenerate conic sections (a point, two intersecting lines, or the empty set) or as one of the non-degenerate conic sections (circle, parabola, ellipse, or hyperbola).
1.5.2. Problems Involving Different Conic Sections
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The following examples require us to use the properties of different conic sections at the same time.
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Example 1.5.1. A circle has center at the focus of the parabola y 2 + 16x + 4y = 44, and is tangent to the directrix of this parabola. Find its standard equation. Solution. The standard equation of the parabola is (y + 2)2 = −16(x − 3). Its vertex is V (3, −2). Since 4c = 16 or c = 4, its focus is F (−1, −2) and its directrix is x = 7. The circle has center at (−1, −2) and radius 8, which is the distance from F to the directrix. Thus, the equation of the circle is 2
D
(x + 1)2 + (y + 2)2 = 64.
Example 1.5.2. The vertices and foci of 5x2 − 4y 2 + 50x + 16y + 29 = 0 are, respectively, the foci and vertices of an ellipse. Find the standard equation of this ellipse. Solution. We first write the equation of the hyperbola in standard form: (x + 5)2 (y − 2)2 − = 1. 16 20
For this hyperbola, using the notations ah , bh , and ch to refer to a, b, and √ c of the standard equation of the hyperbola, respectively, we have a = 4, b = 2 5, h h p 2 2 ch = ah + bh = 6, so we have the following points: 62 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
center: (−5, 2) vertices: (−9, 2) and (−1, 2) foci: (−11, 2) and (1, 2). It means that, for the ellipse, we have these points: center: (−5, 2) vertices: (−11, 2) and (1, 2) foci: (−9, 2) and (−1, 2). p
a2e − c2e =
√
20. The standard
PY
In this case, ce = 4 and ae = 6, so that be = equation of the ellipse is
2
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(x + 5)2 (y − 2)2 + = 1. 36 20
More Solved Examples
1. Identify the graph of each of the following equations. (a) 4x2 − 8x − 49y 2 + 196y − 388 = 0 (c) y 2 − 48x + 6y = −729
(f) x2 + y 2 − 18y − 19 = 0
(d) 49x2 + 196x + 100y 2 + 1400y + 196 = 0
(g) −5x2 + 60x + 7y 2 + 84y + 72 = 0
EP E
D
(b) x2 + 5x + y 2 − y + 7 = 0
(e) 36x2 +360x+64y 2 −512y+1924 = 0
(h) x2 − 16x + 20y = 136
Solution:
D
(a) Since the coefficients of x2 and y 2 have opposite signs, the graph is a hyperbola or a pair of intersecting lines. Completing the squares, we get 4x2 − 8x − 49y 2 + 196y − 388 = 0 4(x2 − 2x) − 49(y 2 − 4y) = 388 4(x2 − 2x + 1) − 49(y 2 − 4y + 4) = 388 + 4(1) − 49(4) (x − 1)2 (y − 2)2 − = 1. 49 4
Thus, the graph is a hyperbola. (b) Since x2 and y 2 have equal coefficients, the graph is a circle, a point, or the empty set. Completing the squares, we get x2 + 5x + y 2 − y + 7 = 0 63 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
25 1 25 1 x2 + 5x + + y 2 − y + = −7 + + 4 4 4 4 2 2 1 1 5 + y− =− . x+ 2 2 2 Since the right hand side is negative, the graph is the empty set. (c) By inspection, the graph is a parabola. (d) Since the coefficients of x2 and y 2 are not equal but have the same sign, the graph is an ellipse, a point, or the empty set. Completing the squares, we get
C O
PY
49x2 + 196x + 100y 2 + 1400y + 196 = 0 49(x2 + 4x) + 100(y 2 + 14y) = −196 49(x2 + 4x + 4) + 100(y 2 + 14y + 49) = −196 + 49(4) + 100(49) (x + 2)2 (y + 7)2 + = 1. 100 49 Thus, the graph is an ellipse.
D
(e) Since the coefficients of x2 and y 2 are not equal but have the same sign, the graph is an ellipse, a point, or the empty set. Completing the squares, we get
EP E
36x2 + 360x + 64y 2 − 512y + 1924 = 0 36(x2 + 10x) + 64(y 2 − 8y) = −1924 36(x2 + 10x + 25) + 64(y 2 − 8y + 16) = −1924 + 36(25) + 64(16) (x + 5)2 (y − 4)2 + = 0. 64 36
Since the right-hand side is 0, the graph is a single point (the point is (−5, 4)).
D
(f) Since x2 and y 2 have equal coefficients, the graph is a circle, a point, or the empty set. Completing the squares, we get x2 + y 2 − 18y − 19 = 0 x2 + y 2 − 18y + 81 = 19 + 81 x2 + (y − 9)2 = 100.
Thus, the graph is a circle. (g) Since the coefficients of x2 and y 2 have opposite signs, the graph is a hyperbola or a pair of intersecting lines. Completing the squares, we get −5x2 + 60x + 7y 2 + 84y + 72 = 0 64 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
−5(x2 − 12x) + 7(y 2 + 12y) = −72 −5(x2 − 12x + 36) + 7(y 2 + 12y + 36) = −72 − 5(36) + 7(36) (x − 6)2 (y + 6)2 − = 0. 7 5 Since the right handrside is 0, the graph is a pair of intersecting lines; 5 these are y + 6 = ± (x − 6). 7 (h) By inspection, the graph is a parabola.
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2. The center of a circle is the vertex of the parabola y 2 + 24x − 12y + 132 = 0. If the circle intersects the parabola’s directrix at a point where y = 11, find the equation of the circle. Solution:
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y 2 − 12y = −24x − 132 y 2 − 12y + 36 = −24x − 132 + 36 (y − 6)2 = −24x − 96 (y − 6)2 = −24(x + 4)
EP E
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The vertex of the parabola is (−4, 6) and its directrix is x = 2. Thus, the circle has center (−4, 6) and p √ contains the point (2, 11). Then its radius is (−4 − 2)2 + (6 − 11)2 = 61. Therefore, the equation of the circle is (x + 4)2 + (y − 6)2 = 61. 3. The vertices of the hyperbola with equation 9x2 − 72x − 16y 2 − 128y − 256 = 0 are the foci of an ellipse that contains the point (8, −10). Find the standard equation of the ellipse. Solution:
D
9x2 − 72x − 16y 2 − 128y − 256 = 0 9(x2 − 8x) − 16(y 2 + 8y) = 256 9(x2 − 8x + 16) − 16(y 2 + 8y + 16) = 256 + 9(16) − 16(16) (x − 4)2 (y + 4)2 − =1 16 9
The vertices of the hyperbola are (0, −4) and (8, −4). Since these are the foci of the ellipse, the ellipse is horizontal with center C(4, −4); also, the focal distance of the ellipse is c = 4. The sum of the distances of the point (8, −10) from the foci is p p (8 − 0)2 + (−10 − (−4))2 + (8 − 8)2 + (−10 − (−4))2 = 16.
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This sum is constant for any point on the ellipse; so 2a = 16 and a = 8. Then b2 = 82 − 42 = 48. Therefore, the equation of the ellipse is (x − 4)2 (y + 4)2 + = 1. 64 48
Supplementary Problems 1.5 1. 9x2 + 72x − 64y 2 + 128y + 80 = 0 2. 49x2 − 490x + 36y 2 + 504y + 1225 = 0 3. y 2 + 56x − 18y + 417 = 0
5. x2 − 10x − 48y + 265 = 0
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4. x2 + 20x + y 2 − 20y + 200 = 0
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For items 1 to 8, identify the graph of each of the following equations.
6. −144x2 − 1152x + 25y 2 − 150y − 5679 = 0
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7. x2 + 4x + 16y 2 − 128y + 292 = 0
EP E
8. x2 − 6x + y 2 + 14y + 38 = 0
9. An ellipse has equation 100x2 − 1000x + 36y 2 − 144y − 956 = 0. Find the standard equations of all circles whose center is a focus of the ellipse and which contains at least one of the ellipse’s vertices. 10. Find all parabolas whose focus is a focus of the hyperbola x2 −2x−3y 2 −2 = 0 and whose directrix contains the top side of the hyperbola’s auxiliary rectangle.
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11. Find the equation of the circle that contains all corners of the auxiliary rectangle of the hyperbola −x2 − 18x + y 2 + 10y − 81 = 0.
12. Find the equations of all horizontal parabolas whose focus is the center of the ellipse 9x2 + 17y 2 − 170y + 272 = 0 and whose directrix is tangent to the same ellipse.
13. Find all values of r 6= 1 so that the graph of (r − 1)x2 + 14(r − 1)x + (r − 1)y 2 − 6(r − 1)y = 60 − 57r is (a) a circle, 66 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(b) a point, (c) the empty set. 14. Find all values of m 6= −7, 0 so that the graph of 2mx2 − 16mx + my 2 + 7y 2 = 2m2 − 18m is (a) a circle. (b) a horizontal ellipse.
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(c) a vertical ellipse. (d) a hyperbola.
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(e) the empty set.
4
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Lesson 1.6. Systems of Nonlinear Equations Learning Outcomes of the Lesson
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At the end of the lesson, the student is able to: (1) illustrate systems of nonlinear equations; (2) determine the solutions of systems of nonlinear equations using techniques such as substitution, elimination, and graphing; and (3) solve situational problems involving systems of nonlinear equations.
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Lesson Outline
(1) Review systems of linear equations (2) Solving a system involving one linear and one quadratic equation (3) Solving a system involving two quadratic equations (4) Applications of systems of nonlinear equations Introduction After recalling the techniques used in solving systems of linear equations in Grade 8, we extend these methods to solving a system of equations to systems in which the equations are not necessarily linear. In this lesson, the equations are restricted to linear and quadratic types, although it is possible to adapt the 67
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methodology to systems with other types of equations. We focus on quadratic equations for two reasons: to include a graphical representation of the solution and to ensure that either a solution is obtained or it is determined that there is no solution. The latter is possible because of the quadratic formula. 1.6.1. Review of Techniques in Solving Systems of Linear Equations Recall the methods we used to solve systems of linear equations.There were three methods used: substitution, elimination, and graphical.
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Example 1.6.1. Use the substitution method to solve the system, and sketch the graphs in one Cartesian plane showing the point of intersection. 4x + y = 6 5x + 3y = 4 Solution. Isolate the variable y in the first equation, and then substitute into the second equation.
EP E
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4x + y = 6 =⇒ y = 6 − 4x
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5x + 3y = 4 5x + 3(6 − 4x) = 4 −7x + 18 = 4 x=2 y = 6 − 4(2) = −2
Example 1.6.2. Use the elimination method to solve the system, and sketch the graphs in one Cartesian plane showing the point of intersection. 2x + 7 = 3y 4x + 7y = 12 Solution. We eliminate first the variable x. Rewrite the first equation wherein only the constant term is on the right-hand side of the equation, then multiply it by −2, and then add the resulting equation to the second equation.
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2x − 3y = −7 (−2)(2x − 3y) = (−2)(−7) −4x + 6y = 14 −4x + 6y = 14 4x + 7y = 12
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13y = 26 y=2 1 x=− 2
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1.6.2. Solving Systems of Equations Using Substitution
We begin our extension with a system involving one linear equation and one quadratic equation. In this case, it is always possible to use substitution by solving the linear equation for one of the variables.
EP E
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Example 1.6.3. Solve the following system, and sketch the graphs in one Cartesian plane. x−y+2=0 y − 1 = x2
Solution. We solve for y in terms of x in the first equation, and substitute this expression to the second equation.
D
x−y+2=0
y − 1 = x2 (x + 2) − 1 = x2 x2 − x − 1 = 0 √ 1± 5 x= 2 Solutions:
=⇒
y =x+2
√ √ √ 1+ 5 1+ 5 5+ 5 x= =⇒ y = +2= 2√ 2√ 2√ 1− 5 1− 5 5− 5 x= =⇒ y = +2= 2 2 2 √ √ ! 1+ 5 5+ 5 , and 2 2
√ √ ! 1− 5 5− 5 , 2 2
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C O
PY
The first equation represents a line with x-intercept −2 and y-intercept 2, while the second equation represents a parabola with vertex at (0, 1) and which opens upward.
1.6.3. Solving Systems of Equations Using Elimination
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Elimination method is also useful in systems of nonlinear equations. Sometimes, some systems need both techniques (substitution and elimination) to solve them.
EP E
Example 1.6.4. Solve the following system: y 2 − 4x − 6y = 11 4(3 − x) = (y − 3)2 . Solution 1. We expand the second equation, and eliminate the variable x by adding the equations.
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4(3 − x) = (y − 3)2 =⇒ 12 − 4x = y 2 − 6y + 9 =⇒ y 2 + 4x − 6y = 3 y 2 − 4x − 6y = 11 y 2 + 4x − 6y = 3
Adding these equations, we get 2y 2 −12y = 14 =⇒ y 2 −6y−7 = 0 =⇒ (y−7)(y+1) = 0 =⇒ y = 7 or y = −1. Solving for x in the second equation, we have x=3−
(y − 3)2 . 4
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y = 7 =⇒ x = −1
and
y = −1 =⇒ x = −1
Solutions: (−1, 7) and (−1, −1)
2
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PY
The graphs of the equations in the preceding example with the points of intersection are shown below.
EP E
D
Usually, the general form is more convenient to use in solving systems of equations. However, sometimes the solution can be simplified by writing the equations in standard form. Moreover, the standard form is best for graphing. Let us again solve the previous example in a different way.
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Solution 2. By completing the square, we can change the first equation into standard form: y 2 − 4x − 6y = 11 =⇒ 4(x + 5) = (y − 3)2 . 4(x + 5) = (y − 3)2 4(3 − x) = (y − 3)2 Using substitution or the transitive property of equality, we get 4(x + 5) = 4(3 − x) =⇒ x = −1. Substituting this value of x into the second equation, we have 4[3 − (−1)] = (y − 3)2 =⇒ 16 = (y − 3)2 =⇒ y = 7 or y = −1. The solutions are (−1, 7) and (−1, −1), same as Solution 1.
2
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Example 1.6.5. Solve the system and graph the curves: (x − 3)2 + (y − 5)2 = 10 x2 + (y + 1)2 = 25. Solution. Expanding both equations, we obtain x2 + y 2 − 6x − 10y + 24 = 0 x2 + y 2 + 2y − 24 = 0.
PY
Subtracting these two equations, we get −6x − 12y + 48 = 0 =⇒ x + 2y − 8 = 0 x = 8 − 2y.
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We can substitute x = 8 − 2y to either the first equation or the second equation. For convenience, we choose the second equation.
D
x2 + y 2 + 2y − 24 = 0 (8 − 2y)2 + y 2 + 2y − 24 = 0 y 2 − 6y + 8 = 0 y = 2 or y = 4 y = 2 =⇒ x = 8 − 2(2) = 4
and
y = 4 =⇒ x = 8 − 2(4) = 0
EP E
The solutions are (4, 2) and (0, 4).
D
√ The graphs of both equations are circles. One has center (3, 5) and radius 10, while the other has center (0, −1) and radius 5. The graphs with the points of intersection are show below.
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1.6.4. Applications of Systems of Nonlinear Equations Let us apply systems of equations to a problem involving modern-day television sets.
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? Example 1.6.6. The screen size of television sets is given in inches. This indicates the length of the diagonal. Screens of the same size can come in different shapes. Wide-screen TV’s usually have screens with aspect ratio 16 : 9, indicating the ratio of the width to the height. Older TV models often have aspect ratio 4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and the width of the screen.
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Solution. Let w represent the width and h the height of the screen. Then, by Pythagorean Theorem, we have the system w2 + h2 = 402 =⇒ w2 + h2 = 1600 w = 16 =⇒ h = 9w h 9 16 2
2
w +
EP E
D
w + h = 1600 =⇒
2
9w 16
2 = 1600
337w2 = 1600 256 r 409 600 w= ≈ 34.86 337
h=
19x 19(34.86) ≈ = 19.61 16 16
D
Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35.86 inches wide and 19.61 inches high. 2
More Solved Examples Solve the system and graph the curves. x2 − y 2 = 21 1. x+y = 7 Solution: We can write y in terms of x using the second equation as y = 7 − x. 73 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Substituting this into the first equation, we have x2 − (7 − x)2 = 21 14x − 70 = 0 x = 5.
EP E
D
C O
PY
Thus, the point of intersection is (5, 2).
x2 + y 2 − x + 6y + 5 = 0 2. x+y+1 = 0
Solution: We can write y in terms of x using the second equation as y = −(x + 1).
D
Substituting this into the first equation, we have x2 + (−(x + 1))2 − x + 6(−(x + 1)) + 5 = 0 2x2 − 5x = 0 x(2x − 5) = 0,
5 which yields x = 0 and x = . Thus, the points of intersection are (0, −1) 2 5 7 and ,− . 2 2
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PY C O
(y − 2)2 = 4(x − 4) 3. (y − 4)2 = x − 5
Solution: We can rewrite the first equation as x−4=
(y − 2)2 , 4
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which can be substituted into the second equation by rewriting it as (y − 2)2 (y − 4) = (x − 4) − 1 = −1 4
EP E
2
which upon expansion yields
3y 2 − 28y + 64 = 0.
D
This equation has roots y = 16/3 and y = 4, giving us the points (5, 4) and 61 16 , . 9 3
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4.
1 (y + 5)2 − 2 2 = −9
x =
y 2 + 10y + (x − 2)2
Solution: We can rewrite the first equation as (y + 5)2 = 2x + 4, which can be substituted into the second equation by completing the square to get
PY
(y 2 + 10y + 25) + (x − 2)2 = −9 + 25 (y + 5)2 + (x − 2)2 = 16 (2x + 4) + (x − 2)2 = 16 x2 − 2x − 8 = 0,
D
EP E
D
C O
This equation has roots x =√−2and x = 4, giving us the points (−2, −5), √ (4, −5 − 12), and 4, −5 + 12 .
Find the system of equations that represents the given problem and solve.
5. The difference of two numbers is 12, and the sum of their squares is 144. Find the numbers. Solution: If x and y are the two numbers, then we have the resulting system ( x−y = 12 x2 + y 2 = 144, where the first equation yields x = y + 12. Combining this with the second equation yields (y + 12)2 + y 2 = 144 or equivalently 2y(y + 12) = 0, giving us the ordered pairs (12, 0) and (0, −12). 76
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C O
1. Solve the system and graph the curves: x2 + 3x − y + 2 = 0 (a) y − 5x = 1 (y − 2)2 = 9(x + 2) (b) 9x2 + 4y 2 + 18x − 16y = 0 (x + 1)2 + 2(y − 4)2 = 12 (c) y 2 − 8y = 4x − 16 x2 − 2x − 4y 2 + 8y − 2 = 0 (d) 5x2 − 10x + 12y 2 + 24y − 58 = 0 x2 + y 2 = 2 (e) x−y = 4
PY
Supplementary Problems 1.6
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2. Ram is speeding along a highway when he sees a police motorbike parked on the side of the road right next to him. He immediately starts slowing down, but the police motorbike accelerates to catch up with him. It is assumed that the two vehicles are going in the same direction in parallel paths.
EP E
The distance that Ram has traveled in meters t seconds after he starts to slow down is given by d (t) = 150 + 75t − 1.2t2 . The distance that the police motorbike travels can be modeled by the equation d (t) = 4t2 . How long will it take for the police motorbike to catch up to Ram?
D
3. The square of a certain number exceeds twice the square of another number 1 5 by . Also, the sum of their squares is . Find possible pairs of numbers 8 16 that satisfy these conditions.
4. Solve the system of equations x2 + y 2 = 41 xy = 20 5. Determine the value(s) of k such that the circle x2 + (y − 6)2 = 36 and the parabola x2 = 4ky will intersect only at the origin.
4
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Topic Test 1 for Unit 1 1. Identify the graph of each of the following equations. (a) x2 − x + y 2 + 3y −
3 2
(c) 3x2 − 42x − 4y 2 − 24y + 99 = 0
=0
(b) x2 + 4x − 14y = 52
(d) 7x2 − 112x + 2y 2 + 448 = 0
2. Determine and sketch the conic with the given equation. Identify the important parts of the conic and include them in the graph.
PY
(a) 25x2 + 7y 2 − 175 = 0 (b) −64x2 + 128x + 36y 2 + 288y − 1792 = 0
3. Find the equation of the conic with the given properties.
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(a) parabola; vertex at (−1, 3); directrix x = −7 (b) hyperbola; asymptotes y = (3, −5)
12 x 5
−
1 5
and y = − 12 x− 5
49 ; 5
one vertex at
EP E
D
4. Solve the following system of equations: (x − 1)2 + (y + 1)2 = 5 y = 2(x − 1)2 − 8
5. A doorway is in the shape of a rectangle capped by a semi-ellipse. If the rectangle is 1 m wide and 2 m high while the ellipse is 0.3 m high at the center, can a cabinet that is 2.26 m high, 0.5 m wide, and 2 m long be pushed through the doorway? Assume that the cabinet cannot be laid down on its side.
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6. A point moves so that its distance from the point (0, −1) is twice its distance from the line x = 3. Derive the equation (in standard form) of the curve that is traced by the point, and identify the curve.
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Topic Test 2 for Unit 1 1. Identify the graph of each of the following equations. (a) y 2 + 8x − 10y = 15 (b) x2 + 10x + y 2 + 18y + 110 = 0 (c) 9x2 + 36x + 4y 2 − 8y + 4 = 0 (d) −11x2 + 132x + 17y 2 − 136y − 124 = 0
PY
2. Determine and sketch the conic with the given equation. Identify the important parts of the conic and include them in the graph. (a) x2 − y 2 = 64
(b) 4x2 + 24x + 49y 2 − 196y + 36
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3. Find the equation of the conic with the given properties. (a) parabola; directrix y = −2; focus at (7, −12)
(b) ellipse; vertical or horizontal major axis; one vertex at (−5, 12); one covertex at (−1, 3)
EP E
D
4. Solve the following system of equations: 9x2 − 4y 2 + 54x + 45 = 0 (x + 3)2 = 4y + 4
D
5. Nikko goes to his garden to water his plants. He holds the water hose 3 feet above the ground, with the hoses opening as the vertex and the water flow following a parabolic path. The water strikes the ground a horizontal distance of 2 feet from where the opening is located. If he were to stand on a 1.5 feet stool, how much further would the water strike the ground? 6. A point moves so that its distance from the point (2, 0) is two-thirds its distance from the line y = 5. Derive the equation (in standard form) of the curve that is traced by the point, and identify the curve.
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Unit 2
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Mathematical Induction
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Batad Rice Terraces in Ifugao, by Ericmontalban, 30 September 2012,
https://commons.wikimedia.org/wiki/File%3ABatad rice terraces in Ifugao.jpg. Public Domain.
Listed as one of the United Nations Educational, Scientific and Cultural Organization (UNESCO) World Heritage sites since 1995, the two-millenniumold Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimony of mankind’s creative engineering to adapt to physically-challenging environment in nature. One of the five clusters of terraces inscribed in the UNESCO list is the majestic Batad terrace cluster (shown above), which is characterized by its amphitheater-like, semicircular terraces with a village at its base.
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Lesson 2.1. Review of Sequences and Series Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) illustrate a series; and (2) differentiate a series from a sequence.
(1) Sequences and series
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Lesson Outline (2) Different types of sequences and series (Fibonacci sequence, arithmetic and geometric sequence and series, and harmonic series)
Introduction
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(3) Difference between sequence and series
In this lesson, we will review the definitions and different types of sequences and series. Lesson Proper
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Recall the following definitions:
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A sequence is a function whose domain is the set of positive integers or the set {1, 2, 3, . . . , n}.
A series represents the sum of the terms of a sequence.
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If a sequence is finite, we will refer to the sum of the terms of the sequence as the series associated with the sequence. If the sequence has infinitely many terms, the sum is defined more precisely in calculus.
A sequence is a list of numbers (separated by commas), while a series is a sum of numbers (separated by “+” or “−” sign). As an illustration, 1, − 21 , 13 , − 41 7 is a sequence, and 1 − 21 + 13 − 41 = 12 is its associated series. The sequence with nth term an is usually denoted by {an }, and the associated series is given by S = a1 + a2 + a3 + · · · + an .
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Example 2.1.1. Determine the first five terms of each defined sequence, and give their associated series. (1) {2 − n} (3) {(−1)n } (2) {1 + 2n + 3n2 }
(4) {1 + 2 + 3 + · · · + n}
Solution. We denote the nth term of a sequence by an , and S = a1 + a2 + a3 + a4 + a5 . (1) an = 2 − n First five terms: a1 = 2 − 1 = 1, a2 = 2 − 2 = 0, a3 = −1, a4 = −2, a5 = −3
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Associated series: S = a1 + a2 + a3 + a4 + a5 = 1 + 0 − 1 − 2 − 3 = −5 (2) an = 1 + 2n + 3n2
First five terms: a1 = 1 + 2 · 1 + 3 · 12 = 6, a2 = 17, a3 = 34, a4 = 57, a5 = 86
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Associated series: S = 6 + 17 + 34 + 57 + 86 = 200 (3) an = (−1)n
First five terms: a1 = (−1)1 = −1, a2 = (−1)2 = 1, a3 = −1, a4 = 1, a5 = −1 Associated series: S = −1 + 1 − 1 + 1 − 1 = −1
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(4) an = 1 + 2 + 3 + · · · + n
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First five terms: a1 = 1, a2 = 1+2 = 3, a3 = 1+2+3 = 6, a4 = 1+2+3+4 = 10, a5 = 1 + 2 + 3 + 4 + 5 = 15 Associated series: S = 1 + 3 + 6 + 10 + 15 = 35
2
The sequence {an } defined by an = an−1 + an−2 for n ≥ 3, where a1 = a2 = 1, is called a Fibonacci sequence. It terms are 1, 1, 2, 3, 5, 8, 13, . . ..
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An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant (called the common difference) to the preceding term.
If the nth term of an arithmetic sequence is an and the common difference is d, then an = a1 + (n − 1)d. The associated arithmetic series with n terms is given by Sn =
n(a1 + an ) n[2a1 + (n − 1)d] = . 2 2 82
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A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant (called the common ratio). If the nth term of a geometric sequence is an and the common ratio is r, then an = a1 rn−1 .
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The associated geometric series with n terms is given by na1 if r = 1 n Sn = a1 (1 − r ) if r = 6 1. (1 − r)
The proof of this sum formula is an example in Lesson 2.3.
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When −1 < r < 1, the infinite geometric series
a1 + a1 r + a1 r2 + · · · + a1 rn−1 + · · · has a sum, and is given by
a1 . 1−r
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S=
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If {an } is an arithmetic sequence, then the sequence with nth term bn = a1n is a harmonic sequence.
More Solved Examples
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1. How many terms are there in an arithmetic sequence with first term 5, common difference −3, and last term −76? Solution: a1 = 5, d = −3, an = −76. Find n.
an = 5 + (n − 1)(−3) = −76 −76 − 5 n−1= = 27, ⇒ −3
n = 28
2. List the first three terms of the arithmetic sequence if the 25th term is 35 and the 30th term is 5. Solution: a24 = a1 + 24d = 35 and a30 = a1 + 29d = 5 Eliminating a1 by subtraction, 5d = −30, or d = −6 This implies that a1 = 179, and the first three terms are 179, 173, 167. 83 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
3. Find the sum of all positive three-digit odd integers. Solution: Find sn if a1 = 101 = 1 + 50(2), an = 999 = 1 + 499(2). There are 450 terms from a1 to an , hence n = 450. 450(101 + 999) = 247 500 sn = 2 4. The seventh term of a geometric sequence is −6 and the tenth term is 162. Find the fifth term.
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Solution: a7 = a1 r6 = −6 and a10 = a1 r9 = 162. a10 162 Eliminating a1 by division: = r3 = = −27. Thus r = −3 a7 −6 −6 2 Since a5 r2 = a7 , a5 = =− . 9 3
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5. Insert three numbers (called geometric means) between 6 and 32/27, so that the five numbers form a geometric sequence.
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Solution: If a1 = 6 and there are three terms between a1 and 32/27, then a5 = 32/27. 16 2 32 a5 = 6(r)4 = ⇒ r4 = ⇒r=± 27 81 3 One possible set of three numbers is 4, 8/3, 16/9, the other is −4, 8/3, −16/9.
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6. A ball dropped from the top of a building 180 m high always rebounds threefourths the distance it has fallen. How far (up and down) will the ball have traveled when it hits the ground for the 6th time? Solution: a1 = 180, r = 3/4, n = 6 h i 3 6 180 1 − 4 s6 = 1 − 34
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The ball traveled 2s6 − 180 ≈ 1003.71 meters. 7. The Cantor set is formed as follows. Divide a segment of one unit into three equal parts. Remove the middle one-third of the segment. From each of the two remaining segments, remove the middle third. From each of the remaining segments, remove the middle third. This process is continued indefinitely. Find the total length of the segments removed. Solution: Let an represent the total length removed in the nth iteration. Hence a1 = 1/3, a2 = 2/9, a3 = 4/27, and so on. 1 2 This means r = . The sum to infinity is s = 3 3 1−
2 3
= 1.
The total length of the segments removed is 1 unit. 84 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
8. The 7th term of an arithmetic sequence is 25. Its first, third, and 21st term form a geometric sequence. Find the first term and the common difference of the sequence. Solution: a7 = a1 + 6d = 25 ⇒ a1 = 25 − 6d a21 a3 = , or a1 a21 = a23 . a1 a3 (25 − 6d) (25 − 6d + 20d) = (25 − 6d + 2d)2 d = 0 and an = 25 for all n, or d = 4 and a1 = 1.
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9. Let {an } be an arithmetic sequence and {bn } an arithmetic sequence of positive integers. Prove that the sequence with nth term abn is arithmetic. Solution: Let the common difference of {an } be d and of {bn } be c. abn+1 − abn = [a1 + (bn+1 − 1) d] − [a1 + (bn − 1) d] = bn+1 − bn = c
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This proves that the difference between any two consecutive terms of {abn } is a constant independent of n. 10. Let {an } be a geometric sequence. Prove that {a3n } is a geometric sequence. Solution: Let r be the common ratio of {an }. 3
n−1
a3n = (a1 rn−1 ) = a31 (r3 )
.
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Thus {a3n } is a geometric sequence with first term a31 and common ratio r3 .
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11. If {an } is a sequence such that its first three terms form both an arithmetic and a geometric sequence, what can be concluded about {an }? Solution: There is a real number r such that a2 = a1 r and a3 = a1 r2 . Since a1 , a2 and a3 form an arithmetic sequence, then a2 − a1 = a3 − a2 , or a3 − 2a2 + a1 = 0. a3 − 2a2 + a1 = a1 r2 − 2a1 r + a1 = a1 (r − 1)2 = 0 ⇒ a1 = 0 or r = 1.
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If a1 = 0, then a1 = a2 = a3 = 0. If r = 1, then a1 = a2 = a3 . In all cases, a1 = a2 = a3 .
Supplementary Problems 2.1 1. Find the 5th term of the arithmetic sequence whose 3rd term is 35 and whose 10th term is 77. 2. Suppose that the fourth term of a geometric sequence is 8 is 81 . Find the first term and the common ratio.
2 9
and the sixth term
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3. The partial sum in the arithmetic series with first term 17 and a common difference 3 is 30705. How many terms are in the series? 4. An arithmetic sequence a1 , a2 , . . . , a100 has a sum of 15,000. Find the first term and the common difference if the sum of the terms in the sequence a3 , a6 , a9 , . . . , a99 is 5016. 5. The sum of an infinite geometric series is 108, while the sum of the first 3 terms is 112. Determine the first term of this series. 32 − 20 33 − 21 3k+1 − 2k−1 + + · · · + + ···. 51 52 5k
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6. Evaluate the infinite series
7. Let n = 0.123 = 0.123123 . . . be a nonterminating repeating decimal. Find a rational number that is equal to n by expressing n as an infinite geometric series. Simplify your answer.
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8. An arithmetic sequence whose first term is 2 has the property that its second, third, and seventh terms are consecutive terms of a geometric sequence. Determine all possible second terms of the arithmetic sequence.
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9. Eighty loaves of bread are to be divided among 4 people so that the amounts they receive form an arithmetic progression. The first two together receive one-third of what the last two receive. How many loaves does each person receive?
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10. Given a and b, suppose that three numbers are inserted between them so that the five numbers form a geometric sequence. If the product of the three inserted numbers between a and b is 27, show that ab = 9. 11. For what values of n will the infinite series (2n − 1) + (2n − 1)2 + . . . + (2n − 1)i + . . . have a finite value?
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4
Lesson 2.2. Sigma Notation Learning Outcomes of the Lesson At the end of the lesson, the student is able to use the sigma notation to represent a series. Lesson Outline (1) Definition of and writing in sigma notation 86
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(2) Evaluate sums written in sigma notation (3) Properties of sigma notation (4) Calculating sums using the properties of sigma notation Introduction The sigma notation is a shorthand for writing sums. In this lesson, we will see the power of this notation in computing sums of numbers as well as algebraic expressions.
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2.2.1. Writing and Evaluating Sums in Sigma Notation
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Mathematicians use the sigma notation to denote a sum. The uppercase Greek letter Σ (sigma) is used to indicate a “sum.” The notation consists of several components or parts. Let f (i) be an expression involving an integer i. The expression f (m) + f (m + 1) + f (m + 2) + · · · + f (n) can be compactly written in sigma notation, and we write it as
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n X
f (i),
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i=m
which is read “the summation of f (i) from i = m to n.” Here, m and n are integers with m ≤ n, f (i) is a term (or summand ) of the summation, and the letter i is the index, m the lower bound, and n the upper bound.
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Example 2.2.1. Expand each summation, and simplify if possible. n 4 X X (3) ai (1) (2i + 3) i=1
i=2
(2)
5 X i=0
i
2
√ 6 X n (4) n+1 n=1
Solution. We apply the definition of sigma notation. (1)
4 X
(2i + 3) = [2(2) + 3] + [2(3) + 3] + [2(4) + 3] = 27
i=2
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(2)
5 X
2i = 20 + 21 + 22 + 23 + 24 + 25 = 63
i=0
(3)
n X
ai = a1 + a2 + a3 + · · · + an
i=1
√ √ √ √ √ 6 X n 1 2 3 2 5 6 = + + + + + (4) n+1 2 3 4 5 6 7 n=1
2
Example 2.2.2. Write each expression in sigma notation. 1 1 1 1 + + + ··· + 2 3 4 100 (2) −1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 − 9 + · · · − 25
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(1) 1 +
(3) a2 + a4 + a6 + a8 + · · · + a20 1 1 1 1 1 1 1 + + + + + + 2 4 8 16 32 64 128
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(4) 1 +
100
X1 1 1 1 1 Solution. (1) 1 + + + + · · · + = 2 3 4 100 n=1 n
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25 X
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(2) −1 + 2 − 3 + 4 − 5 + · · · − 25 = (−1)1 1 + (−1)2 2 + (−1)3 3 + (−1)4 4 + (−1)5 5 + · · · + (−1)25 25 =
(−1)j j
j=1
(3) a2 + a4 + a6 + a8 + · · · + a20 = a2(1) + a2(2) + a2(3) + a2(4) + · · · + a2(10) a2i
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=
10 X i=1
7
X 1 1 1 1 1 1 1 1 (4) 1 + + + + + + + = 2 4 8 16 32 64 128 k=0 2k
2
The sigma notation of a sum expression is not necessarily unique. For example, the last item in the preceding example can also be expressed in sigma notation as follows: 8 X 1 1 1 1 1 1 1 1 1+ + + + + + + = . 2 4 8 16 32 64 128 k=1 2k−1 However, this last sigma notation is equivalent to the one given in the example. 88 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
2.2.2. Properties of Sigma Notation We start with finding a formula for the sum of n X
i = 1 + 2 + 3 + ··· + n
i=1
in terms of n.
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The sum can be evaluated in different ways. One informal but simple approach is pictorial.
i = 1 + 2 + 3 + ··· + n =
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n X i=1
n(n + 1) 2
Another way is to use the formula for an arithmetic series with a1 = 1 and an = n: n(a1 + an ) n(n + 1) S= = . 2 2
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We now derive some useful summation facts. They are based on the axioms of arithmetic addition and multiplication. n X i=m
cf (i) = c
n X
f (i),
c any real number.
i=m
Proof. n X
cf (i) = cf (m) + cf (m + 1) + cf (m + 2) + · · · + cf (n)
i=m
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= c[f (m) + f (m + 1) + · · · + f (n)] n X =c f (i)
2
i=m
n X
[f (i) + g(i)] =
i=m
n X
f (i) +
i=m
n X
g(i)
i=m
n X
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Proof. [f (i) + g(i)]
i=m
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= [f (m) + g(m)] + · · · + [f (n) + g(n)] = [f (m) + · · · + f (n)] + [g(m) + · · · + g(n)] n n X X = f (i) + g(i) i=m
2
i=m
c = c(n − m + 1)
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n X
EP E
i=m
Proof.
n X i=m
c = c| + c + c{z+ · · · + }c n−m+1 terms
2
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= c(n − m + 1)
A special case of the above result which you might encounter more often is the following: n X c = cn. i=1
Telescoping Sum n X
[f (i + 1) − f (i)] = f (n + 1) − f (m)
i=m
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Proof. n X f (i + 1) − f (i) i=m
= [f (m + 1) − f (m)] + [f (m + 2) − f (m + 1)] + [f (m + 3) − f (m + 2)] + · · · + [f (n + 1) − f (n)] Note that the terms, f (m + 1), f (m + 2), . . . , f (n), all cancel out. Hence, we have n X
2
[f (i + 1) − f (i)] = f (n + 1) − f (m).
Example 2.2.3. Evaluate:
30 X
PY
i=m
(4i − 5).
Solution. 30 X
C O
i=1
(4i − 5) =
i=1
30 X
4i −
i=1 30 X
i−
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=4
i=1
30 X
i=1 30 X
5
5
i=1
(30)(31) − 5(30) 2 = 1710
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=4
2
Example 2.2.4. Evaluate:
1 1 1 1 + + + ··· + . 1·2 2·3 3·4 99 · 100
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Solution.
1 1 1 1 + + + ··· + 1·2 2·3 3·4 99 · 100 99 X 1 = i(i + 1) i=1 = =
99 X i+1−i
i(i + 1)
i=1 99 X i=1
i+1 i − i(i + 1) i(i + 1)
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99 X 1
1 = − i i+1 i=1 99 X 1 1 =− − i+1 i i=1 1 and the telescoping-sum property, we get i 99 X 1 1 99 1 =− − = . i(i + 1) 100 1 100 i=1
Example 2.2.5. Derive a formula for
n X
2
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Using f (i) =
i2 using a telescoping sum with terms
i=1
C O
f (i) = i3 .
Solution. The telescoping sum property implies that
n X 3 i − (i − 1)3 = n3 − 03 = n3 . i=1
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On the other hand, using expansion and the other properties of summation, we have
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n n X 3 X i − (i − 1)3 = (i3 − i3 + 3i2 − 3i + 1) i=1
i=1 n X
=3
n n X X i −3 i+ 1 2
i=1
i=1
i=1
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n X n(n + 1) + n. =3 i2 − 3 · 2 i=1
Equating the two results above, we obtain n X 3n(n + 1) 3 i2 − + n = n3 2 i=1 n X 6 i2 − 3n(n + 1) + 2n = 2n3 i=1 n X 6 i2 = 2n3 − 2n + 3n(n + 1) i=1
92 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
= 2n(n2 − 1) + 3n(n + 1) = 2n(n − 1)(n + 1) + 3n(n + 1) = n(n + 1)[2(n − 1) + 3] = n(n + 1)(2n + 1).
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Finally, after dividing both sides of the equation by 6, we obtain the desired formula n X n(n + 1)(2n + 1) i2 = . 2 6 i=1
More Solved Examples
(a)
5 X
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1. Expand the following sums and simplify if possible: (i2 − i + 1)
i=1
i
i=3
(c)
5 X
2
i+1 2
2
x3i y 15−3i
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i=0
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(b)
6 X
9 X x2i+1 (d) (i + 1)2 i=1
(e)
∞ X
3−i+2 2i+1
i=1
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Solution: (a)
5 X
(i2 − i + 1) = (12 − 1 + 1) + (22 − 2 + 1) + . . . + (52 − 5 + 1) = 45 or
i=1
5(5 + 1)(2(5) + 1) 5(5 + 1) − + 5 = 45 6 2 2 2 2 6 X i+1 4 7 2 2 2 (b) =3 + ... + 6 = 802 i 2 2 2 i=3 (c)
5 X
x3i y 15−3i = x3(0) y 15−3(0) + . . . + x3(5) y 15−3(5) = xy 15 + x3 y 12 + x6 y 9 +
i=0
x9 y 6 + x12 y 3 + x15 y 93 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
9 X x2i+1 x3 x5 x19 (d) = + + . . . + = 4x3 + 9x5 + 16x7 + 2 2 2 2 (i + 1) (1 + 1) (2 + 1) (9 + 1) i=1 . . . + 100x19 ∞ ∞ i+1 X X 2 −i+2 i+1 (e) , which is an infinite geometric series with 3 2 = 27 3 i=1 i=1 ∞ X 2 4 4/9 |r| = < 1 and a1 = , giving us = 36. 3−i+2 2i+1 = 27 3 9 1 − (2/3) i=1
[2(i − 1) + 2].
i=1
(2(i − 1) + 2) =
i=1
3. Find a formula for integer n.
20 X
2i = 4
i=1
(20)(21) = 840. 2
1 1 1 1 + + + ··· + given any positive 1(3) 2(4) 3(5) n(n + 2)
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20 X
Solution:
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2. Evaluate
20 X
n
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or equivalently
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X 1 1 1 1 1 Solution: We have + + +· · ·+ = . Rewriting 1(3) 2(4) 3(5) n(n + 2) i(i + 2) i=1 yields (2 − 1) + (i − i) 1 1 1 1 = = − − i(i + 2) i(i + 2) i i + 2 i(i + 2) 1 1 = i+2 2
1 1 − i i+2
.
Expanding the sum term by term, n X 1 1 1 1 = − i(i + 2) 2 i i+2 i=1 1 1 1 1 1 = 1− + − + − ··· 3 2 4 3 5 1 1 1 1 1 1 + − + − + − n−2 n n−1 n+1 n n+2 1 1 1 =1+ − − 2 n+1 n+2 n(3n + 2) = . 2(n + 1)(n + 2)
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n X i=1
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4. Determine the value of N such that
N X i=0
i2
97 1 = . + 3i + 2 98
Solution: Rewrite the sum as N X
1 and use telescoping sums to get i+1 N X
N X 1 1 1 =− − i2 + 3i + 2 i+2 i+1 i=1 1 1 − =− N +2 1 N +1 = . N +2
EP E
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i=1
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Set f (i) =
PY
i=1
N
X 1 1 = 2 i + 3i + 2 (i + 1)(i + 2) i=1 N X 1 1 − = i + 1 i+2 i=1
Since we want the sum to be equal to
97 , N = 96. 98
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Supplementary Problems 2.2 1. Expand the following sums: (a)
10 X √
3·
i=3
(b)
5 X x2i i=1
(c)
i 2
5 X
2i (−1)i xi−1
i=2
2. Write the following in sigma notation. 95 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(a) (x + 5) − (x + 3)2 + (x + 1)3 − (x − 1)4 1 22 32 102 + + + . . . + 33 43 53 113 (c) a3 + a6 + a9 + . . . + a81
(b)
3. Evaluate the following sums (a)
150 X
(4i + 2)
i=1
i(i − 5)
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(b)
120 X i=3
(c)
50 X
(2i − 1)(2i + 1)
i=1 50 X
f (i) = 20 and
i=1
i=1
50 X g(i) + 3f (i) √ ? g(i) = 30, what is the value of 2 i=1
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4. If
50 X
200 200 X X 2 2 5. If s = (i − 1) − i , express i in terms of s. i=1
ai and t =
bi , does it follow that
i=1
n X ai i=1
s = ? bi t
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i=1
n X
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6. If s =
n X
i=1
4
Lesson 2.3. Principle of Mathematical Induction
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Learning Outcomes of the Lesson At the end of the lesson, the student is able to:
(1) illustrate the Principle of Mathematical Induction; and (2) apply mathematical induction in proving identities. Lesson Outline (1) State the Principle of Mathematical Induction (2) Prove summation identities using mathematical induction (3) Prove divisibility statements using mathematical induction (4) Prove inequalities using mathematical induction 96
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Introduction We have derived and used formulas for the terms of arithmetic and geometric sequences and series. These formulas and many other theorems involving positive integers can be proven with the use of a technique called mathematical induction. 2.3.1. Proving Summation Identities The Principle of Mathematical Induction
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Let P (n) be a property or statement about an integer n. Suppose that the following conditions can be proven: (1) P (n0 ) is true (that is, the statement is true when n = n0 ).
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(2) If P (k) is true for some integer k ≥ n0 , then P (k + 1) is true (that is, if the statement is true for n = k, then it is also true for n = k + 1). Then the statement P (n) is true for all integers n ≥ n0 .
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The Principle of Mathematical Induction is often compared to climbing an infinite staircase. First, you need to be able to climb up to the first step. Second, if you are on any step (n = k), you must be able to climb up to the next step (n = k + 1). If you can do these two things, then you will be able to climb up the infinite staircase.
Part 1
Part 2
Another analogy of the Principle of Mathematical Induction that is used is toppling an infinite line of standing dominoes. You need to give the first domino a push so that it falls down. Also, the dominoes must be arranged so that if the kth domino falls down, the next domino will also fall down. These two conditions will ensure that the entire line of dominoes will fall down.
97 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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Standing Domino Tiles, by Nara Cute, 16 October 2015, https://commons.wikimedia.org/wiki/File:Wallpaper kartu domino.png. Public Domain.
EP E
D
There are many mathematical results that can be proven using mathematical induction. In this lesson, we will focus on three main categories: summation identities, divisibility statements, and inequalities. Let us now take a look at some examples on the use of mathematical induction in proving summation identities. Example 2.3.1. Using mathematical induction, prove that 1 + 2 + 3 + ··· + n =
n(n + 1) 2
D
for all positive integers n.
Solution. We need to establish the two conditions stated in the Principle of Mathematical Induction. Part 1. Prove that the identity is true for n = 1.
The left-hand side of the equation consists of one term equal to 1. The righthand side becomes 1(1 + 1) 2 = = 1. 2 2 Hence, the formula is true for n = 1. 98 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Part 2. Assume that the formula is true for n = k ≥ 1: 1 + 2 + 3 + ··· + k =
k(k + 1) . 2
We want to show that the formula is true for n = k + 1; that is, 1 + 2 + 3 + · · · + k + (k + 1) =
(k + 1)(k + 1 + 1) . 2
PY
Using the formula for n = k and adding k + 1 to both sides of the equation, we get k(k + 1) + (k + 1) 2 k(k + 1) + 2(k + 1) = 2 (k + 1)(k + 2) = 2 (k + 1) [(k + 1) + 1] = 2
C O
1 + 2 + 3 + · · · + k + (k + 1) =
D
We have proven the two conditions required by the Principle of Mathematical Induction. Therefore, the formula is true for all positive integers n. 2
EP E
Example 2.3.2. Use mathematical induction to prove the formula for the sum of a geometric series with n terms: Sn =
a1 (1 − rn ) , 1−r
r 6= 1.
Solution. Let an be the nth term of a geometric series. From Lesson 2.1, we know that an = a1 rn−1 .
D
Part 1. Prove that the formula is true for n = 1. a1 (1 − r1 ) = a1 = S 1 1−r
The formula is true for n = 1. Part 2. Assume that the formula is true for n = k ≥ 1: Sk = want to prove that it is also true for n = k + 1; that is, Sk+1 =
a1 (1 − rk ) . We 1−r
a1 (1 − rk+1 ) . 1−r 99
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We know that Sk+1 = a1 + a2 + · · · + ak +ak+1 | {z } Sk
= Sk + ak+1
= = =
PY
=
a1 1 − r k + a1 rk 1−r a1 1 − rk + a1 rk (1 − r) 1−r k a1 1 − r + rk − rk+1 1 −r k+1 a1 1 − r 1−r
C O
By the Principle of Mathematical Induction, we have proven that Sn = for all positive integers n.
a1 (1 − rn ) 1−r
2
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Example 2.3.3. Using mathematical induction, prove that n(n + 1)(2n + 1) 6
EP E
12 + 22 + 32 + · · · + n2 = for all positive integers n.
Solution. We again establish the two conditions stated in the Principle of Mathematical Induction. Part 1
D
1·2·3 1(1 + 1)(2 · 1 + 1) = = 1 = 12 6 6 The formula is true for n = 1.
Part 2 k(k + 1)(2k + 1) . 6 Prove: 12 + 22 + 32 + · · · + k 2 + (k + 1)2 (k + 1)(k + 2) [2(k + 1) + 1] = 6 (k + 1)(k + 2)(2k + 3) = . 6
Assume: 12 + 22 + 32 + · · · + k 2 =
100 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
PY
12 + 22 + 32 + · · · + k 2 + (k + 1)2 k(k + 1)(2k + 1) = + (k + 1)2 6 k(k + 1)(2k + 1) + 6(k + 1)2 = 6 (k + 1) [k(2k + 1) + 6(k + 1)] = 6 (k + 1) (2k 2 + 7k + 6) = 6 (k + 1)(k + 2)(2k + 3) = 6 Therefore, by the Principle of Mathematical Induction,
for all positive integers n.
n(n + 1)(2n + 1) 6
C O
12 + 22 + 32 + · · · + n2 =
2
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2.3.2. Proving Divisibility Statements
We now prove some divisibility statements using mathematical induction.
EP E
Example 2.3.4. Use mathematical induction to prove that, for every positive integer n, 7n − 1 is divisible by 6. Solution. Similar to what we did in the previous session, we establish the two conditions stated in the Principle of Mathematical Induction. Part 1
D
71 − 1 = 6 = 6 · 1
71 − 1 is divisible by 6.
Part 2 Assume: 7k − 1 is divisible by 6. To show: 7k+1 − 1 is divisible by 6. 7k+1 − 1 = 7 · 7k − 1 = 6 · 7k + 7k − 1 = 6 · 7k + (7k − 1) By definition of divisibility, 6 · 7k is divisible by 6. Also, by the hypothesis (assumption), 7k − 1 is divisible by 6. Hence, their sum (which is equal to 7k+1 − 1) is also divisible by 6. 101 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Therefore, by the Principle of Math Induction, 7n − 1 is divisible by 6 for all positive integers n. 2 Note that 70 − 1 = 1 − 1 = 0 = 6 · 0 is also divisible by 6. Hence, a stronger and more precise result in the preceding example is: 7n − 1 is divisible by 6 for every nonnegative integer n. It does not make sense to substitute negative values of n since this will result in non-integer values for 7n − 1. Example 2.3.5. Use mathematical induction to prove that, for every nonnegative integer n, n3 − n + 3 is divisible by 3.
PY
Solution. We again establish the two conditions in the Principle of Mathematical Induction.
03 − 0 + 3 = 3 = 3(1) 03 − 0 + 3 is divisible by 3.
C O
Part 1 Note that claim of the statement is that it is true for every nonnegative integer n. This means that Part 1 should prove that the statement is true for n = 0.
D
Part 2. We assume that k 3 − k + 3 is divisible by 3. By definition of divisibility, we can write k 3 − k + 3 = 3a for some integer a.
EP E
To show: (k + 1)3 − (k + 1) + 3 is divisible by 3.
D
(k + 1)3 − (k + 1) + 3 = k 3 + 3k 2 + 2k + 3 = (k 3 − k + 3) + 3k 2 + 3k = 3a + 3k 2 + 3k = 3(a + k 2 + k)
Since a + k 2 + k is also an integer, by definition of divisibility, (k + 1)3 − (k + 1) + 3 is divisible by 3. Therefore, by the Principle of Math Induction, n3 − n + 3 is divisible by 3 for all positive integers n. 2
? 2.3.3. Proving Inequalities Finally, we now apply the Principle of Mathematical Induction in proving some inequalities involving integers. Example 2.3.6. Use mathematical induction to prove that 2n > 2n for every integer n ≥ 3. 102 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Solution. Just like the previous example, we establish the two conditions in the Principle of Mathematical Induction. Part 1 23 = 8 > 6 = 2(3) This confirms that 23 > 2(3). Part 2 Assume: 2k > 2k, where k is an integer with k ≥ 3
PY
To show: 2k+1 > 2(k + 1) = 2k + 2
C O
We compare the components of the assumption and the inequality we need to prove. On the left-hand side, the expression is doubled. On the right-hand side, the expression is increased by 2. We choose which operation we want to apply to both sides of the assumed inequality. Alternative 1. We double both sides.
Since 2k > 2k, by the multiplication property of inequality, we have 2 · 2k > 2 · 2k. 2k+1 > 2(2k) = 2k + 2k > 2k + 2 if k ≥ 3.
D
Hence, 2k+1 > 2(k + 1).
Alternative 2. We increase both sides by 2.
EP E
Since 2k > 2k, by the addition property of inequality, we have 2k + 2 > 2k + 2. 2(k + 1) = 2k + 2 < 2k + 2 < 2k + 2k if k ≥ 3. The right-most expression above, 2k + 2k , is equal to 2 2k = 2k+1 . Hence, 2(k + 1) < 2k+1 .
D
Therefore, by the Principle of Math Induction, 2n > 2n for every integer n ≥ 3. 2 We test the above inequality for integers less than 3. 20 = 1 > 0 = 2(0)
True
21 = 2 = 2(1)
False
22 = 4 = 2(2)
False
The inequality is not always true for nonnegative integers less than 3. This illustrates the necessity of Part 1 of the proof to establish the result. However, the result above can be modified to: 2n ≥ 2n for all nonnegative integers n. Before we discuss the next example, we review the factorial notation. Recall 103 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
that 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · · n. The factorial also satisfies the property that (n + 1)! = (n + 1) · n!. Example 2.3.7. Use mathematical induction to prove that 3n < (n + 2)! for every positive integer n. Can you refine or improve the result? Solution. We proceed with the usual two-part proof. Part 1 31 = 3 < 6 = 3! = (1 + 2)! =⇒ 31 < (1 + 2)!
PY
Thus, the desired inequality is true for n = 1. Part 2 Assume: 3k < (k + 2)!
C O
To show: 3k+1 < (k + 3)!
Given that 3k < (k + 2)!, we multiply both sides of the inequality by 3 and obtain 3 3k < 3 [(k + 2)!] .
D
This implies that 3 3k < 3 [(k + 2)!] < (k + 3) [(k + 2)!] , and so
since k > 0,
EP E
3k+1 < (k + 3)!.
Therefore, by the Principle of Math Induction, we conclude that 3n < (n + 2)! for every positive integer n. The left-hand side of the inequality is defined for any integer n. The righthand side makes sense only if n + 2 ≥ 0, or n ≥ −2. 1 < 1 = 0! = (−2 + 2)! 9 1 When n = −1: 3−1 = < 1 = 1! = (−1 + 2)! 3 0 When n = 0: 3 = 1 < 2 = 2! = (0 + 2)!
D
When n = −2: 3−2 =
Therefore, 3n < (n + 2)! for any integer n ≥ −2.
2
104 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
More Solved Examples Use mathematical induction to prove the given statements below. 1. 2 · 3 + 2 · 32 + . . . + 2 · 3n−1 = 3n − 3 for n ≥ 1 Solution: Part 1. 2 · 3 = 6 = 32 − 3. The formula is true for n = 1.
PY
Part 2.
Assume: P = 2 · 3 + 2 · 32 + . . . + 2 · 3k−1 = 3k − 3.
C O
To show: 2 · 3 + 2 · 32 + . . . + 2 · 3k = 3k+1 − 3.
2 · 3 + 2 · 32 + . . . + 2 · 3k = P + 2 · 3k
= 3k − 3 + 2 · 3k = 3 · 3k − 3 = 3k+1 − 3.
EP E
Solution:
1 n (4 − 1) for n ≥ 1 3
D
2. 1 + 4 + 42 + . . . + 4n−1 = Part 1.
1=
1 1 (4 − 1). 3
The formula is true for n = 1.
Part 2.
D
1 k 4 −1 . 3 1 To show: 1 + 4 + 42 + . . . + 4k = 4k+1 − 1 . 3 Assume: P = 1 + 4 + 42 + . . . + 4k−1 =
1 + 4 + 42 + . . . + 4k = P + 4k 1 k = 4 − 1 + 4k 3 4 1 = 4k − 3 3 1 k+1 = 4 −1 . 3 105 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
1 1 1 1 n+1 3. 1 − 2 · 1 − 2 · · · 1 − · 1− 2 = for n ≥ 2. 2 2 3 (n − 1) n 2n Solution: Part 1. 1−
1 2+1 3 . = = 2 2 4 2(2)
The formula is true for n = 2.
PY
Part 2.
C O
1 1 1 k+1 1 · 1− 2 = . Assume: P = 1 − 2 · 1 − 2 · · · 1 − 2 2 3 (k − 1) k 2k 1 1 1 k+2 To show: 1 − 2 · · · 1 − 2 · 1 − . = 2 k (k + 1)2 2(k + 1)
EP E
D
1 1 1 1 − 2 ··· 1 − =P · 1− 2 (k + 1)2 (k + 1)2 k + 1 (k 2 + 2k + 1) − 1 · = 2k (k + 1)2 k + 1 k(k + 2) = · 2k (k + 1)2 k+2 = . 2(k + 1)
4. Prove that 4n+1 + 52n−1 is divisible by 21 for all integers n ≥ 1. Solution:
D
Part 1.
41+1 + 52(1)−1 = 21.
The number is divisible by 21 for n = 1.
Part 2. Assume: 4k+1 + 52k−1 is divisible by 21. Prove: 4k+2 + 52(k+1)−1 is divisible by 21. 4k+2 + 52(k+1)−1 = 4 · 4k+1 + 25 · 52k−1 = 4 4k+1 + 52k−1 + 21 · 52k−1 21 · 52k−1 is divisible by 21 and by the hypothesis (assumption), 4k+1 + 52k−1 is divisible by 21. Hence, their sum which is equal to 4k+2 + 52(k+1)−1 is divisible by 21. 106 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
5. n2 > 2n + 3 for n ≥ 4. Solution: Part 1. 24 = 16 > 7 = 2(2) + 3 The inequality is true for n = 4. Part 2 Assume: k 2 > 2k + 3 Prove: (k + 1)2 > 2(k + 1) + 3
PY
We expand (k + 1)2 and use the inequality in the hypothesis to get (k + 1)2 = k 2 + 2k + 1 > (2k + 3) + 2k + 1 = 4(k + 1) > 2(k + 1) + 3 if k > 0.
C O
Therefore, by the principle of math induction, n2 > 2n + 3 for n ≥ 4. 6. Prove that 2n+3 < (n + 3)! for n ≥ 4. Solution: Part 1.
D
24+3 = 27 < 1 · 2 · 3 · · · 7 = (4 + 3)! The inequality is true for n = 1. Part 2
EP E
Assume: 2k+3 < (k + 3)! Prove: 2k+4 < (k + 4)!
Given that 2k+3 < (k + 3)!, we multiply both sides of the inequality by 2 and obtain
D
2 2k+3 < 2 [(k + 3)!].
This implies that 2k+4 < 2 [(k + 3)!] < (k + 4) [(k + 3)!], if k > 0.
Therefore, by the principle of math induction, 2k+3 < (k+3)! for every positive integer n.
107 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Supplementary Problems 2.3 Prove the following by mathematical induction: 1. 2.
2 1 3 n n+2 + 2 + 3 + · · · + n = 2 − n for n ≥ 1 2 2 2 2 2 n X
−(i + 1) = −
i=1
n(n + 3) 2
3. 1(1!) + 2(2!) + . . . + n(n!) = (n + 1)! − 1.
(−1)i i2 =
i=1
(−1)n n(n + 1) 2
C O
6.
n X
PY
4. The sum of the first n odd numbers is equal to n2 . 1 1 1 1 1 1− 1− ... 1 − = . 5. 1 − 2 3 4 n 2n
7. 43n+1 + 23n+1 + 1 is divisible by 7 8. 11n+2 + 122n+1 is divisible by 133
D
9. 52n+1 · 2n+2 + 3n+2 · 22n+1 is divisible by 19 10. 11n − 6 is divisible by 5
EP E
10n 5 11. + + 4n+2 is divisible by 3 3 3 12. n2 < 2n for n ≥ 5.
1 1 1 1 1 + + + . . . + ≤ 2 − for n ≥ 1. 13 23 33 n3 n √ p 14. The sequence an = 2an−1 , a1 = 2 is increasing; that is, an < an+1 .
D
13.
4
Lesson 2.4. The Binomial Theorem Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) illustrate Pascal’s Triangle in the expansion of (x + y)n for small positive integral values of n; 108 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(2) prove the Binomial Theorem; (3) determine any term in (x + y)n , where n is a positive integer, without expanding; and (4) solve problems using mathematical induction and the Binomial Theorem. Lesson Outline (1) Expand (x + y)n for small values of n using Pascal’s Triangle (2) Review the definition of and formula for combination (3) State and prove the Binomial Theorem
PY
(4) Compute all or specified terms of a binomial expansion
(5) Prove some combination identities using the Binomial Theorem Introduction
D
C O
In this lesson, we study two ways to expand (a + b)n , where n is a positive integer. The first, which uses Pascal’s Triangle, is applicable if n is not too big, and if we want to determine all the terms in the expansion. The second method gives a general formula for the expansion of (a + b)n for any positive integer n. This formula is useful especially when n is large because it avoids the process of going through all the coefficients for lower values of n obtained through Pascal’s Triangle.
EP E
2.4.1. Pascal’s Triangle and the Concept of Combination Consider the following powers of a + b:
D
(a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 (a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
We now list down the coefficients of each expansion in a triangular array as follows: n=1: 1 1 n=2:
1
n=3: n=4: n=5: 1
1 1
2 3
4 5
1 3
6 10
1 4
10
1 5
1
109 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
The preceding triangular array of numbers is part of what is called the Pascal’s Triangle, named after the French mathematician, Blaise Pascal (1623-1662). Some properties of the Triangle are the following: (1) Each row begins and ends with 1. (2) Each row has n + 1 numbers. (3) The second and second to the last number of each row correspond to the row number.
PY
(4) There is symmetry of the numbers in each row. (5) The number of entries in a row is one more than the row number (or one more than the number of entries in the preceding row).
C O
(6) Every middle number after first row is the sum of the two numbers above it. It is the last statement which is useful in constructing the succeeding rows of the triangle. Example 2.4.1. Use Pascal’s Triangle to expand the expression (2x − 3y)5 .
D
Solution. We use the coefficients in the fifth row of the Pascal’s Triangle.
EP E
(2x − 3y)5 = (2x)5 + 5(2x)4 (−3y) + 10(2x)3 (−3y)2 + 10(2x)2 (−3y)3 + 5(2x)(−3y)4 + (−3y)5 = 32x5 − 240x4 y + 720x3 y 2 − 1080x2 y 3 + 810xy 4 − 243y 5
2
Example 2.4.2. Use Pascal’s Triangle to expand (a + b)8 .
D
Solution. We start with the sixth row (or any row of the Pascal’s Triangle that we remember). n=6: n=7: n=8: 1
1 1
6 7
8
15 21
28
20 35
56
15 35
70
6 21
56
1 7
28
1 8
1
Therefore, we get (a + b)8 = a8 + 8a7 b + 28a6 b2 + 56a5 b3 + 70a4 b4 + 56a3 b5 + 28a2 b6 + 8ab7 + b8
2
110 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
We observe that, for each n, the expansion of (a + b)n starts with an and the exponent of a in the succeeding terms decreases by 1, while the exponent of b increases by 1. This observation will be shown to be true in general. Let us review the concept of combination. Recall that C(n, k) or nk counts the number of ways of choosing k objects from a set of n objects. It is also useful to know some properties of C(n, k):
PY
(1) C(n, 0) = C(n, n) = 1, (2) C(n, 1) = C(n, n − 1) = n, and (3) C(n, k) = C(n, n − k).
C O
These properties can explain some of the observations we made on the numbers in the Pascal’s Triangle. Also recall the general formula for the number of combinations of n objects taken k at a time: n n! , C(n, k) = = k!(n − k)! k
EP E
D
where 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · · n. 5 8 Example 2.4.3. Compute and . 3 5 5 5! 5! = = = 10 3 (5 − 3)!3! 2!3! 10! 8 8! = = = 56 5 (8 − 5)!5! 3!5!
2
D
Solution.
You may observe that the value of 53 and the fourth coefficient in the fifth 8 row of Pascal’s Triangle are the same. In the same manner, 5 is equal to the sixth coefficient in the expansion of (a + b)8 (see Example 2.4.2). These observed equalities are not coincidental, and they are, in fact, the essence embodied in the Binomial Theorem, as you will see in the succeeding sessions. 2.4.2. The Binomial Theorem
As the power n gets larger, the more laborious it would be to use Pascal’s Triangle (and impractical to use long multiplication) to expand (a + b)n . For example, using Pascal’s Triangle, we need to compute row by row up to the thirtieth row 111 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
to know the coefficients of (a + b)30 . It is, therefore, delightful to know that it is possible to compute the terms of a binomial expansion of degree n without going through the expansion of all the powers less than n. We now explain how the concept of combination is used in the expansion of (a + b)n . (a + b)n = (a + b)(a + b)(a + b) · · · (a + b) {z } | n factors
C O
PY
When the distributive law is applied, the expansion of (a + b)n consists of terms of the form am bi , where 0 ≤ m, i ≤ n. This term is obtained by choosing a for m of the factors and b for the rest of the factors. Hence, m + i = n, or m = n − i. This means that the number of times the term an−i bi will appear in the expansion of (a + b)n equals the number of ways of choosing (n − i) or i factors from the n factors, which is exactly C(n, i). Therefore, we have n X n n−i i n (a + b) = a b. i i=0 To explain the reasoning above, consider the case n = 3.
D
(a + b)3 = (a + b)(a + b)(a + b) = aaa + aab + aba + abb + baa + bab + bba + bbb = a3 + 3a2 b + 3ab2 + b3
EP E
That is, each term in the expansion is obtained by choosing either a or b in each factor. The term a3 is obtained when a is chosen each time, while a2 b is obtained when a is selected 2 times, or equivalently, b is selected exactly once. We will give another proof of this result using mathematical induction. But first, we need to prove a result about combinations.
D
Pascal’s Identity
If n and k are positive integers with k ≤ n, then n+1 n n = + . k k k−1
Proof. The result follows from the combination formula. n n n! n! + = + k k−1 k!(n − k)! (k − 1)!(n − k + 1)! n!(n − k + 1) + n!(k) = k!(n − k + 1)! 112 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
n!(n − k + 1 + k) k!(n + 1 − k)! n!(n + 1) = k!(n + 1 − k)! (n + 1)! = k!(n + 1 − k)! n+1 = k =
2
PY
Pascal’s identity explains the method of constructing Pascal’s Triangle, in which an entry is obtained by adding the two numbers above it. This identity is also an essential part of the second proof of the Binomial Theorem, which we now state.
For any positive integer n,
C O
The Binomial Theorem
n
(a + b) =
n X n i=0
i
an−i bi .
D
Proof. We use mathematical induction. Part 1
EP E
1 X 1 0 1 1 1−i i 1 1 0 a b = ab + a b =a+b i 0 1 i=0
Hence, the formula is true for n = 1. Part 2. Assume that
k
D
(a + b) =
k X k i=0
i
ak−i bi .
We want to show that k+1
(a + b)
k+1 X k + 1 k+1−i i = a b. i i=0
(a + b)k+1 = (a + b)(a + b)k k X k k−i i = (a + b) a b i i=0
113 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
EP E
D
C O
PY
k k X X k k−i i k k−i i =a a b +b a b i i i=0 i=0 k k X k k−i+1 i X k k−i i+1 = a b + a b i i i=0 i=0 k X k k+1 0 k k+1−i i = a b + a b 0 i i=1 k k 1 k k−1 2 k k−2 3 + a b + a b + a b 0 1 2 k k 0 k+1 1 k + ··· + ab + ab k−1 k k X k k+1−i i k+1 =a + a b i i=1 k X k + ak+1−i bi + bk+1 i − 1 i=1 k + 1 k+1 0 = a b 0 k X k k + + ak+1−i bi i i − 1 i=1 k + 1 0 k+1 + ab k+1 k+1 X k + 1 k+1−i i = a b i i=0
The last expression above follows from Pascal’s Identity.
D
Therefore, by the Principle of Mathematical Induction, n X n n−i i (a + b) = a b i i=1 n
2
for any positive integer n. 2.4.3. Terms of a Binomial Expansion We now apply the Binomial Theorem in different examples. Example 2.4.4. Use the Binomial Theorem to expand (x + y)6 .
114 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Solution. 6 X 6 6−k k (x + y) = x y k k=0 6 6 0 6 5 1 6 4 2 = xy + xy + xy 0 1 2 6 3 3 6 2 4 6 1 5 + xy + xy + xy 3 4 5 6 0 6 + xy 6 = x6 + 6x5 y + 15x4 y 2 + 20x3 y 3 + 15x2 y 2 + 6xy 5 + y 6
PY
6
2
EP E
D
C O
Since the expansion of (a + b)n begins with k = 0 and ends with the k = n, n n n expansion has n + 1 terms. The first term in the expansion is 0 a = , the an−1 n n−1 n n=1 second term is 1 a b = na b, the second to the last term is n−1 ab = n n n−1 n nab , and the last term is n b = b . n−k+1 k−1 n The kth term of the expansion is k−1 a b . If n is even, there is a n middle term, which is the 2 +1 th term. If n is odd, there are two middle n+1 + 1 th terms. terms, the n+1 th and 2 2 The general term is often represented by nk an−k bk . Notice that, in any term, the sum of the exponents of a and b is n. The combination nk is the coefficient of the term involving bk . This allows us to compute any particular term without needing to expand (a + b)n and without listing all the other terms. √ 20 Example 2.4.5. Find the fifth term in the expansion of 2x − y .
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Solution. The fifth term in the expansion of a fifth power corresponds to k = 4. 20 √ (2x)20−4 (− y)4 = 4845 65536x16 y 2 4 = 317521920x16 y 2 2
Example 2.4.6. Find the middle term in the expansion of
x 2
+ 3y
6
.
Solution. Since there are seven terms in the expansion, the middle term is the fourth term (k = 3), which is 3 135x3 y 3 x 3 x 6 3 (3y) = 20 27y 3 = . 2 3 2 8 2
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Example 2.4.7. Find the term involving x (with exponent 1) in the expansion 8 2y 2 . of x − x Solution. The general term in the expansion is
PY
k 8 2y 8 16−2k (−2)k y k 2 8−k x − = x · k x k xk 8 = (−2)k x16−2k−k y k k 8 = (−2)k x16−3k y k . k
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The term involves x if the exponent of x is 1, which means 16 − 3k = 1, or k = 5. Hence, the term is 8 (−2)5 xy 5 = −1792xy 5 . 2 5 ? 2.4.4. Approximation and Combination Identities
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We continue applying the Binomial Theorem.
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? Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in the expansion of (1 − 0.2)8 . Compare your answer with the calculator value. (2) Use 5 terms in the binomial expansion to approximate (0.8)8 . Is there an improvement in the approximation? Solution.
8
8
(1)8−k (−0.2)k k k=0 8 X 8 = (−0.2)k k k=0
(0.8) = (1 − 0.2) =
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8 X 8
2 X 8 8 8 8 k (1) (−0.2) = + (−0.2) + (−0.2)2 k 0 1 2 k=0 = 1 − 1.6 + 1.12 = 0.52 The calculator value is 0.16777216, so the error is 0.35222784.
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(2)
4 X 8 k=0
8 8 8 (−0.2) = + (−0.2) + (−0.2)2 k 0 1 2 8 8 3 + (−0.2) + (−0.2)4 3 4 = 0.52 − 0.448 + 0.112 = 0.184 k
The error is 0.01622784, which is an improvement on the previous estimate. 2
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Example 2.4.9. Use the Binomial Theorem to prove that, for any positive integer n, n X n = 2n . k k=0 Solution. Set a = b = 1 in the expansion of (a + b)n . Then n
2 = (1 + 1) =
n X n
n X n (1) = . k k=0
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n
k=0
k
n−k
(1)
k
2
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Example 2.4.10. Use the Binomial Theorem to prove that 100 100 100 100 + + + ··· + 0 2 4 100 100 100 100 100 = + + + ··· + 1 3 5 99 Solution. Let a = 1 and b = −1 in the expansion of (a + b)100 . Then
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100 100 X 100 1 + (−1) = (1)100−k (−1)k . k k=0
100 100 100 100 2 0= + (−1) + (−1) + (−1)3 0 1 2 3 100 100 99 + ··· + (−1) + (−1)100 99 100
If k is even, then (−1)k = 1. If k is odd, then (−1)k = −1. Hence, we have 100 100 100 100 0= − + − 0 1 2 3 100 100 + ··· − + 99 100 117 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Therefore, after transposing the negative terms to other side of the equation, we obtain 100 100 100 100 + + + ··· + 0 2 4 100 100 100 100 100 = + + + ··· + 2 1 3 5 99
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More Solved Examples 5
1. Use the Binomial Theorem to expand (2x4 − 3y 2 ) . Solution: 2x4 − 3y 2
5
=
5 X
2x4
5−k
k
= 32x20 − 240x16 y 2 + 720x12 y 4 −
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k=0
1080x8 y 6 + 810x4 y 8 − 243y 10
3y 2
28
2. Determine the 20th term in the expansion of (x3 − 3y) . 28 27 19 Solution: We see that k = 19 should yield the 20th term, yielding −3 x y . 19 20 x y2 x2 − . 3. Find the term containing 2 in the expansion of y y 2x2
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19
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y2 x Solution: Setting a = , b = − 2 , the (k + 1)th term in the binomial y 2x 20−k 2 k x y (−1)k n 20−3k 3k−20 k 20 expansion is (−1) = x y . To get k y 2x2 2 k k 1 20 2 2 x2 , we get 20 − 3k = 2 ⇒ k = 6, yielding 6 xy . 2 y 2 6 16 2 4. Determine the term not involving x in the expansion of x3 + 5 . x 2 Solution: Setting a = x3 , b = 5 , the (k + 1)th term in the binomial expansion x k 16 2 16 48−8k 16−k is x3 = 2k x . To get the term without x, we get 5 k x k 16 . 48 − 8k = 0 ⇒ k = 6, yielding 26 6 5. Determine the coefficient of x9 in the expansion of (1 + 2x)10 .
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Solution: Setting a = b = 2x, the (k + 1)th term in the binomial expansion 1, 10 k 10 x . To get x9 , we set k = 9, of the first factor is (1)10−k (2x)k = 2k k k 9 10 x9 . yielding 2 9 n X k n 6. Prove that (−1) 3n−k = 2n . k i=0
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Solution: Set a = 3, b = −1. √ √ 5 √ √ 7. If 3 + 2 is written in the form a 3 + b 2 where a, b are integers, what is a + b? 5 √ X √ 5 √ 5−k √ k 5 Solution: We have 3+ 2 = 3 2 . Note that if k k=0 √ 5 − k is odd (or equivalently, k is even), the term has a factor of 3, while √ 5 5 5 + + = 16 and the rest have a factor of 2. Thus, a = 0 2 4 5 5 5 b= + + = 16 yields a + b = 32. 1 3 5
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Supplementary Problems 2.4 1. Use the Binomial Theorem to expand the following:
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(a) (2x − 3y)5 √ 4 2 x (b) − 2 3 x √ 4 (c) (1 + x)
2. Without expanding completely, find the indicated value(s) in the expansion of the following: (a) (2 + x)9 , two middle terms 10 p 2 (b) + , 3rd term 2 q 21
(c) (x2 + y 4 ) , last 2 terms 20 1 (d) √ , middle term x
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15 2y 4 x5 , term not involving y (e) + x3 4y 13 1 2 (f) −x , term involving x2 2 2x
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(g) (1 − 2x)6 , coefficient of x3 30 1 1 (h) 2y 7/3 − 5/3 , coefficient of 2 2y y √ 8 (i) ( x − 3) , coefficient of x7/2 √ 6 (j) ( x + 2) , coefficient of x3/2 3. Approximate (2.1)10 by using the first 5 terms in the expansion of (2 + 0.1)20 . Compare your answer with the calculator result.
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4. In the expansion of (4x + 3)34 , the kth value and the (k + 1)st terms have equal coefficients. What is the value of k?
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5. Determine the value of 19 19 19 19 19 2 19 3 19 18 − 3+ 3 − 3 + ... + 3 − 3 19 0 1 2 3 18
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4
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Topic Test 1 for Unit 2 1. Determine if the given sequence is arithmetic, geometric, or neither by writing A, G, or O, respectively. 1 1 3 9 27 , , , , ,... 3 2 4 16 32 1 1 1 1 1 (b) , , , , , . . . 2 7 12 17 21 (c) 0, 3, 8, 15, 24, . . . (a)
i=1
i2 + 3i + 2
.
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3. Evaluate the sum
50 X 2i3 + 9i2 + 13i + 6
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2. Three numbers form an arithmetic sequence, the common difference being 5. If the last number is increased by 1, the second by 2, and the first by 4, the resulting numbers form a geometric sequence. Find the numbers.
4. Find the indicated terms in the expansion of the given expression. 8 28 1 (b) (n3 − 3m) , 20th term 2 , term involving x8 (a) x − 2
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5. Prove the statement below for all positive integers n by mathematical induction. 1 1 1 n + + ··· + = 1·3 3·5 (2n − 1)(2n + 1) 2n + 1
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6. On his 20th birthday, Ian deposited an amount of 10,000 pesos to a timedeposit scheme with a yearly interest of 4%. Ian decides not to withdraw any amount of money or earnings and vows to keep it in the same time-deposit scheme year after year. Show that the new amounts in Ian’s time-deposit account in each succeeding birthday represent a geometric sequence, and use this to determine the value of the money during Ian’s 60th birthday.
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Topic Test 2 for Unit 2 1. Determine if the given sequence is arithmetic, geometric, or neither by writing A, G, or O, respectively. 2 8 32 128 512 , , , , ,... 3 15 75 375 1875 1 2 3 4 5 (b) , , , , , . . . 2 3 4 5 6 11 21 (c) 3, , 8, , 13, . . . 2 2 (a)
PY
2. The sum of the first two terms of an arithmetic sequence is 9 and the sum of the first three terms is also 9. How many terms must be taken to give a sum of −126?
(a)
50 X
(2i + 1)(i − 3)
i=1
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3. Evaluate the following sums. (b)
30 X
r
i=1
i2 − 2i + 1 4
8 .
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4. Find the term not involving x in the expansion of
1 x + x 3
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5. Prove that the following statements are true for all positive integers n by mathematical induction. n(3n − 1) 2 n n−1 (b) 3 + 7 + 8 is divisible by 12.
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(a) 1 + 4 + 7 + . . . + (3n − 2) =
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Unit 3
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Trigonometry
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Puerto Princesa Subterranean River National Park, by Giovanni G. Navata, 12 November 2010, https://commons.wikimedia.org/wiki/File%3AUnderground River.jpg. Public Domain
Named as one of the New Seven Wonders of Nature in 2012 by the New7Wonders Foundation, the Puerto Princesa Subterranean River National Park is worldfamous for its limestone karst mountain landscape with an underground river. The Park was also listed as UNESCO World Heritage Site in 1999. The underground river stretches about 8.2 km long, making it one of the world’s longest rivers of its kind.
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Lesson 3.1. Angles in a Unit Circle Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) illustrate the unit circle and the relationship between the linear and angular measures of arcs in a unit circle. (2) convert degree measure to radian measure, and vice versa.
Lesson Outline (1) Linear and angular measure of arcs
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(3) illustrate angles in standard position and coterminal angles.
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(2) Conversion of degree to radian, and vice versa (3) Arc length and area of the sector
(4) Angle in standard position and coterminal angles Introduction
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Angles are being used in several fields like engineering, medical imaging, electronics, astronomy, geography and many more. Added to that, surveyors, pilots, landscapers, designers, soldiers, and people in many other professions heavily use angles and trigonometry to accomplish a variety of practical tasks. In this lesson, we will deal with the basics of angle measures together with arc length and sectors. 3.1.1. Angle Measure
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An angle is formed by rotating a ray about its endpoint. In the figure shown below, the initial side of ∠AOB is OA, while its terminal side is OB. An angle is said to be positive if the ray rotates in a counterclockwise direction, and the angle is negative if it rotates in a clockwise direction.
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An angle is in standard position if it is drawn in the xy-plane with its vertex at the origin and its initial side on the positive x-axis. The angles α, β, and θ in the following figure are angles in standard position.
To measure angles, we use degrees, minutes, seconds, and radians.
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A central angle of a circle measures one degree, written 1◦ , if it inter1 of the circumference of the circle. One minute, written 10 , is cepts 360 1 1 ◦ of 1 , while one second, written 100 , is 60 of 10 . 60
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For example, in degrees, minutes, and seconds, 0 18 ◦ 0 00 ◦ 10 30 18 = 10 30 + 60 ◦ 0 = 10 30.3 ◦ 30.3 = 10 + 60 ◦ = 10.505
and
79.251◦ = 79◦ (0.251 × 60)0 = 79◦ 15.060 = 79◦ 150 (0.06 × 60)00 = 79◦ 150 3.600 . Recall that the unit circle is the circle with center at the origin and radius 1 unit. 125 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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A central angle of the unit circle that intercepts an arc of the circle with length 1 unit is said to have a measure of one radian, written 1 rad. See Figure 3.1.
Figure 3.1
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In trigonometry, as it was studied in Grade 9, the degree measure is often used. On the other hand, in some fields of mathematics like calculus, radian measure of angles is preferred. Radian measure allows us to treat the trigonometric functions as functions with the set of real numbers as domains, rather than angles.
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Example 3.1.1. In the following figure, identify the terminal side of an angle in standard position with given measure. (1) degree measure: 135◦ , −135◦ , −90◦ , 405◦ π 4
rad, − 3π rad, 4
3π 2
rad, − π2 rad
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(2) radian measure:
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Solution.
−→ −−→ −−→ −−→ (1) 135◦ : OC; −135◦ : OD; −90◦ : OE; and 405◦ : OB
(2) radian measure: −−→ OE
π 4
−−→ −−→ rad: OB; − 3π rad: OD; 4
3π 2
−−→ rad: OE; and − π2 rad: 2
Since a unit circle has circumference 2π, a central angle that measures 360◦ has measure equivalent to 2π radians. Thus, we obtain the following conversion rules. Converting degree to radian, and vice versa π . 180 180 . π
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1. To convert a degree measure to radian, multiply it by 2. To convert a radian measure to degree, multiply it by
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Figure 3.2 shows some special angles in standard position with the indicated terminal sides. The degree and radian measures are also given.
Figure 3.2
Example 3.1.2. Express 75◦ and 240◦ in radians. Solution. 75
π 5π = 180 12
=⇒
75◦ =
5π rad 12
240
π 4π = 180 3
=⇒
240◦ =
4π rad 3
2
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Example 3.1.3. Express Solution.
π 8
11π 6
π 8
180 π
rad and
11π 6
rad in degrees.
180 π
= 22.5
=⇒
π rad = 22.5◦ 8
= 330
=⇒
11π rad = 330◦ 6
2
PY
3.1.2. Coterminal Angles Two angles in standard position that have a common terminal side are called coterminal angles. Observe that the degree measures of coterminal angles differ by multiples of 360◦ .
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Two angles are coterminal if and only if their degree measures differ by 360k, where k ∈ Z. Similarly, two angles are coterminal if and only if their radian measures differ by 2πk, where k ∈ Z.
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As a quick illustration, to find one coterminal angle with an angle that measures 410◦ , just subtract 360◦ , resulting in 50◦ . See Figure 3.3.
Figure 3.3 Example 3.1.4. Find the angle coterminal with −380◦ that has measure (1) between 0◦ and 360◦ , and 128 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(2) between −360◦ and 0◦ . Solution. A negative angle moves in a clockwise direction, and the angle −380◦ lies in Quadrant IV. (1) −380◦ + 2 · 360◦ = 340◦ 2
(2) −380◦ + 360◦ = −20◦ 3.1.3. Arc Length and Area of a Sector
θ θ × circumference of circle = (2πr) = rθ. 2π 2π
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s=
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In a circle, a central angle whose radian measure is θ subtends an arc that is the θ fraction 2π of the circumference of the circle. Thus, in a circle of radius r (see Figure 3.4), the length s of an arc that subtends the angle θ is
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Figure 3.4
In a circle of radius r, the length s of an arc intercepted by a central angle with measure θ radians is given by s = rθ.
Example 3.1.5. Find the length of an arc of a circle with radius 10 m that subtends a central angle of 30◦ . Solution. Since the given central angle is in degrees, we have to convert it into radian measure. Then apply the formula for an arc length. π π 30 = rad 180 6 129 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
π
5π m 2 6 3 Example 3.1.6. A central angle θ in a circle of radius 4 m is subtended by an arc of length 6 m. Find the measure of θ in radians. s = 10
Solution. θ=
=
6 3 s = = rad r 4 2
2
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A sector of a circle is the portion of the interior of a circle bounded by the initial and terminal sides of a central angle and its intercepted arc. It is like a “slice of pizza.” Note that an angle with measure 2π radians will define a sector that corresponds to the whole “pizza.” Therefore, if a central angle of a sector θ of a complete has measure θ radians, then the sector makes up the fraction 2π circle. See Figure 3.5. Since the area of a complete circle with radius r is πr2 , we have θ 1 Area of a sector = (πr2 ) = θr2 . 2π 2
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Figure 3.5
In a circle of radius r, the area A of a sector with a central angle measuring θ radians is 1 A = r2 θ. 2
Example 3.1.7. Find the area of a sector of a circle with central angle 60◦ if the radius of the circle is 3 m. Solution. First, we have to convert 60◦ into radians. Then apply the formula for computing the area of a sector. π π 60 = rad 180 3 130 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
1 π 3π 2 A = (32 ) = m 2 3 2
2
Example 3.1.8. A sprinkler on a golf course fairway is set to spray water over a distance of 70 feet and rotates through an angle of 120◦ . Find the area of the fairway watered by the sprinkler. Solution.
π 2π = rad 180 3 1 2π 4900π A = (702 ) = ≈ 5131 ft2 2 3 3 120
5 48
radians.
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1. Find the equivalent degree measure of 5 5 180 75 ◦ Solution: 48 rad = 48 = 4π π
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More Solved Examples
2
2. Find the equivalent angle measure in degrees and in radians of an angle tracing 2 53 revolutions.
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Solution: One revolution around a circle is equivalent to tracing 360◦ . 3 3 360 2 rev = 2 rev = 936◦ 5 5 1 rev π 26π ◦ = rad 936 = 936 180 5 3. Find the smallest positive angle coterminal with −2016◦ .
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Solution: Add 6 complete revolutions or 6(360◦ ) = 2160◦ to the given angle (or keep on adding 360◦ until you get a positive angle). −2016◦ + 2160◦ = 144◦
4. Find the largest negative angle coterminal with
137π . 5
Solution: Subtract 14 complete revolutions or 14(2π) = 28π to the given angle (or keep on subtracting 2π until you get a negative angle). 137π 3π − 28π = − rad 5 5
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5. Find the length of the arc of a circle with radius 15 cm that subtends a central angle of 84◦ . Solution: π 7π = rad 84◦ = 84 15 180 7π s = 15 = 7π cm 15 6. A central angle θ in a circle of radius 12 inches is subtended by an arc of length 27 inches. Find the measure of θ in degrees.
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Solution:
s = rθ =⇒ θ =
s r
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12 9 θ= = rad 27 4 ◦ 9 9 180 405 rad = = 4 4 π π
7. Find the area of a sector of a circle with central angle of 108◦ if the radius of the circle is 15 cm. Solution:
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π 3π = rad 180 5 135π 1 3π = cm2 A = (15)2 2 5 2
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108◦ = 108
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8. Given isosceles right triangle ABC with AC as the hypotenuse (as shown below), a circle with center at A and radius AB intersects AC at D. What is the ratio of the area of sector BAD to the area of the region BCD?
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Solution: Let r be the radius of the circle; that is, r = AB. 1 2 π πr2 =⇒ Area of sector BAD = r = 2 4 8 4r2 − πr2 Area of region BCD = Area of 4ABC − Area of sector BAD = 8 πr2 area of sector BAD π 8 = 4r2 −πr 2 = area of the region BCD 4−π 8
Supplementary Problems 3.1
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π ∠A = rad 4
2. How many radians is
11 5
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1. How many degrees is 1 15 of a complete revolution? of a complete revolution?
3. What is the length of an arc of a circle with radius 4 cm that subtends a central angle of 216◦ ? 6 π
cm that subtends a central
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4. Find the length of an arc of a circle with radius angle of 99◦ .
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5. What is the smallest positive angle coterminal with 2110◦ ? 6. Find the largest negative angle coterminal with
107π . 6
7. Find the area of a sector of a circle with central angle of the circle is 9 cm.
7π 6
if the diameter of
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8. Find the area of a sector of a circle with central angle of 108◦ if the radius of the circle is 15 cm. 9. What is the radius of a circle in which a central angle of 150◦ determines a sector of area 15 in2 ?
10. Find the radius of a circle in which a central angle of of area 32 in2 .
5π 4
determines a sector
? 11. A central angle of a circle of radius 6 inches is subtended by an arc of length 6 inches. What is the central angle in degrees (rounded to two decimal places)? ? 12. An arc of length π cm subtends a central angle θ of a circle with radius 5 What is θ in degrees?
2 3
cm.
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13. Two overlapping circles of radii 1 cm are drawn such that each circle passes through the center of the other. What is the perimeter of the entire region? 14. The length of arc AB of a circle with center at O is equal to twice the length of the radius r of the circle. Find the area of sector AOB in terms of r. 15. The angle of a sector in a given circle is 20◦ and the area of the sector is equal to 800 cm2 . Find the arc length of the sector. 16. In Figure 3.6, AE and BC are arcs of two concentric circles with center at D. If AD = 2 cm, BD = 8 cm, and ∠ADE = 75◦ , find the area of the region AECB.
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17. In Figure 3.7, AB and DE are diameters. If AB = 12 cm and ∠AOD = 126◦ , find the area of the shaded region.
Figure 3.7
D
Figure 3.6
Figure 3.8
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18. A point moves outside an equilateral triangle of side 5 cm such that its distance from the triangle is always 2 cm. See Figure 3.8. What is the length of one complete path that the point traces?
Figure 3.9 19. The segment of a circle is the region bounded by a chord and the arc subtended by the chord. See Figure 3.9. Find the area of a segment of a circle with a central angle of 120◦ and a radius of 64 cm. 134 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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Figure 3.10
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20. In Figure 3.10, diameter AB of circle O measures 12 cm and arc BC measures 120◦ . Find the area of the shaded region.
4
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Lesson 3.2. Circular Functions
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Learning Outcomes of the Lesson
At the end of the lesson, the student is able to: (1) illustrate the different circular functions; and (2) use reference angles to find exact values of circular functions. Lesson Outline
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(1) Circular functions (2) Reference angles Introduction
We define the six trigonometric function in such a way that the domain of each function is the set of angles in standard position. The angles are measured either in degrees or radians. In this lesson, we will modify these trigonometric functions so that the domain will be real numbers rather than set of angles.
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3.2.1. Circular Functions on Real Numbers Recall that the sine and cosine functions (and four others: tangent, cosecant, secant, and cotangent) of angles measuring between 0◦ and 90◦ were defined in the last quarter of Grade 9 as ratios of sides of a right triangle. It can be verified that these definitions are special cases of the following definition. Let θ be an angle in standard position and P (θ) = P (x, y) the point on its terminal side on the unit circle. Define 1 csc θ = , y 6= 0 y
cos θ = x
sec θ =
1 , x 6= 0 x x cot θ = , y 6= 0 y
y , x 6= 0 x
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tan θ =
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sin θ = y
Example 3.2.1. Find the values of cos 135◦ , tan 135◦ , sin(−60◦ ), and sec(−60◦ ).
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Solution. Refer to Figure 3.11(a).
(a)
(b) Figure 3.11
From properties of 45◦ -45◦ and 30◦ -60◦ right triangles (with hypotenuse 1 unit), we obtain the lengths of the legs as in Figure 3.11(b). Thus, the coordinates of A and B are √ √ ! √ ! 2 2 1 3 A= − , and B = ,− . 2 2 2 2 136 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Therefore, we get √
2 , 2
◦
cos 135 = − √ 3 sin(−60◦ ) = − , 2
tan 135◦ = −1, and
sec(−60◦ ) = 2.
2
PY
From the last example, we may then also say that √ √2 π π 3 rad = , sin − rad = − , cos 4 2 3 2 and so on.
C O
From the above definitions, we define the same six functions on real numbers. These functions are called trigonometric functions.
sin s = sin θ
csc s = csc θ
cos s = cos θ
sec s = sec θ
D
Let s be any real number. Suppose θ is the angle in standard position with measure s rad. Then we define
cot s = cot θ
EP E
tan s = tan θ
From the last example, we then have cos
and
π 4
√ 2 ◦ = cos rad = cos 45 = 4 2 π
D
√ π π 3 sin − = sin − rad = sin(−60◦ ) = − . 3 3 2 In the same way, we have tan 0 = tan(0 rad) = tan 0◦ = 0.
Example 3.2.2. Find the exact values of sin 3π , cos 3π , and tan 3π . 2 2 2 Solution. Let P 3π be the point on the unit circle and on the terminal side of 2 3π 3π the angle in the standard position with measure 2 rad. Then P 2 = (0, −1), and so 3π 3π sin = −1, cos = 0, 2 2 but tan 3π is undefined. 2 2 137 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Example 3.2.3. Suppose s is a real number such that sin s = − 43 and cos s > 0. Find cos s. Solution. We may consider s as the angle with measure s rad. Let P (s) = (x, y) be the point on the unit circle and on the terminal side of angle s. Since P (s) is on the unit circle, we know that x2 + y 2 = 1. Since sin s = y = we get
− 34 ,
2
2
7 = 16
√
7 . 4
2
D
EP E
D
C O
Since cos s = x > 0, we have cos s =
=⇒
√ 7 x=± . 4
PY
3 x =1−y =1− − 4 2
Let P (x1 , y1 ) and Q(x, y) be points on the terminal side of an angle θ in standard position, where P is on the unit circle and Q on the circle of radius r (not necessarily 1) with center also at the origin, as shown above. Observe that we can use similar triangles to obtain cos θ = x1 =
x1 x = 1 r
and
sin θ = y1 =
y1 y = . 1 r
We may then further generalize the definitions of the six circular functions.
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Let θ be an angle in standard position, Q(x, y) any point on the terp minal side of θ, and r = x2 + y 2 > 0. Then sin θ =
y r
r csc θ = , y 6= 0 y
cos θ =
x r
sec θ =
tan θ =
r , x 6= 0 x x cot θ = , y 6= 0 y
y , x 6= 0 x
3.2.2. Reference Angle
C O
PY
We then have a second solution for Example 3.2.3 as follows. With sin s = − 43 and sin s = yr , we may choose y = −3 and r = 4 (which is always positive). In this case, we can solve for x, which is positive since cos s = x4 is given to be positive. √ p √ 7 4 = x2 + (−3)2 =⇒ x = 7 =⇒ cos s = 4
D
EP E
D
We observe that if θ1 and θ2 are coterminal angles, the values of the six circular or trigonometric functions at θ1 agree with the values at θ2 . Therefore, in finding the value of a circular function at a number θ, we can always θ to a number reduce 14π 14π 2π between 0 and 2π. For example, sin 3 = sin 3 − 4π = sin 3 . Also, observe = sin π3 . from Figure 3.12 that sin 2π 3
Figure 3.12
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C O
PY
In general, if θ1 , θ2 , θ3 , and θ4 are as shown in Figure 3.13 with P (θ1 ) = (x1 , y1 ), then each of the x-coordinates of P (θ2 ), P (θ3 ), and P (θ4 ) is ±x1 , while the y-coordinate is ±y1 . The correct sign is determined by the location of the angle. Therefore, together with the correct sign, the value of a particular circular function at an angle θ can be determined by its value at an angle θ1 with radian measure between 0 and π2 . The angle θ1 is called the reference angle of θ.
EP E
D
Figure 3.13
D
The signs of the coordinates of P (θ) depends on the quadrant or axis where it terminates. It is important to know the sign of each circular function in each quadrant. See Figure 3.14. It is not necessary to memorize the table, since the sign of each function for each quadrant is easily determined from its definition. We note that the signs of cosecant, secant, and cotangent are the same as sine, cosine, and tangent, respectively.
Figure 3.14
Using the fact that the unit circle is symmetric with respect to the x-axis, the y-axis, and the origin, we can identify the coordinates of all the points using the 140 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
C O
PY
coordinates of corresponding points in the Quadrant I, as shown in Figure 3.15 for the special angles.
Figure 3.15
EP E
D
Example 3.2.4. Use reference angle and appropriate sign to find the exact value of each expression. and cos 11π (3) sin 150◦ (1) sin 11π 6 6 (2) cos − 7π (4) tan 8π 6 3 is π6 , and it lies in Quadrant IV wherein Solution. (1) The reference angle of 11π 6 sine and cosine are negative and positive, respectively. 11π π 1 = − sin = − 6 6 2 √ π 3 11π = cos = cos 6 6 2
D
sin
(2) The angle − 7π lies in Quadrant II wherein cosine is negative, and its refer6 π ence angle is 6 . √ 7π π 3 cos − = − cos = − 6 6 2 (3) sin 150◦ = sin 30◦ = (4) tan
8π 3
= − tan
π 3
=
1 2 π − cos 3π 3 sin
√
=−
3 2 1 2
√ =− 3
2
141 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
More Solved Examples 1. If P (θ) is a point on the unit circle and θ = P (θ)?
17π , 3
what are the coordinates of
5π is coterminal with which terminates in QIV. The reference 3 √ angle is π3 , therefore P 17π = 21 , − 23 . 3
Solution:
17π 3
2. If P (θ) is a point on the unit circle and θ = − 5π , find the values of the six 6 trigonometric functions of θ.
PY
terminates in QIII, the reference angle is π6 , therefore Solution: The angle −5π 6 √ P − 5π = − 23 , − 12 . 6
C O
√ √ 5π 3 5π 2 2 3 cos − =− sec − = −√ = − 6 2 6 3 3 5π 1 5π sin − =− csc − = −2 6 2 6 √ √ 5π 5π 1 3 3 √ cot − = 3 tan − =√ = = 6 3 6 1 3
EP E
D
3. Find the six trigonometric functions of the angle θ if the terminal side of θ in standard position passes through the point (5, −12). p Solution: x = 5, y = −12, r = (5)2 + (−12)2 = 13. x 5 = r 13 y 12 sin θ = = − r 13 12 y tan θ = = − x 5
r 13 = x 5 r 13 csc θ = = − y 12 x 5 cot θ = = − y 12
D
cos θ =
4. Given sec θ = − 25 and π ≤ θ ≤ 24
sec θ =
3π , 2
find sin θ + cos θ. p Solution: r = 25, x = −24, y = (25)2 − (−24)2 = ±7.
Since θ is in QIII, y = −7. sin θ + cos θ = 5. If tan A = 45 , determine
−7 −24 31 + =− . 25 25 25
2 sin A−cos A . 3 cos A
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Solution: sin A 4 =⇒ = tan A = 5 cos A 2 sin A − cos A 2 sin A 1 cos A = − = 3 cos A 3 cos A 3 cos A
4 5 2 4 1 1 − (1) = 3 5 3 5
6. What is the reference angle of − 29π . ? Find the value of tan − 29π 6 6 Solution: − 29π is coterminal with 6
in QIII, so its reference angle is π6 . π √3 29π tan − = tan = 6 6 3
PY
7π 6
7. For what angle θ in the third quadrant is cos θ = sin 5π ? 3 Solution: √
3 and θ in QIII 2
D
cos θ = −
5π = cos θ 3
C O sin
=⇒
θ=
7π 6
Supplementary Problems 3.2
EP E
1. In what quadrant is P (θ) located if θ =
33π ? 4
2. In what quadrant is P (θ) located if θ = − 17π ? 6 3. In what quadrant is P (θ) located if sec θ > 0 and cot θ < 0? 4. In what quadrant is P (θ) located if tan θ > 0 and cos θ < 0 ?
D
5. If P (θ) is a point on the unit circle and θ = P (θ)?
5π , 6
what are the coordinates of
6. If P (θ) is a point on the unit circle and θ = − 11π , what are the coordinates 6 of P (θ)? 7. If cos θ > 0 and tan θ = − 23 , find
8. If tan θ =
3 5
sec θ+tan θ sec θ−tan θ
and θ is in QIII, what is sec θ?
9. If csc θ = 2 and cos θ < 0, find sec θ. 10. Find the values of the other trigonometric functions of θ if cot θ = − 43 and sin θ < 0. 143 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
11. Find the values of the other trigonometric functions of θ if csc θ = −4 and θ does not terminate in QIII. 12. The terminal side of an angle θ in standard position contains the point (7, −1). Find the values of the six trigonometric functions of θ. 13. The terminal side of an angle θ in standard position contains the point (−2, 4). Find the values of the six trigonometric functions of θ. 14. If the terminal point of an arc of length θ lies on the line joining the origin and the point (−3, −1), what is cos2 θ − sin2 θ?
16. Determine the reference angle of 3π 2
and find cos 35π . 4
< θ < 2π, find θ if cos θ = sin 2π . 3
C O
17. If
35π , 4
PY
15. If the terminal point of an arc of length θ lies on the line joining the origin and the point (2, −6), what is sec2 θ − csc2 θ?
18. Evaluate the sum of sin 30◦ + sin 60◦ + sin 90◦ + · · · + sin 510◦ + sin 540◦ . 19. If f (x) = sin 2x + cos 2x + sec 2x + csc 2x + tan 2x + cot 2x, what is f
7π 8
?
D
+ sec 25π + · · · + sec 109π . 20. Evaluate the sum of sec π6 + sec 13π 6 6 6
EP E
4
Lesson 3.3. Graphs of Circular Functions and Situational Problems Learning Outcomes of the Lesson
D
At the end of the lesson, the student is able to:
(1) determine the domain and range of the different circular functions; (2) graph the six circular functions with its amplitude, period, and phase shift; and (3) solve situational problems involving circular functions. Lesson Outline (1) Domain and range of circular functions (2) Graphs of circular functions (3) Amplitude, period, and phase shift 144
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Introduction There are many things that occur periodically. Phenomena like rotation of the planets and comets, high and low tides, and yearly change of the seasons follow a periodic pattern. In this lesson, we will graph the six circular functions and we will see that they are periodic in nature. 3.3.1. Graphs of y = sin x and y = cos x
PY
Recall that, for a real number x, sin x = sin θ for an angle θ with measure x radians, and that sin θ is the second coordinate of the point P (θ) on the unit circle. Since each x corresponds to an angle θ, we can conclude that (1) sin x is defined for any real number x or the domain of the sine function is R, and
C O
(2) the range of sine is the set of all real numbers between −1 and 1 (inclusive). From the definition, it also follows that sin(x+2π) = sin x for any real number x. This means that the values of the sine function repeat every 2π units. In this case, we say that the sine function is a periodic function with period 2π.
EP E
D
Table 3.16 below shows the values of y = sin x, where x is the equivalent radian measure of the special angles and their multiples from 0 to 2π. As commented above, these values determine the behavior of the function on R. x
0
π 6
y
0
1 2
0
0.5
7π 6
D
x y
− 12
−0.5
π 4 √ 2 2
5π 4 √
−
2 2
π 3 √ 3 2
π 2
1
0.71 0.87 4π 3 √
−
3 2
2π 3 √ 3 2
1
0.87 0.71
3π 2
−1
−0.71 −0.87 −1
3π 4 √ 2 2
5π 3 √
−
3 2
7π 4 √
−
2 2
5π 6
π
1 2
0
0.5
0
11π 6
2π
− 12
0
−0.87 −0.71 −0.5
0
Table 3.16
From the table, we can observe that as x increases from 0 to π2 , sin x also increases from 0 to 1. Similarly, as x increases from 3π to 2π, sin x also increases 2 from −1 to 0. On the other hand, notice that as x increases from π2 to π, sin x decreases from 1 to 0. Similarly, as x increases from π to 3π , sin x decreases from 2 0 to −1. 145 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
D
C O
Figure 3.17
PY
To sketch the graph of y = sin x, we plot the points presented in Table 3.16, and join them with a smooth curve. See Figure 3.17. Since the graph repeats every 2π units, Figure 3.18 shows periodic graph over a longer interval.
EP E
Figure 3.18
We can make observations about the cosine function that are similar to the sine function.
D
• y = cos x has domain R and range [−1, 1]. • y = cos x is periodic with period 2π. The graph of y = cos x is shown in Figure 3.19.
Figure 3.19
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From the graphs of y = sin x and y = cos x in Figures 3.18 and 3.19, respectively, we observe that sin(−x) = − sin x and cos(−x) = cos x for any real number x. In other words, the graphs of y = cos(−x) and y = cos x are the same, while the graph of y = sin(−x) is the same as that of y = − sin x. In general, if a function f satisfies the property that f (−x) = f (x) for all x in its domain, we say that such function is even. On the other hand, we say that a function f is odd if f (−x) = −f (x) for all x in its domain. For example, the functions x2 and cos x are even, while the functions x3 − 3x and sin x are odd.
PY
3.3.2. Graphs of y = a sin bx and y = a cos bx
EP E
D
C O
Using a table of values from 0 to 2π, we can sketch the graph of y = 3 sin x, and compare it to the graph of y = sin x. See Figure 3.20 wherein the solid curve belongs to y = 3 sin x, while the dashed curve to y = sin x. For instance, if x = π2 , then y = 1 when y = sin x, and y = 3 when y = 3 sin x. The period, x-intercepts, and domains are the same for both graphs, while they differ in the range. The range of y = 3 sin x is [−3, 3].
Figure 3.20
D
In general, the graphs of y = a sin x and y = a cos x with a > 0 have the same shape as the graphs of y = sin x and y = cos x, respectively. If a < 0, there is a reflection across the x-axis. In the graphs of y = a sin x and y = a cos x, the number |a| is called its amplitude. It dictates the height of the curve. When |a| < 1, the graphs are shrunk vertically, and when |a| > 1, the graphs are stretched vertically.
Now, in Table 3.21, we consider the values of y = sin 2x on [0, 2π].
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0
π 6 √ 3 2
0
0.87
x
0
y
x y
π 4
1
π 3 √ 3 2
0
−
1
0.87
0 3π 2
0
−
0
−0.87
7π 6 √ 3 2
5π 4
1
4π 3 √ 3 2
0.87
1
0.87
π 2
2π 3 √
3π 4
3 2
5π 6 √
3 2
π
−1
−
−0.87
−1
−0.87
5π 3 √
7π 4
2π
−1
11π 6 √ − 23
−1
−0.87
0
3 2
0 0
0
C O
PY
Table 3.21
EP E
D
Figure 3.22
Figure 3.22 shows the graphs of y = sin 2x (solid curve) and y = sin x (dashed curve) over the interval [0, 2π]. Notice that, for sin 2x to generate periodic values similar to [0, 2π] for y = sin x, we just need values of x from 0 to π. We then expect the values of sin 2x to repeat every π units thereafter. The period of y = sin 2x is π. 2π . |b| If 0 < |b| < 1, the graphs are stretched horizontally, and if |b| > 1, the graphs are shrunk horizontally.
D
If b 6= 0, then both y = sin bx and y = cos bx have period given by
To sketch the graphs of y = a sin bx and y = a cos bx, a, b 6= 0, we may proceed with the following steps: (1) Determine the amplitude |a|, and find the period 2π . To draw one cycle |b| of the graph (that is, one complete graph for one period), we just need to complete the graph from 0 to 2π . |b| 148 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(2) Divide the interval into four equal parts, and get five division points: x1 = 0, , where x3 is the midpoint between x1 and x5 (that x2 , x3 , x4 , and x5 = 2π |b| 1 is, 2 (x1 + x5 ) = x3 ), x2 is the midpoint between x1 and x3 , and x4 is the midpoint between x3 and x5 . (3) Evaluate the function at each of the five x-values identified in Step 2. The points will correspond to the highest point, lowest point, and x-intercepts of the graph.
PY
(4) Plot the points found in Step 3, and join them with a smooth curve similar to the graph of the basic sine curve. (5) Extend the graph to the right and to the left, as needed.
Example 3.3.1. Sketch the graph of one cycle of y = 2 sin 4x. (1) The period is
2π 4
= π2 , and the amplitude is 2.
C O
Solution.
(2) Dividing the interval [0, π2 ] into 4 equal parts, we get the following x, and π2 . coordinates: 0, π8 , π4 , 3π 8 (3) When x = 0, π4 , and π2 , we get y = 0. On the other hand, when x = π8 , we . have y = 2 (the amplitude), and y = −2 when x = 3π 8
D
EP E
D
(4) Draw a smooth curve by connecting the points. There is no need to proceed to Step 5 because the problem only asks for one cycle.
Example 3.3.2. Sketch the graph of y = −3 cos x2 . Solution.
(1) The amplitude is | − 3| = 3, and the period is
2π 1 2
= 4π.
(2) We divide the interval [0, 4π] into four equal parts, and we get the following x-values: 0, π, 2π, 3π, and 4π.
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(3) We have y = 0 when x = π and 3π, y = −3 when x = 0 and 4π, and y = 3 when x = 2π. (4) We trace the points in Step 3 by a smooth curve.
PY
(5) We extend the pattern in Step 4 to the left and to the right.
C O
Example 3.3.3. Sketch the graph of two cycles of y = 21 sin − 2x . 3
Solution. Since the sine function is odd, the graph of y = 12 sin − 2x is the same 3 1 2x as that of y = − 2 sin 3 . 2π 2 3
= 3π.
D
(1) The amplitude is 12 , and the period is
EP E
(2) Dividing the interval [0, 3π] into four equal parts, we get the x-coordinates of the five important points: 0 + 3π 3π = , 2 2
(3) We get y = 0 when x = 0, 9π . 4
0 + 3π 3π 2 = , 2 4
3π , 2
3π 2
+ 3π 9π = . 2 4
and 3π, y = − 21 when
3π , 4
and y =
1 2
when
D
(4) We trace the points in Step 3 by a smooth curve. (5) We extend the pattern in Step 4 by one more period to the right.
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3.3.3. Graphs of y = a sin b(x − c) + d and y = a cos b(x − c) + d We first compare the graphs of y = sin x and y = sin x − values and the 5-step procedure discussed earlier.
π 3
using a table of
x−
π 3
sin x −
π 3
π 3
5π 6
4π 3
0
π 2
π
1
0
11π 6
7π 3
3π 2
2π
−1
0
C O
x
PY
, the value of the expression x− π3 runs from 0 to 2π. So As x runs from π3 to 7π 3 for one cycle of the graph of y = sin x − π3 , we then expect to have the graph of y = sin x starting from x = π3 . This is confirmed by the values in Table 3.23. We then apply a similar procedure to complete one cycle of the graph; that is, divide the interval [ π3 , 7π ] into four equal parts, and then determine the key values of 3 x in sketching the graphs as discussed earlier. The one-cycle graph of y = sin x (dashed curve) and the corresponding one-cycle graph of y = sin x − π3 (solid curve) are shown in Figure 3.24.
0
EP E
D
Table 3.23
D
Figure 3.24
Observe that the graph of y = sin x − π3 shifts π3 units to the right of y = sin x. Thus, they have the same period, amplitude, domain, and range. The graphs of y = a sin b(x − c) and y = a cos b(x − c) have the same shape as y = a sin bx and y = a cos bx, respectively, but shifted c units to the right when c > 0 and shifted |c| units to the left if c < 0. The number c is called the phase shift of the sine or cosine graph. 151
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Example 3.3.4. In the same Cartesian plane, sketch one cycle of the graphs of π y = 3 sin x and y = 3 sin x + 4 . Solution. We have sketched the graph of y = 3 sin x earlier at the start of the π lesson. We consider y = 3 sin x + 4 . We expect that it has the same shape as that of y = 3 sin x, but shifted some units. Here, we have a = 3, b = 1, and c = − π4 . From these constants, we get the amplitude, the period, and the phase shift, and these are 3, 2π, and − π4 , respectively.
7π 4
=
3π , 4
x
3π 4
=
π 4
We now compute
π , 4
3π 4
+ 2
7π 4
=
5π 4
− π4
π 4
3π 4
5π 4
7π 4
0
3
0
−3
0
D
EP E
D
y = 3 sin x +
− π4 + 2
C O
− π4 + 2
7π . 4
PY
One cycle starts at x = − π4 and ends at x = − π4 + 2π = the important values of x.
While the effect of c in y = a sin b(x − c) and y = a cos b(x − c) is a horizontal shift of their graphs from the corresponding graphs of y = a sin bx and y = a cos bx, the effect of d in the equations y = a sin b(x − c) + d and y = a cos b(x − c) + d is a vertical shift. That is, the graph of y = a sin b(x − c) + d has the same amplitude, period, and phase shift as that of y = a sin b(x − c), but shifted d units upward when d > 0 and |d| units downward when d < 0.
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Example 3.3.5. Sketch the graph of π y = −2 cos 2 x − − 3. 6
π 6
Start of one cycle:
π 6
+ 7π 2π 6 = , 2 3
π 6
π 6
+π =
7π 6
+ 2π 5π 3 = , 2 12 x
2π 3
+ 2
7π 6
=
11π 12
C O
End of the cycle:
PY
Solution. Here, a = −2, b = 2, c= π6 , and d = −3. We first sketch one cycle of the graph of y = −2 cos 2 x − π6 , and then extend this graph to the left and to the right, and then move the resulting graph 3 units downward. The graph of y = −2 cos 2 x − π6 has amplitude 2, period π, and phase shift π . 6
5π 12
2π 3
11π 12
7π 6
−2
0
2
0
−2
−5
−3
−1
−3
−5
D
EP E
D
y = −2 cos 2 x − π6 y = −2 cos 2 x − π6 − 3
π 6
Before we end this sub-lesson, we make the following observation, which will be used in the discussion on simple harmonic motion (Sub-Lesson 3.3.6). 153 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Different Equations, The Same Graph 1. The graphs of y = sin x and y = sin(x + 2πk), k any integer, are the same. 2. The graphs of y = sin x, y = − sin(x + π), y = cos(x − π2 ), and y = − cos(x + π2 ) are the same. 3. In general, the graphs of y = a sin b(x − c) + d,
y = a cos[b(x − c) −
π 2
+ 2πk] + d,
and π 2
+ 2πk] + d,
C O
y = −a cos[b(x − c) +
PY
y = −a sin[b(x − c) + π + 2πk] + d,
where k is any integer, are all the same.
Similar observations are true for cosine.
1 sin x
if sin x 6= 0. Using this relationship, we can sketch the
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We know that csc x = graph of y = csc x.
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3.3.4. Graphs of Cosecant and Secant Functions
First, we observe that the domain of the cosecant function is {x ∈ R : sin x 6= 0} = {x ∈ R : x 6= kπ, k ∈ Z}.
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Table 3.25 shows the key numbers (that is, numbers where y = sin x crosses the x-axis, attain its maximum and minimum values) and some neighboring points, where “und” stands for “undefined,” while Figure 3.26 shows one cycle of the graphs of y = sin x (dashed curve) and y = csc x (solid curve). Notice the asymptotes of the graph y = csc x. x
0
π 6
π 2
5π 6
π
y = sin x
0
1 2
1
1 2
0
y = csc x
und
2
1
2
7π 6
3π 2
11π 6
− 12 −1 − 12
2π 0
und −2 −1 −2 und
Table 3.25
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Figure 3.26
We could also sketch the graph of csc x directly from the graph of y = sin x by observing the following facts:
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(1) If sin x = 1 (or −1), then csc x = 1 (or −1).
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(2) At each x-intercept of y = sin x, y = csc x is undefined; but a vertical asymptote is formed because, when sin x is close to 0, the value of csc x will have a big magnitude with the same sign as sin x.
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Refer to Figure 3.27 for the graphs of y = sin x (dashed curve) and y = csc x (solid curve) over a larger interval.
Figure 3.27
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Like the sine and cosecant functions, the cosine and secant functions are also reciprocals of each other. Therefore, y = sec x has domain {x ∈ R : cos x 6= 0} = {x ∈ R : x 6=
kπ , k odd integer}. 2
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Similarly, the graph of y = sec x can be obtained from the graph of y = cos x. These graphs are shown in Figure 3.28.
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Figure 3.28
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Example 3.3.6. Sketch the graph of y = 2 csc x2 .
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Solution. First, we sketch the graph of y = 2 sin x2 , and use the technique discussed above to sketch the graph of y = 2 csc x2 .
The vertical asymptotes of y = 2 csc x2 are the x-intercepts of y = 2 sin x2 : x = 0, ±2π, ±4π, . . .. After setting up the asymptotes, we now sketch the graph of y = 2 csc x2 as shown below. 156 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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Example 3.3.7. Sketch the graph of y = 2 − sec 2x.
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Solution. Sketch the graph of y = − cos 2x (note that it has period π), then sketch the graph of y = − sec 2x (as illustrated above), and then move the resulting graph 2 units upward to obtain the graph of y = 2 − sec 2x.
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3.3.5. Graphs of Tangent and Cotangent Functions sin x , where cos x = 6 0. From this definition of the tangent We know that tan x = cos x function, it follows that its domain is the same as that of the secant function, which is
{x ∈ R : cos x 6= 0} = {x ∈ R : x 6=
kπ , k odd integer}. 2
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We note that tan x = 0 when sin x = 0 (that is, when x = kπ, k any integer), and that the graph of y = tan x has asymptotes x = kπ , k odd integer. Furthermore, 2 by recalling the signs of tangent from Quadrant I to Quadrant IV and its values, we observe that the tangent function is periodic with period π. To sketch the graph of y = tan x, it will be enough to know its one-cycle π π graph on the open interval − 2 , 2 . See Table 3.29 and Figure 3.30.
y = tan x x
− π3 √ − 3
− π4
− π6
−1
−
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− π2
x
und
π 6 √ 3 3
y = tan x
π 4
1
π 3
√
3
0
√
3 3
0
π 2
und
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Table 3.29
Figure 3.30
In the same manner, the domain of y = cot x =
cos x sin x
is
{x ∈ R : sin x = 6 0} = {x ∈ R : x = 6 kπ, k ∈ Z}, and its period is also π. The graph of y = cot x is shown in Figure 3.31. 158 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
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Figure 3.31
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In general, to sketch the graphs of y = a tan bx and y = a cot bx, a 6= 0 and b > 0, we may proceed with the following steps: π π we draw one cycle of the graph on − 2b , 2b (1) Determine the period πb . Then π for y = a tan bx, and on 0, b for y = a cot bx.
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(2) Determine the two adjacent vertical asymptotes. For y = a tan bx, these π vertical asymptotes are given by x = ± 2b . For y = a cot bx, the vertical asymptotes are given by x = 0 and x = πb .
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(3) Divide the interval formed by the vertical asymptotes in Step 2 into four equal parts, and get three division points exclusively between the asymptotes. (4) Evaluate the function at each of these x-values identified in Step 3. The points will correspond to the signs and x-intercept of the graph.
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(5) Plot the points found in Step 3, and join them with a smooth curve approaching to the vertical asymptotes. Extend the graph to the right and to the left, as needed.
Example 3.3.8. Sketch the graph of y = 21 tan 2x.
Solution. The period of the function is π2 , and the adjacent asymptotes are x = ± π4 , ± 3π , . . .. Dividing the interval − π4 , π4 into four equal parts, the key x-values 4 are − π8 , 0, and π8 . x
− π8
0
π 8
y = 21 tan 2x
− 21
0
1 2
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Example 3.3.9. Sketch the graph of y = 2 cot x3 on the interval (0, 3π).
Solution. The period of the function is 3π, and the adjacent asymptotes are x = 0 and x = 3π. We now divide the interval (0, 3π) into four equal parts, and the key x-values are 3π , 3π , and 9π . 4 2 4 3π 4
y = 2 cot x3
2
3π 2
9π 4
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x
−2
D
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0
3.3.6. Simple Harmonic Motion Repetitive or periodic behavior is common in nature. As an example, the timetelling device known as sundial is a result of the predictable rising and setting of the sun everyday. It consists of a flat plate and a gnomon. As the sun moves across the sky, the gnomon casts a shadow on the plate, which is calibrated to tell the time of the day.
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Sundial, by liz west, 29 March 2007,
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https://commons.wikimedia.org/wiki/File:Sundial 2r.jpg. Public Domain.
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Some motions are also periodic. When a weight is suspended on a spring, pulled down, and released, the weight oscillates up and down. Neglecting resistance, this oscillatory motion of the weight will continue on and on, and its height is periodic with respect to time.
t = 0 sec
t = 2.8 sec
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t = 9 sec
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t = 6.1 sec
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Periodic motions are usually modeled by either sine or cosine function, and are called simple harmonic motions. Unimpeded movements of objects like oscillation, vibration, rotation, and motion due to water waves are real-life occurrences that behave in simple harmonic motion. Equations of Simple Harmonic Motion
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The displacement y (directed height or length) of an object behaving in a simple harmonic motion with respect to time t is given by one of the following equations:
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y = a sin b(t − c) + d
or
y = a cos b(t − c) + d.
In both equations, we have the following information:
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• amplitude = |a| = 21 (M − m) - the maximum displacement above and below the rest position or central position or equilibrium, where M is the maximum height and m is the minimum height;
• period = 2π - the time required to complete one cycle (from one |b| highest or lowest point to the next); • frequency =
|b| 2π
- the number of cycles per unit of time;
• c - responsible for the horizontal shift in time; and • d - responsible for the vertical shift in displacement. Example 3.3.10. A weight is suspended from a spring and is moving up and down in a simple harmonic motion. At start, the weight is pulled down 5 cm below the resting position, and then released. After 8 seconds, the weight reaches its 162 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
highest location for the first time. Find the equation of the motion. Solution. We are given that the weight is located at its lowest position at t = 0; that is, y = −5 when t = 0. Therefore, the equation is y = −5 cos bt. Because it took the weight 8 seconds from the lowest point to its immediate highest point, half the period is 8 seconds.
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1 2π π πt · = 8 =⇒ b = =⇒ y = −5 cos 2 2 b 8 8 ? Example 3.3.11. Suppose you ride a Ferris wheel. The lowest point of the wheel is 3 meters off the ground, and its diameter is 20 m. After it started, the Ferris wheel revolves at a constant speed, and it takes 32 seconds to bring you back again to the riding point. After riding for 150 seconds, find your approximate height above the ground.
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Solution. We ignore first the fixed value of 3 m off the ground, and assume that the central position passes through the center of the wheel and is parallel to the ground.
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Let t be the time (in seconds) elapsed that you have been riding the Ferris wheel, and y is he directed distance of your location with respect to the assumed central position at time t. Because y = −10 when t = 0, the appropriate model is y = −10 cos bt for t ≥ 0.
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Given that the Ferris wheel takes 32 seconds to move from the lowest point to the next, the period is 32. π 2π = 32 =⇒ b = =⇒ b 16 When t = 150, we get y = 10 cos 150π ≈ 3.83. 16
y = −10 cos
πt 16
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Bringing back the original condition given in the problem that the riding point is 3 m off the ground, after riding for 150 seconds, you are approximately located 3.83 + 13 = 16.83 m off the ground. 2 In the last example, the central position or equilibrium may be vertically shifted from the ground or sea level (the role of the constant d). In the same way, the starting point may also be horizontally shifted (the role of the constant c). Moreover, as observed in Sub-Lesson 3.3.3 (see page 154), to find the function that describes a particular simple harmonic motion, we can either choose y = a sin b(t − c) + d or y = a cos b(t − c) + d, and determine the appropriate values of a, b, c, and d. In fact, we can assume that a and b are positive numbers, and c is the smallest such nonnegative number. 163 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Example 3.3.12. A signal buoy in Laguna Bay bobs up and down with the height h of its transmitter (in feet) above sea level modeled by h(t) = a sin bt + d at time t (in seconds). During a small squall, its height varies from 1 ft to 9 ft above sea level, and it takes 3.5 seconds from one 9-ft height to the next. Find the values of the constants a, b, and d. Solution. We solve the constants step by step. • The minimum and maximum values of h(t) are 1 ft and 9 ft, respectively. Thus, the amplitude is a = 12 (M − m) = 12 (9 − 1) = 4.
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• Because it takes 3.5 seconds from one 9-ft height to the next, the period is 3.5. Thus, we have 2π = 3.5, which gives b = 4π . b 7
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• Because the lowest point is 1 ft above the sea level and the amplitude is 4, it follows that d = 5. 2
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Solution.
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Example 3.3.13. A variable star is a star whose brightness fluctuates as observed from Earth. The magnitude of visual brightness of one variable star ranges from 2.0 to 10.1, and it takes 332 days to observe one maximum brightness to the next. Assuming that the visual brightness of the star can be modeled by the equation y = a sin b(t − c) + d, t in days, and putting t = 0 at a time when the star is at its maximum brightness, find the constants a, b, c, and d, where a, b > 0 and c the least nonnegative number possible. 10.1 − 2.0 M −m = = 4.05 2 2 π 2π = 332 =⇒ b = b 166 d = a + m = 4.05 + 2.0 = 6.05
a=
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For the (ordinary) sine function to start at the highest point at t = 0, the least possible horizontal movement to the right (positive value) is 3π units. 2 bc =
3π 2
=⇒
c=
3π 3π = π = 249 2b 2 · 166
2
? Example 3.3.14. The path of a fast-moving particle traces a circle with equation (x + 7)2 + (y − 5)2 = 36. It starts at point (−1, 5), moves clockwise, and passes the point (−7, 11) for the first time after traveling 6 microseconds. Where is the particle after traveling 15 microseconds?
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Solution. As described above, we may choose sine or cosine function. Here, we choose the sine function to describe both x and y in terms of time t in microseconds; that is, we let x = a sin b(t − c) + d and y = e sin f (t − g) + h, where we appropriately choose the positive values for a, b, e, and f , and the least nonnegative values for c and g. The given circle has radius 6 and center (−7, 5). Defining the central position of the values of x as the line x = −7 and that of the values of y as the line y = 5, we get a = e = 6, d = −7, and h = 5.
=⇒
b=f =
π 4
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3 2π 3 2π · = · =6 4 b 4 f
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From the point (−1, 5) to the point (−7, 11) (moving clockwise), the particle has traveled three-fourths of the complete cycle; that is, three-fourths of the period must be 2.
As the particle starts at (−1, 5) and moves clockwise, the values of x start at its highest value (x = −1) and move downward toward its central position (x = −7) and continue to its lowest value (x = −13). Therefore, the graph of = 6 units to the right, and so we get c = 6. a sin bt + d has to move 3π 2b
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As to the value of g, we observe the values of y start at its central position (y = 5) and go downward to its lowest value (y = −1). Similar to the argument used in determining c, the graph of y = e sin f t + h has to move πb = 4 units to the right, implying that g = 4. Hence, We have the following equations of x and y in terms of t: x = 6 sin π4 (t − 6) − 7 and y = 6 sin π4 (t − 4) + 5.
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When t = 15, we get
and
√ x = 6 sin π4 (15 − 6) − 7 = −7 + 3 2 ≈ −2.76 √ y = 6 sin π4 (15 − 4) + 5 = 5 + 3 2 ≈ 9.24.
That is, after traveling for 15 microseconds, the particle is located near the point (−2.76, 9.24). 2
More Solved Examples 1. Find the period of the function y = 4 sin x−π − 3. 4 x−π 1 Solution: y = 4 sin 4 − 3 =⇒ y = 4 sin 3 (x − π) − 3 =⇒ P =
2π 1 3
= 6π
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2. In the function y = 3 tan(2kx − π), the period is 4π. Find the value of k and the phase shift of the graph of the function. Solution: The period of the tangent function is P = πb . π π y = 3 tan(2kx − π) =⇒ y = 3 tan 2k x − =⇒ P = = 4π. 2k 2k 1 1 = 4 =⇒ k = 2k 8
and Phase shift = π 6
+2 over one period. Determine
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3. Sketch the graph of function y = 12 sin 21 x + the domain and range of the function.
π π = 4π = 2k 2 81
Solution: The graph is a vertical translation of y = 12 sin 12 x + π6 by 2 units upward. The period of the given function is 2π 1 = 4π. One complete cycle may start at x = − π6 and end at x = − π6 + 4π =
23π . 6
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The critical points for the graph are
2
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π 5π 11π 17π 23π x=− , x=− , x= , x= , and x = . 6 6 6 6 6
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The domain of the function is R and its range is
5
3 , . 2 2
4. Sketch the graph of the function y = −2 cos(x − π2 ) + 3 over two periods. Find the domain and range of the function. Solution: The graph of the given function is a vertical translation of y = −2 cos(x − π2 ) by 3 units upward. The period of the function is 2π. One complete cycle may start and end at x = π2 and x = 5π , respectively. The 2 9π next complete cycle starts at x = 5π and ends at x = . 2 2 π 3π 5π 7π 9π critical points: , π, , 2π, , 3π, , 4π, 2 2 2 2 2
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The domain of the given cosine function is R, and its range is [1, 5]. 5. Sketch the graph of the function y = 41 tan x − π4 over three periods. Find the domain and range of the function.
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Solution: The period of the function is π. One complete cycle may start at x = π4 and end at x = 5π . 4
The domain of the function is {x|x 6=
3π 4
+ kπ, k ∈ Z}, and its range is R. π 6. Sketch the graph of the function y = −3 cot 12 x + 12 + 2 over three periods. Find the domain and range of the function. π + 2 = −3 cot 21 x + π6 + 2 =⇒ P = 2π Solution: y = −3 cot 21 x + 12 One complete cycle may start at x = − π6 and end at x =
11π . 6
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The domain of the function is {x|x 6= − π6 + 2kπ, k ∈ Z}, and its range is R. 7. The graph of the function g(x) is the same as that of f (x) = 3 sin x − π3 but shifted 2 units downward and π2 units to the right. What is g(−π)? Solution: The function f (x) = 3 sin x − π3 when shifted 2 units downward and π2 units to the right is π π 5π g(x) = 3 sin x − − − 2 = 3 sin x − − 2. 3 2 6 5π 1 g(−π) = 3 sin −π − −2=− 6 2
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8. The graph of the function h(x) is the same as that of f (x) = 3 sin(2x − 3π) + 1 but shifted 3 units upward and π2 units to the left. What is h( 5π )? 6 Solution: h(x) = 3 sin 2 x + π2 − 3π + 1 + 3 = 3 sin(2x − 2π) + 4 h
5π 6
√ 5π 8−3 3 = 3 sin 2 − 2π + 4 = 6 2
9. Sketch the graph of y = 2 sec 12 x − and range of the function.
π 4
over two periods. Find the domain
Solution: The period of the function is 4π. One complete cycle may start at x = π4 and end at x = 17π . 4
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+ 2kπ, k ∈ Z}, and its range is The domain of the function is {x|x 6= 5π 4 (−∞, −2] ∪ [2, ∞). 10. Sketch the graph of y = − csc x + π3 + 2 over two periods. Find the domain and range of the function.
D
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Solution: The period of the function is 2π. One complete cycle may start and end at x = − π3 and x = 5π , respectively. 3
The domain of the function is {x|x 6= − π3 + kπ, k ∈ Z}, and its range is (−∞, 1] ∪ [3, ∞).
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Supplementary Problems 3.3 1. What is the period of the function y = −2 cos 14 x −
π 2
2. The amplitude and period of the function y = 4 − a2 cos respectively. Find |a| + b.
? bx 3
− π are 3 and 4π,
3. In the function y = 2π − 3π cot 4π (x − 2), the period is 2. Find the value of k. k
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− 4. What are the minimum and maximum values of the function y = 3 sin 34 x + 2π 3 5? 5. Given the function y = 3 sin 43 x + 2π − 5, find the value of y when x = 8π . 3 9 6. Given the function y = −2 cot 34 x − π6 + 3, find the value of y when x = 7π . 6 7. Find the domain and range of the function y = − 32 sin 13 x − 3π + 2? 4 8. Find the range of the function y = 3 sec 2x . 3 9. Find the equation of the secant function whose graph is the graph of y = 3 sec 2x shifted π units to the right and 3 units downward.
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10. Find the equation of the sine function whose graph is the graph of y = π −2 sin 2 x − 4 + 1 shifted π2 units to the left and 3 units upward. 11. Given the tangent function y = 1 − 3 tan 2x−π , find the equations of all its 4 vertical asymptotes. 12. Given the cosecant function y = csc x2 − π3 , find the equations of all its vertical asymptotes.
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13. Sketch the graph over one period, and indicate the period, phase shift, domain, and range for each. (c) y = 12 csc 34 (2x − π) − 1 (a) y = 2 sin 14 x + π4 − 1 +2 (b) y = tan 12 2x + π3 − 2 (d) y = sec 21 4x + 2π 3 14. A point P in simple harmonic motion has a frequency of 21 oscillation per minute and amplitude of 4 ft. Express the motion of P by means of an equation in the form d = a sin bt.
? 15. A mass is attached to a spring, and then pulled and released 8 cm below its resting position at the start. If the simple harmonic motion is modeled by 1 y = a cos 10 (t − c), where a > 0, c the least nonnegative such number, and t in seconds, find the location of the mass 10 seconds later.
4
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Lesson 3.4. Fundamental Trigonometric Identities Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) determine whether an equation is an identity or a conditional equation; (2) derive the fundamental trigonometric identities; (3) simplify trigonometric expressions using fundamental trigonometric identities; and
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(4) prove other trigonometric identities using fundamental trigonometric identities. Lesson Outline
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(1) Domain of an equation (2) Identity and conditional equation
(3) Fundamental trigonometric identities (4) Proving trigonometric identities Introduction
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In previous lessons, we have defined trigonometric functions using the unit circle and also investigated the graphs of the six trigonometric functions. This lesson builds on the understanding of the different trigonometric functions by discovery, deriving, and working with trigonometric identities. 3.4.1. Domain of an Expression or Equation
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Consider the following expressions: √ x2 − 1, 2x + 1,
x2
x , − 3x − 4
√
x . x−1
What are the real values of the variable x that make the expressions defined in the set of real numbers? In the first expression, every real value of x when substituted to the expression makes it defined in the set of real numbers; that is, the value of the expression is real when x is real. In the second expression, not every real value of x makes √ the expression defined in R. For example, when x = 0, the expression becomes −1, which is not a real number. √ x2 − 1 ∈ R ⇐⇒ x2 − 1 ≥ 0 ⇐⇒ x ≤ −1 or x ≥ 1 171
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Here, for
√
x2 − 1 to be defined in R, x must be in (−∞, −1] ∪ [1, ∞).
In the third expression, the values of x that make the denominator zero make the entire expression undefined. x2 − 3x − 4 = (x − 4)(x + 1) = 0
⇐⇒
x = 4 or x = −1
x is real when x 6= 4 and x 6= −1. − 3x − 4 √ In the fourth expression, because the expression x − 1 is in the denominator, x must be greater than 1. Although the value of the entire expression is 0 when x = 0, we do not include 0 as allowed value of x because part of the expression is not real when x = 0. x2
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Hence, the expression
In the expressions above, the allowed values of the variable x constitute the domain of the expression.
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The domain of an expression (or equation) is the set of all real values of the variable for which every term (or part) of the expression (equation) is defined in R.
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In the expressions above, the domains of the first, second, third, and fourth expressions are R, (−∞, −1] ∪ [1, ∞), R \ {−1, 4}, and (1, ∞), respectively. Example 3.4.1. Determine the domain of the expression/equation.
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√ x2 − 1 x+1 − (a) 3 2 x + 2x − 8x 1−x (b) tan θ − sin θ − cos 2θ (c) x2 −
√
1 + x2 = √ 3
2
x2
−1
cos2 z = 4 sin z − 1 1 + sin z
D (d) z −
Solution. (a) x3 + 2x2 − 8x = x(x + 4)(x − 2) = 0 −4, or x = 2 √ x + 1 ∈ R ⇐⇒ x + 1 ≥ 0 ⇐⇒ x ≥ −1 1−x=0
⇐⇒
⇐⇒
x = 0, x =
x=1
Domain = [−1, ∞) \ {−4, 0, 1, 2} = [−1, 0) ∪ (0, 1) ∪ (1, 2) ∪ (2, ∞)
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(b) tan θ − sin θ − cos 2θ = cos θ = 0
⇐⇒
θ=
sin θ − sin θ − cos 2θ cos θ kπ , k odd integer 2
Domain = R \ { kπ | k odd integer} 2 √ (c) The expression 1+x2 is always positive, and so 1 + x2 is defined in R. On √ 3 2 the other hand, the expression x − 1 is also defined in R, but it cannot be zero because it is in the denominator. Therefore, x should not be −1 and 1. Domain = R \ {−1, 1} z=
3π 2
+ 2kπ, k ∈ Z
Domain = R \ { 3π + 2kπ|k ∈ Z} 2
2
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3.4.2. Identity and Conditional Equation
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⇐⇒
(d) 1 + sin z = 0
Consider the following two groups of equations: Group A
Group B
(A1) x2 − 1 = 0
(B2) (x + 7)2 = x2 + 14x + 49 x2 − 4 (B3) =x+2 x−2
EP E
D
(A2) (x + 7)2 = x2 + 49 x2 − 4 (A3) = 2x − 1 x−2
(B1) x2 − 1 = (x − 1)(x + 1)
D
In each equation in Group A, some values of the variable that are in the domain of the equation do not satisfy the equation (that is, do not make the equation true). On the other hand, in each equation in Group B, every element in the domain of the equation satisfies the given equation. The equations in Group A are called conditional equations, while those in Group B are called identities. An identity is an equation that is true for all values of the variable in the domain of the equation. An equation that is not an identity is called a conditional equation. (In other words, if some values of the variable in the domain of the equation do not satisfy the equation, then the equation is a conditional equation.)
Example 3.4.2. Identify whether the given equation is an identity or a conditional equation. For each conditional equation, provide a value of the variable in the domain that does not satisfy the equation. (1) x3 − 2 = x −
√ √ √ 3 2 x2 + 3 2x + 3 4 173
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(2) sin2 θ = cos2 θ + 1 (3) sin θ = cos θ − 1 √ √ 1− x 1−2 x+x √ = (4) 1−x 1+ x Solution. (1) This is an identity because this is simply factoring of difference of two cubes. (2) This is a conditional equation. If θ = 0, then the left-hand side of the equation is 0, while the right-hand side is 2.
PY
(3) This is also a conditional equation. If θ = 0, then both sides of the equation are equal to 0. But if θ = π, then the left-hand side of the equation is 0, while the right-hand side is −2.
C O
(4) This is an identity because the right-hand side of the equation is obtained by rationalizing the denominator of the left-hand side. 2 3.4.3. The Fundamental Trigonometric Identities
EP E
D
Recall that if P (x, y) is the terminal point on the unit circle corresponding to θ, then we have y 1 tan θ = sin θ = y csc θ = y x 1 x cos θ = x sec θ = cot θ = . x y From the definitions, the following reciprocal and quotient identities immediately follow. Note that these identities hold if θ is taken either as a real number or as an angle.
D
Reciprocal Identities
csc θ =
1 sin θ
sec θ =
1 cos θ
cot θ =
1 tan θ
Quotient Identities tan θ =
sin θ cos θ
cot θ =
cos θ sin θ
We can use these identities to simplify trigonometric expressions.
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Example 3.4.3. Simplify: tan θ cos θ (1) sin θ Solution. (2)
(2)
tan θ cos θ = (1) sin θ
cos θ cot θ
sin θ cos θ
cos θ =1 sin θ
cos θ cos θ = cos θ = sin θ cot θ sin θ
2
If P (x, y) is the terminal point on the unit circle corresponding to θ, then x + y 2 = 1. Since sin θ = y and cos θ = x, we get
PY
2
sin2 θ + cos2 θ = 1.
By dividing both sides of this identity by cos2 θ and sin2 θ, respectively, we obtain and 1 + cot2 θ = csc2 θ.
C O
tan2 θ + 1 = sec2 θ
Pythagorean Identities sin2 θ + cos2 θ = 1
1 + cot2 θ = csc2 θ
D
tan2 θ + 1 = sec2 θ
EP E
Example 3.4.4. Simplify: (1) cos2 θ + cos2 θ tan2 θ Solution.
1 + tan2 θ 1 + cot2 θ
(1) cos2 θ + cos2 θ tan2 θ = (cos2 θ)(1 + tan2 θ) = cos2 θ sec2 θ =1
1 + tan2 θ sec2 θ = = 1 + cot2 θ csc2 θ
D (2)
(2)
1 cos2 θ 1 sin2 θ
=
sin2 θ = tan2 θ cos2 θ
2
In addition to the eight identities presented above, we also have the following identities. Even-Odd Identities sin(−θ) = − sin θ
cos(−θ) = cos θ
tan(−θ) = − tan θ
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The first two of the negative identities can be obtained from the graphs of the sine and cosine functions, respectively. (Please review the discussion on page 147.) The third identity can be derived as follows: tan(−θ) =
− sin θ sin(−θ) = = − tan θ. cos(−θ) cos θ
The reciprocal, quotient, Pythagorean, and even-odd identities constitute what we call the fundamental trigonometric identities. We now solve Example 3.2.3 in a different way.
PY
Example 3.4.5. If sin θ = − 34 and cos θ > 0. Find cos θ.
C O
Solution. Using the identity sin2 θ + cos2 θ = 1 with cos θ > 0, we have s 2 √ p 3 7 2 = . cos θ = 1 − sin θ = 1 − − 4 4
2
Example 3.4.6. If sec θ = 52 and tan θ < 0, use the identities to find the values of the remaining trigonometric functions of θ.
2 1 = sec θ 5
EP E
cos θ =
D
Solution. Note that θ lies in QIV.
√ sin θ = − 1 − cos2 θ = −
√ 2 2 21 1− =− 5 5
s
√ 1 5 21 csc θ = =− sin θ 21
D
√ √ − 521 sin θ 21 = 2 =− tan θ = cos θ 2 5 √ 1 2 21 cot θ = =− tan θ 21
2
3.4.4. Proving Trigonometric Identities We can use the eleven fundamental trigonometric identities to establish other identities. For example, suppose we want to establish the identity csc θ − cot θ =
sin θ . 1 + cos θ
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To verify that it is an identity, recall that we need to establish the truth of the equation for all values of the variable in the domain of the equation. It is not enough to verify its truth for some selected values of the variable. To prove it, we use the fundamental trigonometric identities and valid algebraic manipulations like performing the fundamental operations, factoring, canceling, and multiplying the numerator and denominator by the same quantity. Start on the expression on one side of the proposed identity (preferably the complicated side), use and apply some of the fundamental trigonometric identities and algebraic manipulations, and arrive at the expression on the other side of the proposed identity. Explanation
csc θ − cot θ cos θ 1 − = sin θ sin θ
Start on one side.
1 − cos2 θ (sin θ)(1 + cos θ) sin2 θ = (sin θ)(1 + cos θ)
EP E =
Add the quotients. Multiply the numerator and denominator by 1 + cos θ. Multiply.
D
=
Apply some reciprocal and quotient identities.
C O
1 − cos θ sin θ 1 − cos θ 1 + cos θ = · sin θ 1 + cos θ =
PY
Expression
sin θ 1 + cos θ
Apply a Pythagorean identity. Reduce to lowest terms.
D
Upon arriving at the expression of the other side, the identity has been established. There is no unique technique to prove all identities, but familiarity with the different techniques may help. Example 3.4.7. Prove: sec x − cos x = sin x tan x. Solution. 1 − cos x cos x sin2 x sin x 1 − cos2 x = = = sin x · = sin x tan x cos x cos x cos x
sec x − cos x =
Example 3.4.8. Prove:
2
1 + sin θ 1 − sin θ − = 4 sin θ sec2 θ 1 − sin θ 1 + sin θ 177
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Solution.
2
PY
1 + sin θ 1 − sin θ (1 + sin θ)2 − (1 − sin θ)2 − = 1 − sin θ 1 + sin θ (1 − sin θ)(1 + sin θ) 1 + 2 sin θ + sin2 θ − 1 + 2 sin θ − sin2 θ = 1 − sin2 θ 4 sin θ = cos2 θ = 4 sin θ sec2 θ
More Solved Examples Solution: 1 cos θ
C O
1. Express each of the other circular functions of θ in terms of cos θ. •
sec θ =
•
√ sin2 θ + cos2 θ = 1 =⇒ sin2 θ = 1 − cos2 θ =⇒ sin θ = ± 1 − cos2 θ
•
csc θ =
•
cot θ =
D
1 1 = √ sin θ ± 1 − cos2 θ
EP E
cos θ cos θ = √ sin θ ± 1 − cos2 θ √ sin θ ± 1 − cos2 θ tan θ = = cos θ cos θ
•
2. If tan θ = a, express cos2 θ in terms of a.
D
Solution:
a=
sin θ sin2 θ 1 − cos2 θ 2 =⇒ a2 = =⇒ a = cos θ cos2 θ cos2 θ 2 2 2 a cos θ = 1 − cos θ
a2 cos2 θ + cos2 θ = 1 =⇒ cos2 θ(a2 + 1) = 1 =⇒ cos2 θ =
1 a2 + 1
3. Given a = cos x, simplify and express sin4 x − cos4 x in terms of a. Solution: sin4 x − cos4 x = (sin2 x + cos2 x)(sin2 x − cos2 x) = sin2 x − cos2 x = 1 − cos2 x − cos2 x = 1 − 2 cos2 x = 1 − 2a2 178 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
4. Simplify (csc x − sec x)2 + (csc x + sec x)2 .
5. Verify the identity
csc θ = cos θ. tan θ + cot θ
Solution:
PY
Solution: (csc x − sec x)2 + (csc x + sec x)2 = (csc2 x − 2 csc x sec x + sec2 x) + (csc2 x + 2 csc x sec x + sec2 x) = 2 csc2 x + 2 sec2 x 2 2 = + 2 sin x cos2 x 2 2(cos2 x + sin2 x) = = 2 csc2 x sec2 x = 2 2 2 2 sin x cos x sin x cos x
C O
1 1 1 cos θ sin θ csc θ sin θ sin θ = = · = cos θ = 2 2 sin θ cos θ tan θ + cot θ sin θ 1 sin θ + cos θ + cos θ sin θ cos θ sin θ csc θ + cot θ − 1 1 + cos θ = . cot θ − csc θ + 1 sin θ csc θ + cot θ − 1 Solution: cot θ − csc θ + 1 csc θ + cot θ − 1 csc θ + cot θ · = cot θ − csc θ + 1 csc θ + cot θ (csc θ + cot θ − 1)(csc θ + cot θ) = (cot θ − csc θ)(csc θ + cot θ) + (csc θ + cot θ) (csc θ + cot θ − 1)(csc θ + cot θ) = cot2 θ − csc2 θ + csc θ + cot θ (csc θ + cot θ − 1)(csc θ + cot θ) = −1 + csc θ + cot θ 1 cos θ 1 + cos θ = csc θ + cot θ = + = sin θ sin θ sin θ
D
EP E
D
6. Establish the identity
Supplementary Problems 3.4 1. Using fundamental identities, simplify the expression
tan x − sin x . sin x
2. Using fundamental identities, simplify the expression
1 . csc x − cot x
3. Simplify sin A +
cos2 A . 1 + sin A 179
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4. Simplify (1 − cos2 A)(1 + cot2 A). csc x + sec x in terms of sine and cosine. cot x + tan x
6. Express
tan x − cot x in terms of sine and cosine. tan x + cot x
7. Express
tan x + sin x in terms cosine only. csc x + cot x
8. Express
1 in terms sine only. 1 + tan2 x
9. If cot θ = a, express sin θ cos θ in terms of a.
PY
5. Express
10. If sec θ = a > 0 and sin θ > 0, express sin θ cos θ in terms of a.
11.
csc a + 1 1 + sin a = csc a − 1 1 − sin a
C O
For numbers 11 - 20, establish the identities.
1 + sin a 1 − sin a − = 4 tan a sec a 1 − sin a 1 + sin a cos a 13. = 1 − sin a sec a + tan a
D
12.
16.
sin3 α − cos3 α = 1 + sin α cos α sin α − cos α
tan α sin α cos α = 2 1 − tan α 2 cos2 α − 1
D
17.
EP E
csc a + sec a tan a = tan a sec a csc2 a 1 1 15. + = 2 csc2 a 1 − cos a 1 + cos a
14.
18.
tan2 α + sec α + 1 = tan α + sin α tan α + cot α
19.
cot α − sin α sec α = cos2 α − sin2 α sec α csc α
20. tan2 α sec2 α − sec2 α + 1 = tan4 α
4
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Lesson 3.5. Sum and Difference Identities Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) derive trigonometric identities involving sum and difference of two angles; (2) simplify trigonometric expressions using fundamental trigonometric identities and sum and difference identities;
PY
(3) prove other trigonometric identities using fundamental trigonometric identities and sum and difference identities; and (4) solve situational problems involving trigonometric identities. Lesson Outline (2) Cofunction identities (3) More trigonometric identities Introduction
C O
(1) The sum and difference identities for cosine, sine, and tangent functions
EP E
D
In previous lesson, we introduced the concept of trigonometric identity, presented the fundamental identities, and proved some identities. In this lesson, we derive the sum and difference identities for cosine, sine, and tangent functions, establish the cofunction identities, and prove more trigonometric identities. 3.5.1. The Cosine Difference and Sum Identities
D
Let u and v be any real numbers with 0 < v ≤ u < 2π. Consider the unit circle with points A = (1, 0), P1 , P2 , P3 , and u and v with corresponding angles as shown below. Then P1 P2 = AP3 .
181 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Recall that P1 = P (u) = (cos u, sin u), P2 = P (v) = (cos v, sin v), and P3 = P (u − v) = (cos(u − v), sin(u − v)), so that p P1 P2 = (cos u − cos v)2 + (sin u − sin v)2 , while AP3 =
p [cos(u − v) − 1]2 + [sin(u − v) − 0]2 .
Equating these two expressions and expanding the squares, we get
PY
(cos u − cos v)2 + (sin u − sin v)2 = [cos(u − v) − 1]2 + sin2 (u − v)
C O
cos2 u − 2 cos u cos v + cos2 v + sin2 u − 2 sin u sin v + sin2 v = cos2 (u − v) − 2 cos(u − v) + 1 + sin2 (u − v)
Applying the Pythagorean identity cos2 θ+sin2 θ = 1 and simplifying the resulting equations, we obtain (cos2 u + sin2 u) + (cos2 v + sin2 v) − 2 cos u cos v − 2 sin u sin v = [cos2 (u − v) + sin2 (u − v)] − 2 cos(u − v) + 1
D
1 + 1 − 2 cos u cos v − 2 sin u sin v = 1 − 2 cos(u − v) + 1
EP E
cos(u − v) = cos u cos v + sin u sin v. We have thus proved another identity.
D
Although we assumed at the start that 0 < v ≤ u < 2π, but because cos(−θ) = cos θ (one of the even-odd identities), this new identity is true for any real numbers u and v. As before, the variables can take any real values or angle measures. Cosine Difference Identity
cos(A − B) = cos A cos B + sin A sin B
Replacing B with −B, and applying the even-odd identities, we immediately get another identity. Cosine Sum Identity cos(A + B) = cos A cos B − sin A sin B π Example 3.5.1. Find the exact values of cos 105◦ and cos 12 .
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Solution. cos 105◦ = cos(60◦ + 45◦ ) = cos 60◦ cos 45◦ − sin 60◦ sin 45◦ √ √ √ 1 2 3 2 = · − · 2√ 2 √ 2 2 2− 6 = 4
Example 3.5.2. Given cos α = QI, find cos(α + β).
PY
π π π = cos − 12 4 6 π π π π = cos cos + sin sin 4 6 √ 4√ 6 √ 2 3 2 1 · + · = 2 2 2 √ √2 6+ 2 = 4
C O
cos
3 5
and sin β =
12 , 13
2
where α lies in QIV and β in
D
EP E
D
Solution. We will be needing sin α and cos β. s 2 √ 3 4 =− sin α = − 1 − cos2 α = − 1 − 5 5 s 2 q 12 5 2 cos β = 1 − sin β = 1 − = 13 13 cos(α + β) = cos α cos β − sin α sin β 3 5 4 12 = · − − 5 13 5 13 63 = 65
2
3.5.2. The Cofunction Identities and the Sine Sum and Difference Identities In the Cosine Difference Identity, if we let A = π2 , we get π π π cos − B = cos cos B + sin sin B 2 2 2 183 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
= (0) cos B + (1) sin B = sin B.
C O
PY
From this identity, if we replace B with π2 − B, we have hπ π i π cos − − B = sin −B 2 2 2 π cos B = sin −B . 2 As for the tangent function, we have sin π − B π 2 −B = tan 2 cos π2 − B cos B = sin B = cot B. We have just derived another set of identities.
Cofunction Identities
2
π
− B = sin B sin − B = cos B 2 π − B = cot B tan 2
EP E
D
cos
π
Using the first two cofunction identities, we now derive the identity for sin(A+
B).
D
hπ
i − (A + B) h2 π i = cos − A − B) π2 π = cos − A cos B + sin − A sin B 2 2 = sin A cos B + cos A sin B
sin(A + B) = cos
Sine Sum Identity sin(A + B) = sin A cos B + cos A sin B In the last identity, replacing B with −B and applying the even-odd identities yield sin(A − B) = sin[A + (−B)] 184 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
= sin A cos(−B) + cos A sin(−B) = sin A cos B − cos A sin B. Sine Difference Identity sin(A − B) = sin A cos B − cos A sin B Example 3.5.3. Find the exact value of sin
5π 12
.
Solution.
π + 4 6 π π π π cos + cos sin = sin 4 6 √ 4√ √6 2 3 2 1 = · + · 2 2 2 2 √ √ 6+ 2 = 4 = sin
π
C O
sin
5π 12
PY
π 2
and
π 2
< β < π,
D
3 and sin β = 21 , where 0 < α < Example 3.5.4. If sin α = 13 find sin(α + β) and sin(β − α).
2
D
EP E
Solution. We first compute cos α and cos β. s √ 2 p 3 4 10 2 cos α = 1 − sin α = 1 − = 13 13 s √ 2 q 1 3 cos β = − 1 − sin2 β = − 1 − =− 2 2 sin(α + β) = sin α cos β + cos α sin β √ ! √ 3 4 10 1 3 − + = · 13 2 13 2 √ √ 4 10 − 3 3 = 26
sin(β − α) = sin β cos α − cos β sin α √ √ ! 1 4 10 3 3 = · − − 2 13 2 13 √ √ 4 10 + 3 3 = 26
2
185 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Example 3.5.5. Prove: sin(x + y) = (1 + cot x tan y) sin x cos y. Solution.
2
PY
(1 + cot x tan y) sin x cos y = sin x cos y + cot x tan y sin x cos y cos x sin y = sin x cos y + sin x cos y sin x cos y = sin x cos y + cos x sin y = sin(x + y) 3.5.3. The Tangent Sum and Difference Identities
C O
Recall that tan x is the ratio of sin x over cos x. When we replace x with A + B, we obtain sin(A + B) tan(A + B) = . cos(A + B) Using the sum identities for sine and cosine, and then dividing the numerator and denominator by cos A cos B, we have sin A cos B + cos A sin B cos A cos B − sin A sin B sin A cos B cos A sin B + cos cos A cos B A cos B = cos A cos B sin A sin B − cos A cos B cos A cos B tan A + tan B . = 1 − tan A tan B
EP E
D
tan(A + B) =
We have just established the tangent sum identity.
D
In the above identity, if we replace B with −B and use the even-odd identity tan(−θ) = − tan θ, we get tan(A − B) = tan[A + (−B)] =
tan A + tan(−B) tan A − tan B = . 1 − tan A tan(−B) 1 + tan A tan B
This is the tangent difference identity. Tangent Sum and Difference Identities tan A + tan B 1 − tan A tan B tan A − tan B tan(A − B) = 1 + tan A tan B tan(A + B) =
186 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
More Solved Examples
PY
7 , and tan B = 24 , B in QI, find: (a) sin(A + B), 1. Given cos A = − 35 , π < A < 3π 2 (b) cos(A + B), and (c) tan(A + B). 3 4 Solution: cos A = − and A in QIII =⇒ sin A = − 5 5 7 7 24 tan B = and B in QIII =⇒ sin B = − and cos B = − 24 25 25 (a) sin(A + B) = sin A cos B + cos A sin B 4 24 3 7 117 = − − + − − = 5 25 5 25 125
(b) cos(A + B) = cos A cos B − sin A sin B 24 4 7 44 3 − − − − = = − 5 25 5 25 125 sin(A + B) = cos(A + B)
117 125 44 125
117 44
C O
(c) tan(A + B) =
=
EP E
D
5π 2. Find the exact value of cos . 12 5π 2π π 2π π 2π π Solution: cos = cos − = cos cos + sin sin 12 3 4 3 4 3 4 √ ! √ ! √ ! 2 3 2 1 + = − 2 2 2 2 √ √ 6− 2 = 4 π 2
+ 2kπ, k ∈ Z, prove that sin A = cos B. π Solution: sin A = sin + 2kπ − B 2 π π = sin + 2kπ cos B − cos + 2kπ sin B = cos B 2 2
D
3. If A + B =
4. Find the value of (tan 10◦ )(tan 15◦ )(tan 20◦ )(tan 15◦ ) · · · (tan 65◦ )(tan 70◦ )(tan 75◦ )(tan 80◦ ). Solution: From the previous item, we know that sin θ = cos(90◦ − θ). We write each tangent in terms of sine and cosine. (tan 10◦ )(tan 15◦ )(tan 20◦ )(tan 15◦ ) · · · (tan 65◦ )(tan 70◦ )(tan 75◦ )(tan 80◦ ) sin 10 sin 15 sin 20 sin 70 sin 75 sin 80 ··· = cos 10 cos 15 cos 20 cos 70 cos 75 cos 80 =1 187
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5. If A, B and C are the angles of a triangle and √ (tan A)(tan B)(tan C) = −
3 , 3
find tan A + tan B + tan C. Solution:
6. Establish the identity
C O
PY
A + B + C = 180◦ =⇒ tan(A + B + C) = tan 180◦ = 0 tan A + tan(B + C) = 0 =⇒ tan A + tan(B + C) = 0 1 − tan A tan(B + C) tan B + tan C tan A + =0 1 − tan B tan C tan A − tan A tan B tan C + tan B + tan C =0 1 − tan B tan C =⇒ tan A − tan A tan B tan C + tan B + tan C = 0 √ 3 tan A + tan B + tan C = tan A tan B tan C = − 3 tan A + tan B sin(A + B) = . cos(A − B) 1 + tan A tan B
D
Solution:
D
EP E
sin A cos B + cos A sin B sin(A + B) = cos(A − B) cos A cos B + sin A sin B 1 sin A cos B + cos A sin B cos A cos B · = 1 cos A cos B + sin A sin B cos A cos B sin A cos B cos A sin B + cos A cos B cos A cos B = cos A cos B sin A sin B + cos A cos B cos A cos B tan A + tan B = 1 + tan A tan B
Supplementary Problems 3.5 1. If
3π 2
< θ < 2π, find the radian measure of θ if cos θ = sin 2π . 3
2. For what angle θ in QIV is sin θ = cos 4π ? 3 3. If A + B =
π 2
+ kπ, k ∈ Z, prove that tan A = cot B. 188
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4. What is the exact value of cot − 5π ? 12 5. What is the exact value of sin 105◦ − cos 15◦ ? 6. What is the exact value of tan 1875◦ ? 7. Let α and β be acute angles such that cot α = 7 and csc β = cos(α + β).
√ 10. Find
9. If 3 sin x = 2, find sin(x − π) + sin(x + π). 10. Simplify: cos x + π2 + cos π2 − x .
PY
8 8. Given sin α = − 17 and sin β = − 12 , find cos(α + β) if both α and β are in QIV.
13. Given sin α =
4 5
and cos β =
C O
11. Given sin A = 45 , π2 ≤ A ≤ π, and cos B = 45 , B not in QI, find: (a) sin(A − B), (b) cos(A − B), and (c) tan(A − B). Also, determine the quadrant in which A − B terminate. √ √ 12. Given csc A = 3, A in QI, and sec B = 2, sin B < 0, find: (a) sin(A − B), (b) cos(A − B), and (c) tan(A − B). Also, determine the quadrant in which A − B terminate. 5 , 13
find sin(α + β) + sin(α − β).
EP E
D
14. Given sin α = 23 , α in QII, and cos β = 34 , find cos(α + β) + cos(α − β). √ 15. If A and B are acute angles (in degrees) such that csc A = 17 and csc B = √ 34 , what is A + B? 3 16. If tan(x + y) =
1 3
and tan y = 12 , what is tan x?
tan π9 + tan 23π 36 17. Evaluate: . π 1 − tan 9 tan 23π 36
D
18. Establish the identity:
sin(A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C.
19. Prove: sin 2θ = 2 sin θ cos θ. 20. Prove: cot 2θ =
cot2 θ − 1 . 2 cot θ
4
189 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Topic Test 1 for Unit 3 1. A central angle in a circle of radius 6 cm measures 37.5◦ . Find: (a) length of the intercepted arc and (b) the area of the sector. 2. The point (−1, −2) lies on the terminal side of the angle θ in standard position. Find sin θ + cos θ + tan θ.
tan 57◦ + tan 78◦ 4. Find the exact value of . 1 − tan 57◦ tan 78◦ 5. If sin x = a and cos x ≥ 0, express
PY
, where A is not in QI, and csc B = − 53 , where B is not in 3. Given sin A = 12 13 QIII, find: (a) cos(A − B) and (b) tan(A − B).
cos x tan x + sin x in terms of a. tan x
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6. Prove the identity cos6 x + sin6 x = 3 cos4 x − 3 cos2 x − 1.
7. A regular hexagon of side length 1 unit is inscribed in a unit circle such that two of its vertices are located on the x-axis. Determine the coordinates of the hexagon.
D
8. Determine the amplitude, period and phase shift of the graph of x π + − 1, y = 2 sin 2 3
D
EP E
and sketch its graph over one period. Find the range of the function.
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Topic Test 2 for Unit 3 1. The area of a sector of a circle formed by a central angle of 30◦ is d π3 cm2 . Find the length of the intercepted arc. 2. The point (8, −6) lies on the terminal side of the angle θ in standard position. Find (sin θ + cos θ)2 . π π 8 and cos A > 0, evaluate sin − A + cos −A . 3. Given sin A = − 17 2 2 4. Find the exact value of sin 160◦ cos 35◦ − sin 70◦ cos 55◦ . 7π . 12
PY
5. Find the exact value of tan
7. Establish the identity
24 , 7
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6. Given cos A = − 53 , where A is not in QII, and tan B = QI, find: (a) sin(A + B) and (b) cot(A + B).
where B is not in
tan2 x = sin x cos x. tan x + tan3 x
1 sin x 8. If sin x − cos x = , find . 3 sec x
x − + 2, 9. Determine the period and phase shift of the graph of y = tan 18 3 and sketch its graph over two periods.
D
EP E
D
π
191 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Lesson 3.6. Double-Angle and Half-Angle Identities Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) derive the double-angle and half-angle identities; (2) simplify trigonometric expressions using known identities; (3) prove other trigonometric identities using known identities; and (4) solve situational problems involving trigonometric identities.
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Lesson Outline
(1) The double-angle and half-angle identities for cosine, sine, and tangent (2) More trigonometric identities
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Introduction
D
Trigonometric identities simplify the computations of trigonometric expressions. In this lesson, we continue on establishing more trigonometric identities. 1 In particular, we derive the formulas for f (2θ) and f 2 θ , where f is the sine, cosine, or tangent function. 3.6.1. Double-Angle Identities
EP E
Recall the sum identities for sine and cosine. sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
D
When A = B, these identities becomes sin 2A = sin A cos A + cos A sin A = 2 sin A cos A
and
cos 2A = cos A cos A − sin A sin A = cos2 A − sin2 A. Double-Angle Identities for Sine and Cosine sin 2A = 2 sin A cos A
cos 2A = cos2 A − sin2 A
The double-identity for cosine has other forms. We use the Pythagorean identity sin2 θ + cos2 θ = 1. cos 2A = cos2 A − sin2 A 192 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
= cos2 A − (1 − cos2 A) = 2 cos2 A − 1 cos 2A = cos2 A − sin2 A = (1 − sin2 A) − sin2 A = 1 − 2 sin2 A
Other Double-Angle Identities for Cosine
Example 3.6.1. Given sin t =
3 5
and
cos 2A = 1 − 2 sin2 A π 2
PY
cos 2A = 2 cos2 A − 1
< t < π, find sin 2t and cos 2t.
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Solution. We first find cos t using the Pythagorean identity. Since t lies in QII, we have s 2 p 4 3 2 =− . cos t = − 1 − sin t = − 1 − 5 5
EP E
D
sin 2t = 2 sin t cos t 3 4 =2 − 5 5 24 =− 25
cos 2t = 1 − 2 sin2 t 2 3 =1−2 5 7 = 25
2
In the last example, we may compute cos 2t using one of the other two doubleangle identities for cosine. For the sake of answering the curious minds, we include the computations here. cos 2t = 2 cos2 t − 1 2 4 =2 − −1 5 7 = 25
D
cos 2t = cos2 t − sin2 t 2 2 3 4 − = − 5 5 7 = 25
In the three cosine double-angle identities, which formula to use depends on the convenience, what is given, and what is asked. Example 3.6.2. Derive an identity for sin 3x in terms of sin x.
193 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Solution. We use the sum identity for sine, the double-angle identities for sine and cosine, and the Pythagorean identity. sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x = (2 sin x cos x) cos x + (1 − 2 sin2 x) sin x = 2 sin x cos2 x + sin x − 2 sin3 x = 2(sin x)(1 − sin2 x) + sin x − 2 sin3 x = 3 sin x − 4 sin3 x
2
tan(A + B) =
PY
For the double-angle formula for tangent, we recall the tangent sum identity: tan A + tan B . 1 − tan A tan B
When A = B, we obtain
2 tan A tan A + tan A = . 1 − tan A tan A 1 − tan2 A
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tan(A + A) =
Tangent Double-Angle Identity 2 tan A 1 − tan2 A
D
tan 2A =
EP E
Example 3.6.3. If tan θ = − 13 and sec θ > 0, find sin 2θ, cos 2θ, and tan 2θ. Solution. We can compute immediately tan 2θ. 2 − 13 3 2 tan θ tan 2θ = =− = 2 2 1 − tan θ 4 1 − − 13
D
From the given information, we deduce that θ lies in QIV. Using one Pythagorean identity, we compute cos θ through sec θ. (We may also use the technique discussed in Lesson 3.2 by solving for x, y, and r.) Then we proceed to find cos 2θ. s 2 √ p 1 10 2 = sec θ = 1 + tan θ = 1 + − 3 3 √ 3 10 1 1 cos θ = = √ = 10 sec θ 10 3 √ !2 3 10 4 cos 2θ = 2 cos2 θ − 1 = 2 −1= 10 5 tan 2θ =
sin 2θ 3 =⇒ sin 2θ = tan 2θ cos 2θ = − cos 2θ 5
2
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3.6.2. Half-Angle Identities Recall two of the three double-angle identities for cosine: cos 2A = 2 cos2 A − 1 and
cos 2A = 1 − 2 sin2 A.
From these identities, we obtain two useful identities expressing sin2 A and cos2 A in terms of cos 2A as follows: 1 − cos 2A 1 + cos 2A and sin2 A = . cos2 A = 2 2
cos2 A =
1 + cos 2A 2
sin2 A = A , 2
and
1 − cos 2A 2
we get
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From these identities, replacing A with cos2
PY
Some Useful Identities
1 + cos 2 A = 2 2
A 2
=
1 + cos A 2
1 − cos 2 A2 1 − cos A A sin = = . 2 2 2 These are the half-angle identities for sine and cosine.
D
2
EP E
Half-Angle Identities for Sine and Cosine A 1 + cos A A 1 − cos A 2 2 cos = sin = 2 2 2 2
D
Because of the “square” in the formulas, we get r r A 1 + cos A A 1 − cos A cos = ± and sin = ± . 2 2 2 2
The appropriate signs of cos A2 and sin A2 depend on which quadrant
A 2
lies.
Example 3.6.4. Find the exact values of sin 22.5◦ and cos 22.5◦ .
Solution. Clearly, 22.5◦ lies in QI (and so sin 22.5◦ and cos 22.5◦ are both positive), and 22.5◦ is the half-angle of 45◦ . s √ p r √ 2 ◦ 1 − 1 − cos 45 2 − 2 2 sin 22.5◦ = = = 2 2 2 s √ p r √ 2 ◦ 1 + 1 + cos 45 2 + 2 2 cos 22.5◦ = = = 2 2 2 2 195 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
θ tan θ + sin θ Example 3.6.5. Prove: cos = . 2 2 tan θ 2
Solution. θ 1 + cos θ = cos 2 2 1 + cos θ tan θ = 2 tan θ tan θ + cos θ tan θ = 2 tan θ sin θ tan θ + cos θ · cos θ = 2 tan θ tan θ + sin θ = 2 tan θ
PY
2
2
tan
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We now derive the first version of the half-angle formula for tangent. sin A2 A = 2 cos A2 sin A2 = cos A2
!
2 sin2 2 2 sin A2 cos A2
D =
2 sin A2 2 sin A2 A
EP E
A 2 · 1−cos 2 sin 2 · A2 1 − cos A = sin A
=
D
There is another version of the tangent half-angle formula, and we can derive it from the first version. tan
A 1 − cos A = 2 sin A 1 − cos A 1 + cos A = sin A 1 + cos A 2 1 − cos A = (sin A)(1 + cos A) sin2 A = (sin A)(1 + cos A) sin A = 1 + cos A 196
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Tangent Half-Angle Identities tan
1 − cos A A = 2 sin A
tan
sin A2 A = 2 cos A2
A sin A = 2 1 + cos A A 1 − cos A = tan2 2 1 + cos A tan
π . Example 3.6.6. Find the exact value of tan 12
1 − cos π6 1− π tan = = π 1 12 sin 6 2
√
3 2
√ 3
PY
Solution.
=2−
2
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Example 3.6.7. If sin θ = − 25 , cot θ > 0, and 0 ≤ θ < 2π, find sin 2θ , cos 2θ , and tan 2θ . 3π . 2
It follows
D
Solution. Since sin θ < 0 and cot θ > 0, we conclude the π < θ < that θ 3π π < < , 2 2 4 which means that 2θ lies in QII.
EP E
s √ 2 p 2 21 2 cos θ = − 1 − sin θ = − 1 − − =− 5 5
θ sin = 2
r
1 − cos θ = 2
v √ u u 1 − − 21 t 5
√ 50 + 10 21 10
p
D
= 2 v √ u p r √ u 1 + − 21 t 5 θ 1 + cos θ 50 − 10 21 cos = − =− =− 2 2 2 10 √ √ 1 − − 521 θ 1 − cos θ 5 + 21 tan = = =− 2 sin θ 2 − 25
2
More Solved Examples 5 1. If cos θ = − 13 with 0 < θ < π, find sin 2θ and cos 2θ.
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Solution: In this problem, we use the Pythagorean identity sin2 θ + cos2 θ = 1. 5 , we must have Because cos θ = − 13 sin2 θ = 1 − cos2 θ = 1 −
25 144 = . 169 169
PY
Moreover, since 0 < θ < π, we take the square root of both sides of the above equation to get 12 sin θ = . 13 Now, using the double-angle identities we get sin 2θ = 2 sin θ cos θ and cos 2θ = cos2 θ − sin2 θ 5 25 = 2 12 − 13 = 169 − 144 13 169 = − 119 . 169
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120 = − 169
2. Derive an identity for cos 3x in terms of cos x.
Solution: We use the sum identity for cosine, the double-angle identities for sine and cosine, and the Pythagorean identity.
EP E
D
cos 3x = cos(2x + x) = cos 2x cos x − sin 2x sin x = (2 cos2 x − 1) cos x − (2 sin x cos x) sin x = 2 cos3 x − cos x − 2 sin2 x cos x = 2 cos3 x − cos x − 2(1 − cos2 x) cos x = 4 cos3 x − 3 cos x.
3. Derive the identity for tan 3t in terms of tan t.
D
Solution: Using the sum identity for tangent, we obtain tan 3t = tan(2t + t) =
tan 2t + tan t . 1 − tan 2t tan t
Now, using the tangent double-angle identity, we have tan 3t =
2 tan t + tan t 1−tan2 t . 2 tan t 1 − 1−tan 2 t tan t
Upon simplifying the terms on the right side of the equation, we finally obtain tan 3t =
3 tan t − tan3 t . 1 − 3 tan2 t 198
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π 4. Find the exact value of sin 12 . π Solution: To find the value of sin 12 , we use the identity sin2 21 y = With y = π6 , we obtain √ √ 3 π 1 − cos π 1 − 1 2 − π 3 6 2 = sin2 = = = . sin2 12 2 6 2 2 4 π 12
π < π, sin 12 must be positive, and so p √ π 2− 3 sin = . 12 2
PY
Now, since 0 <
1−cos y . 2
θ sec θ − 1 5. Prove: sin = . 2 2 sec θ Solution: θ 1 − cos θ 2 sin = 2 2 1 − cos θ sec θ = 2 sec θ sec θ − cos θ sec θ = 2 sec θ sec θ − cos θ · cos1 θ = 2 sec θ sec θ − 1 = . 2 sec θ
EP E
D
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2
6. Use the half-angle identity to find the exact value of tan 75◦ . √ √ 1−cos 150◦ 1+ 23 1 ◦ ◦ Solution: tan 75 = tan 2 · 150 = sin 150◦ = 1 = 2 + 3. 2
D
7. A ball is thrown following a projectile motion. It is known that the horizontal distance (range) the ball can travel is given by R=
v02 sin 2θ, g
where R is the range (in feet), v0 is the initial speed (in ft/s), θ is the angle of elevation the ball is thrown, and g = 32 f t/s2 is the acceleration due to gravity. (a) Express the new range in terms of the original range when an angle θ (0 < θ ≤ 45◦ ) is doubled? (b) If a ball travels a horizontal√distance of 20 ft when kicked at an angle of α with initial speed of 20 2 ft/s, find the horizontal distance it can travel when you double α. Hint: Use the result of item (a) 199 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Solution: (a) Let R =
v02 g
sin 2θ be the original range. When the angle is doubled, the v2
new range will become R0 = g0 sin 4θ. Now, we solve sin 4θ in terms of the original range. Note that sin 2θ = gR . So, as a consequence of the fundamental identity, v02 we obtain s p v04 − g 2 R2 g 2 R2 . cos 2θ = 1 − 4 = v0 v02
PY
Since sin 4θ = 2 sin 2θ cos 2θ, it follows that ! p p 2 2 4 2 R2 − g v 2R v04 − g 2 R2 v v gR 0 0 0 0 R = sin 4θ = 2· 2 · = . g g v0 v02 v02
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(b) Using the result in (a), if α is doubled, then the new range is given by p √ 2R v04 − g 2 R2 40 640000 − 409600 0 R = = = 24. v02 800
D
Therefore, the new horizontal distance is 24 ft.
EP E
Supplementary Problems 3.6 1. Let θ be an angle in the first quadrant and sin θ = 13 . Find (a) sin 2θ
(c) tan 2θ
(e) csc 2θ
(b) cos 2θ
(d) sec 2θ
(f) cot 2θ
D
2. Find the approximate value of csc 46◦ and sec 46◦ , given that sin 23◦ ≈ 0.3907. 3. If cos t = 43 , what is cos 2t?
4. Derive a formula for sin 4x in terms of sin x and cos x. 5. Let − π4 < x < 0. Given that tan 2x = −2, solve for tan x. 6. Obtain an identity for tan 4θ in terms of tan θ. 7. Solve for the exact value of cot 4θ if tan θ = 12 . 8. Use half-angle identities to find the exact value of (a) sin2 15◦ and (b) cos2 15◦ . 9. Use half-angle identities to find the exact value of (a) sin2
5π 8
and (b) cos2
5π . 8
200 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
10. Find the exact value of cos π8 . 11. Prove that
tan 12 y − 1 sin y − cos y − 1 = . 1 sin y + cos y + 1 tan 2 y + 1
12. Verify that the following equation is an identity: cot 21 t = cot t(sec t + 1). 13. Use half-angle identities to find the exact value of (a) cos 105◦ and (b) tan 22.5◦ .
PY
4
Lesson 3.7. Inverse Trigonometric Functions
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Learning Outcomes of the Lesson
At the end of the lesson, the student is able to:
(1) graph the six basic inverse trigonometric functions;
(2) illustrate the domain and range of the inverse trigonometric functions; (3) evaluate inverse trigonometric expressions; and
EP E
Lesson Outline
D
(4) solve situational problems involving inverse trigonometric functions.
(1) Definitions of the six inverse trigonometric functions (2) Graphs of inverse trigonometric functions (3) Domain and range of inverse trigonometric functions (4) Evaluation of inverse trigonometric expressions
D
Introduction
In the previous lessons on functions (algebraic and trigonometric), we computed for the value of a function at a number in its domain. Now, given a value in the range of the function, we reverse this process by finding a number in the domain whose function value is the given one. Observe that, in this process, the function involved may or may not give a unique number in the domain. For example, each of the functions f (x) = x2 and g(x) = cos x do not give a unique number in their respective domains for some values of each function. Given f (x) = 1, the function gives x = ±1. If g(x) = 1, then x = 2kπ, where k is an integer. Because of this possibility, in order for the reverse process to produce a function, we restrict this process to one-to-one functions or at least restrict the domain of a non-one-to-one function to make it one-to-one so that the process 201
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works. Loosely speaking, a function that reverses what a given function f does is called its inverse function, and is usually denoted by f −1 . More formally, two functions f and g are inverse functions if g(f (x)) = x for any x in the domain of f , and f (g(x)) = x for any x in the domain of g. We denote the inverse function of a function f by f −1 . The graphs of a function and its inverse function are symmetric with respect to the line y = x.
PY
In this lesson, we first restrict the domain of each trigonometric function because each of them is not one-to-one. We then define each respective inverse function and evaluate the values of each inverse trigonometric function.
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3.7.1. Inverse Sine Function
D
All the trigonometric functions that we consider are periodic over their entire domains. This means that all trigonometric functions are not one-to-one if we consider their whole domains, which implies that they have no inverses over those sets. But there is a way to make each of the trigonometric functions one-to-one. This is done by restricting their respective domains. The restrictions will give us well-defined inverse trigonometric functions.
EP E
The domain of the sine function is the set R of real numbers, and its range is the closed interval [−1, 1]. As observed in the previous lessons, the sine function is not one-to-one, and the first step is to restrict its domain (by agreeing what the convention is) with the following conditions: (1) the sine function is one-to-one in that restricted domain, and (2) the range remains the same.
D
The inverse of the (restricted) sine function π f π(x) = sin x, where the domain is restricted to the closed interval − 2 , 2 , is called the inverse sine function or arcsine function, denoted by f −1 (x) = sin−1 x or f −1 (x) = arcsin x. Here, the domain of f −1 (x) = arcsin x is [−1, 1], and its range is − π2 , π2 . Thus, y = sin−1 x or y = arcsin x if and only if sin y = x, where −1 ≤ x ≤ 1 and − π2 ≤ y ≤ π2 .
Throughout the lesson, we interchangeably use sin−1 x and arcsin x to mean the inverse sine function. 202 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Example 3.7.1. Find the exact value of each expression. (3) arcsin 0 (1) sin−1 21 (2) arcsin(−1) (4) sin−1 − 12 Solution. (1) Let θ = sin−1 21 . This is equivalent to sin θ = 12 . This means that π π we are looking for the number θ in the closed interval − 2 , 2 whose sine is 1 . We get θ = π6 . Thus, we have sin−1 21 = π6 . 2 (2) arcsin(−1) = − π2 because sin − π2 = −1 and − π2 ∈ − π2 , π2 .
PY
(3) arcsin 0 = 0 (4) sin−1 − 12 = − π6
2
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As emphasized ≤ 1, sin−1 x is that π π in the last example, as long as −1 ≤ x−1 number y ∈ − 2 , 2 such that sin y = x. If |x| > 1, then sin x is not defined in R. We can sometimes find the exact value of sin−1 x (that is, we can find a value in terms of π), but if no such special value exists, then we leave it in the form sin−1 x. For example, as shown above, sin−1 12 is equal to π6 . However, as studied in Lesson 3.2, no special number θ satisfies sin θ = 23 , so we leave sin−1 32 as is.
EP E
D
Example 3.7.2. Find the exact value of each expression. (3) arcsin(sin π) (1) sin sin−1 12 π (2) arcsin sin 3 (4) sin sin−1 − 12 Solution. (1) sin sin−1 12 = sin π6 = √ (2) arcsin sin π3 = arcsin 23 = π3
1 2
(3) arcsin(sin π) = arcsin 0 = 0 (4) sin sin−1 − 12 = sin − π6 = − 12
D
2
From the last example, we have the following observations:
1. sin(arcsin x) = x for any x ∈ [−1, 1]; and 2. arcsin(sin θ) = θ if and only if θ ∈ − π2 , π2 , and if θ 6∈ − π2 , π2 , then arcsin(sin θ) = ϕ, where ϕ ∈ − π2 , π2 such that sin ϕ = sin θ. To sketch the graph of y = sin−1 x, Table 3.32 presents the tables of values for y = sin x and y = sin−1 x. Recall that the graphs of y = sin x and y = sin−1 x are symmetric with respect to the line y = x. This means that if a point (a, b) is on y = sin x, then (b, a) is on y = sin−1 x. 203 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
y = sin x
x
− π2
− π3
y
−1
−
− π4
√
√
3 2
−
2 2
√
y = sin−1 x
3 2
x
−1
−
y
− π2
− π3
√
−
2 2
− π4
− π6
0
π 6
− 12
0
1 2
− 21
0
1 2
− π6
0
π 6
π 4 √ 2 2
π 3 √ 3 2
√
√
2 2
π 4
π 2
1
3 2
1
π 3
π 2
Table 3.32
EP E
D
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PY
The graph (solid thick curve) of the restricted sine function y = sin x is shown in Figure 3.33(a), while the graph of inverse sine function y = arcsin x is shown in Figure 3.33(b).
(b) y = sin−1 x
D
(a) y = sin x
Figure 3.33
Example 3.7.3. Sketch the graph of y = sin−1 (x + 1). Solution 1. In this solution, we use translation of graphs. Because y = sin−1 (x + 1) is equivalent to y = sin−1 [x − (−1)], the graph of y = sin−1 (x + 1) is 1-unit to the left of y = sin−1 x. The graph below shows y = sin−1 (x + 1) (solid line) and y = sin−1 x (dashed line).
204 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
PY
Solution 2. In this solution, we graph first the corresponding sine function, and then use the symmetry with respect to y = x to graph the inverse function.
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y = sin−1 (x + 1) ⇐⇒ sin y = x + 1 ⇐⇒ x = sin y − 1
D
EP E
D
The graph below shows the process of graphing of y = sin−1 (x + 1) from y = sin x − 1 with − π2 ≤ x ≤ π2 , and then reflecting it with respect to y = x.
3.7.2. Inverse Cosine Function The development of the other inverse trigonometric functions is similar to that of the inverse sine function.
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y = cos−1 x or y = arccos x means cos y = x, where −1 ≤ x ≤ 1 and 0 ≤ y ≤ π.
D
C O
PY
The graph (solid thick curve) of the restricted cosine function y = cos x is shown in Figure 3.34(a), while the graph of inverse cosine function y = arccos x is shown in Figure 3.34(b).
(b) y = cos−1 x
EP E
(a) y = cos x
Figure 3.34
D
Example 3.7.4. Find the exact value of each expression. (1) cos−1 0 (4) cos−1 cos 3π 4 √ (2) arccos − 23 (5) arccos cos 7π 6 √ (3) cos cos−1 − 23
Solution. (1) cos−1 0 = √ (2) arccos − 23 = 5π 6
π 2
(6) sin cos−1 because cos π2 = 0 and
π 2
√
2 2
∈ [0, π].
√ √ √ (3) cos cos−1 − 23 = − 23 because − 23 ∈ [−1, 1] (4) cos−1 cos 3π = 4
3π 4
because
3π 4
∈ [0, π].
206 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
√ = arccos − 23 = (5) arccos cos 7π 6 (6) sin cos−1
√
2 2
5π 6
√
=
2
2 2
Example 3.7.5. Simplify: sin arcsin 23 + arccos 21 .
PY
Solution. We know that arccos 12 = π3 . Using the Sine Sum Identity, we have sin arcsin 32 + arccos 12 = sin arcsin 23 + π3 = sin arcsin 23 cos π3 + cos arcsin 23 sin π3 √ = 23 · 21 + cos arcsin 23 · 23 √ = 13 + 23 cos arcsin 23 .
C O
We compute cos arcsin 23 . Let θ = arcsin 23 . By definition, sin θ = 23 , where θ lies in QI. Using the Pythagorean identity, we have p √ cos arcsin 32 = cos θ = 1 − sin2 θ = 35 .
D
Going back to the original computations above, we have √ sin arcsin 32 + arccos 12 = 13 + 23 cos arcsin 32
EP E
=
=
Example 3.7.6. Simplify: sin 2 cos−1 − 45
√ 3 1 + 3 2 √ 2+ 15 . 6
√
·
5 3
2
.
Solution. Let θ = cos−1 − 45 . Then cos θ = − 54 . Because cos θ < 0 and range of inverse cosine function is [0, π], we know that θ must be within the interval π , π . Using the Pythagorean Identity, we get sin θ = 53 . 2
D
Using the Sine Double-Angle Identity, we have sin 2 cos−1 − 54 = sin 2θ = 2 sin θ cos θ = 2 · 35 − 54 = − 24 . 25
2
Example 3.7.7. Sketch the graph of y = 41 cos−1 (2x). Solution. y=
1 1 cos−1 (2x) ⇐⇒ 4y = cos−1 (2x) ⇐⇒ x = cos(4y) 4 2 207
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C O
PY
We graph first y = 12 cos(4x). The domain of this graph comes from the restriction of cosine as follows: π 0 ≤ 4x ≤ π =⇒ 0 ≤ x ≤ . 4 Then reflect this graph with respect to y = x, and we finally obtain the graph of y = 41 cos−1 (2x) (solid line).
D
EP E
D
In the last example, we may also use the following technique. In graphing y = 14 cos−1 (2x), the horizontal length of cos−1 x is reduced to half, while the vertical height is reduced to quarter. This comparison technique is shown in the graph below with the graph of y = cos−1 x in dashed line and the graph of y = 41 cos−1 (2x) in solid line.
3.7.3. Inverse Tangent Function and the Remaining Inverse Trigonometric Functions The inverse tangent function is similarly defined as inverse sine and inverse cosine functions.
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y = tan−1 x or y = arctan x means tan y = x, where x ∈ R and − π2 < y < π2 .
D
C O
PY
The graph (solid thick curve) of the restricted function y = tan x is shown in Figure 3.35(a), while the graph of inverse function y = arctan x is shown in Figure 3.35(b).
(b) y = tan−1 x
EP E
(a) y = tan x
Figure 3.35
D
Example 3.7.8. Find the exact value of each expression. (1) tan−1 1 (4) tan−1 tan − π6 √ (2) arctan − 3 (5) tan−1 tan 7π 6 (3) tan tan−1 − 52 (6) arctan tan − 19π 6
Solution. Note the range of arctan is the open interval − π2 , π2 . (1) tan−1 1 =
π 4
√ (2) arctan − 3 = − π3 (3) tan tan−1 − 52 = − 25 (4) tan−1 tan − π6 = − π6 because − π6 ∈ − π2 , π2 .
209 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(5) Here, note that = tan π6 . tan 7π 6
7π 6
6∈ − π2 , π2 . Use the idea of reference angle, we know that
−1
tan
7π π π tan = tan−1 tan = 6 6 6
PY
(6) Here, we cannot use the idea of reference angle, but the idea can help in a way. The number (or angle) − 19π is in QII, wherein tangent is negative, and 6 its reference angle is π6 . π 19π arctan tan − = arctan tan − 6 6 π 2 =− 6
C O
Example 3.7.9. Findthe exact value of each expression. (1) sin 2 tan−1 − 38 (2) tan sin−1 53 − tan−1 14
Solution. (1) Let θ = tan−1 − 38 . Then tan θ = − 83 . Following the notations in Lesson 3.2 and the definition of inverse tangent know that θ lies p function, we √ in QIV, and x = 3 and y = −8. We get r = 32 + (−8)2 = 73.
D
EP E
D
Applying the Sine Double-Angle Identity (page 192) gives 8 −1 = sin 2θ sin 2 tan − 3 = 2 sin θ cos θ y x =2· · r r 8 3 √ = 2 −√ 73 73 48 =− . 73
(2) Using the Tangent Difference Identity, we obtain −1 3 −1 1 tan sin − tan 5 4 −1 3 tan sin 5 − tan tan−1 41 = 1 + tan sin−1 35 tan tan−1 41 tan sin−1 53 − 14 . = 1 + tan sin−1 35 · 41 We are left to compute tan sin−1 35 . We proceed as in (1) above. Let θ = sin−1 35 . Then sin θ = 53 . From the definition of inverse sine function and 210
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the notations used √ in Lesson 3.2, we know that θ lies in QI, and y = 3 and r = 5. We get x = 52 − 32 = 4, so that tan θ = xy = 34 . 1 −1 3 −4 tan sin 3 1 5 tan sin−1 − tan−1 = −1 3 5 4 1 + tan sin 5 · 14 − 41 1 + 34 · 14 8 = 19
2
C O
D
? Example 3.7.10. A student is viewing a painting in a museum. Standing 6 ft from the painting, the eye level of the student is 5 ft above the ground. If the painting is 10 ft tall, and its base is 4 ft above the ground, find the viewing angle subtended by the painting at the eyes of the student.
3 4
PY
=
EP E
Solution. Let θ be the viewing angle, and let θ = α + β as shown below. We observe that tan α =
1 6
and
9 tan β = . 6
Using the Tangent Sum Identity, we have tan α + tan β 1 − tan α tan β 1 +9 = 6 169 1− 6 · 6 20 = . 9 Using a calculator, the viewing angle is θ = tan−1
D
tan θ = tan(α + β) =
20 9
≈ 65.8◦ .
2
We now define the remaining inverse trigonometric functions.
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Define
π − tan−1 x. 2 It follows that the domain of y = cot−1 x is R and its range is (0, π). cot−1 x =
y = sec−1 x or y = arcsec x
where |x| ≥ 1 and y ∈ 0,
π 2
sec y = x, ∪ π, 3π . 2
Define
PY
means
π − sec−1 x. 2 This means that thedomainof y = csc−1 x is (−∞, −1] ∪ [1, ∞) and its range is −π, − π2 ∪ 0, π2 .
C O
csc−1 x =
D
EP E
D
The graphs of these last three inverse trigonometric functions are shown in Figures 3.36, 3.37, and 3.38, respectively.
(b) y = cot−1 x
(a) y = cot x Figure 3.36
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PY (b) y = sec−1 x
(a) y = sec x
EP E
D
C O
Figure 3.37
(b) y = csc−1 x
(a) y = csc x
D
Figure 3.38
Observe that the process in getting the value of an inverse function is the same to all inverse functions. That is, y = f −1 (x) is the same as f (y) = x. We need to remember the range of each inverse trigonometric function. Table 3.39 summarizes all the information about the six inverse trigonometric functions.
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Function
Domain
Range
Graph
sin−1 x
[−1, 1]
− π2 , π2
Figure 3.33(b)
cos−1 x
[−1, 1]
[0, π]
tan−1 x
R
− π2 , π2
cot−1 x
R
(0, π)
sec−1 x
{x : |x| ≥ 1}
π 3π 0, 2 ∪ π, 2
Figure 3.37(b)
csc−1 x
{x : |x| ≥ 1}
−π, − π2 ∪ 0, π2
Figure 3.38(b)
Figure 3.34(b) Figure 3.35(b)
Figure 3.36(b)
C O
PY
Table 3.39
EP E
D
Example 3.7.11. Find the exact value of each expression. √ −1 (1) sec−1 (−2) − 3 (3) cot √ √ 2 3 −1 −1 − 3 (2) csc − 32 − csc−1 − 2 3 3 (4) sin sec Solution. (1) sec−1 (−2) = √ (2) csc−1 − 2 3 3 = − 2π 3 √ (3) cot−1 − 3 =
4π 3
because sec 4π = −2 and 3
4π 3
∈ π, 3π 2
5π 6
√ 2 3 − 3
3 −1 = − 2π . Let θ = sec − . Then 3 2
D
(4) From (2), we know that csc
−1
sec θ = − 32 . From defined range of inverse secant function and the notations in Lesson and r = 3 and x = −2.√ Solving for y, we get p 3.2, θ lies in QIII, √ y = − 32 − (−2)2 = − 5. It follows that sin θ = − 35 and cos θ = − 23 . We now use the Sine Sum Identity. 3 −1 sin sec − − csc−1 2 2π = sin θ − − 3 2π = sin θ + 3
√ !! 2 3 − 3
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2π 2π = sin θ cos + cos θ sin 3 3 √ ! 5 1 2 − + − = − 3 2 3 √ √ 5−2 3 = 6
√ ! 3 2 2
PY
More Solved Examples 1. Find the exact values of the following, if they exist. √ (b) arcsin − 21 (c) sin−1 2 (a) sin−1 22
(a) sin−1
√
2 2
= π4 , (b) arcsin − 12 = − π6 , and
C O
Solution: Note that the range of f (x) = sin−1 x is [− π2 , π2 ]. Thus, if we let y = sin−1 x, then we are looking for y ∈ [− π2 , π2 ] such that sin y = x. Hence,
D
(c) sin−1 2 is undefined because sin y ≤ 1.
EP E
2. Find the exact value of each expression. √ (b) cos arcsin − 12 (a) sin sin−1 22
(c) sin−1 sin 11π 2
D
Solution: √ √ (a) sin sin−1 22 = sin( π4 ) = 22 √ (b) cos arcsin − 12 = cos(− π6 ) = 23 (c) sin−1 sin 11π = sin−1 (−1) = − π2 2
3. Answer the following. (a) What is the domain of y = sin−1 2x? (b) What is the range of y = sin−1 2x? (c) What is the x−intercept of y = sin−1 2x? Solution: (a) Consider the function f (θ) = sin−1 θ. The domain of sin−1 θ is [−1, 1]. So, θ = 2x ∈ [−1, 1]. Therefore, the domain of sin−1 2x is [−1/2, 1/2]. 215
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(b) [−π/2, π/2] (c) (0, 0) 4. From the concept of projectile motion, if an object is directed at an angle θ v2 (with θ ∈ [0, π/2]), then the range will be R = g0 sin 2θ (in feet) where v0 (in f t/s) is the initial speed and g = 32 f t/s2 is the acceleration due to gravity. At what angle shall the object be directed so that the range will be 100 f t, given that the initial speed is v0 = 80 f t/s? Solution: From the formula of the range, we get 1 802 sin 2θ =⇒ = sin 2θ 32 2
PY
100 =
Since θ must be from 0 to π2 (i.e. 0 ≤ 2θ ≤ π), this is equivalent to finding 2θ such that 2θ = sin−1 12 . Hence, π π =⇒ θ = . 6 12
C O
2θ =
Therefore, the object must be directed at an angle of a projectile range of 100 f t.
π 12
rad (or 15◦ ), to have
(a) cos−1
√
2 2 −1
1
−2
EP E
(b) cos cos
D
5. Find the exact values of the following, if they exist.
(c) arccos(cos π) (d) arccos π
Solution:
(a) cos−1
(b) cos
√
2 = π4 2 cos−1 − 12
= cos 2π = − 21 3
D
(c) arccos(cos π) = arccos(−1) = π
(d) Let y = arccos π. Since cos y ≤ 1, we have y is undefined because π > 3. √ −1 3 −1 1 6. Simplify: (a) cos cos − cos 3 2 √
Solution: We know that cos−1 23 = π6 . Let θ = cos−1 13 . Which is equivalent to cos θ = 31 with 0 ≤ θ ≤ π. Using the Cosine Difference Identity, we have ! √ π π 3 1 π cos cos−1 − cos−1 = cos − θ = cos cos θ + sin sin θ 2 3 6 6 6 √ 3 1 1 = · + · sin θ. 2 3 2 216 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Now, we solve for sin θ using the Pythagorean Identity which gives sin2 θ = 1 − (1/3)2 = 1 − (1/9) = 8/9. √ 2 2 3
because θ ∈ [0, π]. Finally, we obtain ! √ √ √ 1 3 3 1 1 2 2 −1 −1 − cos = · + · cos cos 2 3 2 3 2 3 √ √ 3 2 2 + = √6 √6 3+2 2 = . 6 7. Simplify: (a) cos 2 cos−1 52 ; (b) sin cos−1 25
PY
Thus, sin θ =
C O
Solution: Let θ = cos−1 52 . Which is equivalent to cos θ = 25 with 0 ≤ θ ≤ π. Using the Double-Angle Identity for Cosine and one of the Fundamental Idenity, we have 2 2 17 8 −1 2 2 −1=− cos 2 cos = cos(2θ) = 2 cos θ − 1 = 2 −1= 5 5 25 25 and sin cos−1
2 5
√
r
D
= sin θ =
1 − cos2 θ =
4 = 1− 25
√
21 . 5
EP E
Here, sin θ ≥ 0 because θ ∈ [0, π]. 8. Graph: y = 1 + cos−1 x
D
Solution: The graph can be obtained by translating the graph of the inverse cosine function one unit upward.
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9. Find the exact value of each expression. (b) tan(tan−1 54 )
) (a) arctan(tan 4π 3 Solution: (a) arctan(tan 4π ) = arctan 3 (b) tan(tan−1 45 ) =
√ 3 = π/3
4 5
Solution:
PY
10. Find the exact value of tan(tan−1 76 + tan−1 12 ). tan(tan−1 76 ) + tan(tan−1 12 ) −1 7 −1 1 tan tan + tan = 6 2 1 − tan(tan−1 76 ) · tan(tan−1 12 ) + 12 1 − 76 · 12 7 6
C O =
=4
D
11. Find the exact values of the following, if they exist. √ √ (a) sec−1 2 (e) cos(arccsc−1 2) (c) cot−1 33 (d) arcsec−1 (cot(− π4 ))
EP E
(b) csc−1 1
(f) arccot−1 (sin 20π ) 3
Solution:
(a) sec−1
√
2=
(b) csc−1 1 = √
3 3
D
(c) cot−1
π 4
π 2
=
π 3
(d) arcsec(cot(− π4 )) = arcsec(−1) = −π (e) cos(arccsc(2)) = cos π6 =
√
√
3 2
√
(f) arccot(sin 20π ) = arccot 23 . Let θ = arccot 23 . Then, 3 √ 3 2 cot θ = =⇒ tan θ = √ 2 3 2 =⇒ θ = tan−1 √ (≈ 0.8571). 3 Here, we needed to use a calculator to solve for the approximate value, since √23 is not a special value for tangent function. 218 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Supplementary Problems 3.7 1. Find the exact value of the following. (a) (b) (c) (d)
√ (e) cos[arccos( 2)]
sin[sin−1 (1/2)] √ cos[cos−1 (− 2/2)] √ tan[tan−1 (− 3)] √ sin[arctan( 3)]
(f) tan[arcsin(1/4)] √ (g) cos[sin−1 ( 3/2)]
PY
2. Find the exact value of the following. (a) sin−1 [sin(25π/6)]
(c) tan−1 [tan(−1)]
(e) cos−1 [sec(23π)]
(b) arccos[cos(23π/4)]
(d) arcsin[cos(13π/4)]
(f) arctan[sin(−π/12)]
(a) sin[2 cos−1 (−4/5)] (b) cos[2 sin−1 (5/13)]
C O
3. Solve the exact value of the following.
(c) sin(sin−1 (3/5) + cos−1 (−5/13))
D
(d) cos[sin−1 (1/2) − cos−1 (8/17)]
4. Consider the function f (x) = tan−1 (x + 1). Do the following.
EP E
(a) Find the domain of f .
(b) Find the range of f .
(c) Find the x− and y−intercept of f , if there are any.
(d) Graph f .
D
5. Evaluate and simplify the following, if they exist. √ (a) arcsec(− 2) (d) [sec−1 (−1)] · [cos−1 (−1)] √ (b) arccsc(−2) (e) 2 cot−1 3 + 3 csc−1 2 √ (c) arccot 3 (f) csc−1 0 6. Evaluate and simplify the following, if they exist. −1 2 sin (1/2) cos(sec−1 3 + tan−1 2) (c) cos (a) 2 cos(tan−1 2) (b) tan(2 arcsin(1/6)) (d) arcsec(sin(100π/3))
219 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
C O
PY
7. A trough is in the shape of an inverted triangular prism whose cross section has the shape of an inverted isosceles√triangle (see Figure 3.40). If the length of the √ base of the cross section is 2 3 m. and the length of the trough is 100 3 m., find the size of the vertex angle so that the volume is 900 m3 . Hint: V = bhl/2.
D
Figure 3.40
EP E
4
Lesson 3.8. Trigonometric Equations Learning Outcomes of the Lesson At the end of the lesson, the student is able to:
D
(1) solve trigonometric equations; and (2) solve situational problems involving trigonometric equations. Lesson Outline (1) Definition of a trigonometric equation (2) Solution to a trigonometric equation (3) Techniques of solving a trigonometric equation Introduction We have studied equations in Lesson 3.4. We differentiated an identity from a conditional equation. Recall that an identity is an equation that is true for all 220
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values of the variable in the domain of the equation, while a conditional equation is an equation that is not an identity. In this lesson, we mostly study conditional trigonometric equations. Though not explicitly, we have started it in the preceding lesson. For example, the equation x = 12 has the unique solution x = sin−1 12 = π6 in the closed interval π sin − 2 , π2 . However, if we consider the entire domain (not the restricted domain) of the sine function, which is the set R of real numbers, there are solutions (other than π6 ) of the equation sin x = 12 . This current lesson explores the techniques of solving (conditional) trigonometric equations.
PY
We divide the lesson into two groups of equations: the ones using a basic way of solving, and those using more advanced techniques. 3.8.1. Solutions of a Trigonometric Equation
C O
Any equation that involves trigonometric expressions is called a trigonometric equation. Recall that a solution or a root of an equation is a number in the domain of the equation that, when substituted to the variable, makes the equation true. The set of all solutions of an equation is called the solution set of the equation.
EP E
D
Technically, the basic method to show that a particular number is a solution of an equation is to substitute the number to the variable and see if the equation becomes true. However, we may use our knowledge gained from the previous lessons to do a quicker verification process by not doing the manual substitution and checking. We use this technique in the example. , 3π , 5π , π, 2π are Example 3.8.1. Which numbers in the set 0, π6 , π4 , π3 , π2 , 2π 3 4 6 solutions to the following equations? (1) sin x = 12 (7) cos2 x = cos 2x + sin2 x (2) tan x = 1
(8) sin x + cos 2x = 0
(10) sin2 x + cos2 x = 2
(5) sec2 x − tan2 x = 1
(11) sin 2x = sin x
(6) sin x + cos x = 0
(12) 2 tan x + 4 sin x = 2 + sec x
D
√ (3) 3 sec x = −2 3 √ (4) 3| cot x| = 1
(9) 2 sin x + tan x − 2 cos x = 2
Solution. Note that the choices (except 2π) are numbers within the interval [0, π]. To quickly determine which numbers among the choices are solutions to a particular equation, we use some distinctive properties of the possible solutions. (1) The sine function is positive on (0, π). From Lesson 3.2, we recall that π6 is an obvious solution. We may imagine the graph of y = sin x. We may also use the idea of reference angle. Thus, among the choices, only π6 and 5π are 6 1 the only solutions of sin x = 2 . 221 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(2) Since tan x = 1 > 0, any solution of the equation among the choices must π be in the interval 0, 2 (that is, in QI). Again, among the choices, the only solution to tan x = 1 is π4 . √
(3) Here, the given equation is equivalent to sec x = − 2 3 3 . Among the choices, √ the only solution of the equation 3 sec x = −2 3 is 5π . 6
PY
(4) Eliminating the absolute value sign, the given equation is equivalent to√cot x = √ √ 3 3 or cot x = − 3 . Among the choices, the only solution of cot x = 33 is π3 , 3 √ 3| cot x| = 1 while the other equation has 2π . Thus, the only solutions of 3 from the given set are π3 and 2π . 3
C O
(5) The given equation is one of the Pythagorean Identities (page 175). It means that all numbers in the domain of the equation are solutions. The domain of the equation is R \ {x : cos x = 0}. Thus, all except π2 are solutions of sec2 x − tan2 x = 1. (6) For the sum of sin x and cos x to be 0, they must have equal absolute values satisfies these properties, and but different signs. Among the choices, only 3π 4 it is the only solution of sin x + cos x = 0.
D
(7) This equations is one of the Double-Angle Identities for Cosine. This means that all numbers in the domain of the equation are its solutions. Because the domain of the given equation is R, all numbers in the given set are solutions of cos2 x = cos 2x + sin2 x.
EP E
(8) We substitute each number in the choices to the expression on the left-side of the equation, and select those numbers that give resulting values equal to 1. x = 0: sin 0 + cos 2(0) = 0 + 1 = 1 x = π6 : sin π6 + cos 2( π6 ) =
D
x = π4 : sin π4 + cos 2( π4 ) = x = π3 : sin π3 + cos 2( π3 ) =
1 + 12 = 1 2 √ √ 2 2 + 0 = 2 2 √ √ 3 3−1 1 − = 2 2 2
x = π2 : sin π2 + cos 2( π2 ) = 1 − 1 = 0 x=
2π : 3
sin 2π + cos 2( 2π )= 3 3
x=
3π : 4 5π : 6
sin 3π + cos 2( 3π )= 4 4
x=
sin 5π + cos 2( 5π )= 6 6
√
√ 3 3−1 1 − = 2 2 2 √ √ 2 2 +0= 2 2 1 + 12 = 1 2
x = π: sin π + cos 2π = 0 + 1 = 1 x = 2π: sin 2π + cos 2(2π) = 0 + 1 = 1
222 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
From these values, the only solution of sin x + cos 2x = 0 among the choices is π2 . (9) We again substitute the numbers in the given set one by one, and see which resulting values are equal to 1. x = 0: 2 sin 0 + tan 0 − 2 cos 0 = −2 x = π6 : 2 sin π6 + tan π6 − 2 cos π6 =
√ 3−2 3 3
x = π4 : 2 sin π4 + tan π4 − 2 cos π4 = 1 √ x = π3 : 2 sin π3 + tan π3 − 2 cos π3 = 2 3 − 1
2π : 3
2 sin 2π + tan 2π − 2 cos 2π =1 3 3 3 √ : 2 sin 3π + tan 3π − 2 cos 3π =2 2−1 x = 3π 4 4 4 4 x=
5π : 6
√ 3+2 3 3
C O
x=
PY
x = π2 : Since tan π2 is undefined, this value of x cannot be a solution of the equation.
2 sin 5π + tan 5π − 2 cos 5π = 6 6 6
x = π: 2 sin π + tan π − 2 cos π = 2
x = 2π: 2 sin 2π + tan 2π − 2 cos 2π = −2
D
Thus, the only solution of 2 sin x + tan x − 2 cos x = 2 from the given set is π.
EP E
(10) This equation has no solution because one of the Pythagorean Identities says sin2 x + cos2 x = 1. (11) We substitute each number in the given set to the expression of each side of the equation, and see which resulting values are equal. x = 0: sin 2(0) = 0; sin 0 = 0 x = π6 : sin 2( π6 ) =
√
3 ; 2
sin π6 =
D
x = π4 : sin 2( π4 ) = 1; sin π4 = x = π3 : sin 2( π3 ) =
√
3 ; 2
√
sin π3 =
1 2
2 2 √
3 2
x = π2 : sin 2( π2 ) = 0; sin π2 = 1 x=
3π : 4
sin 2( 3π ) = −1; sin 3π = 4 4
x=
5π : 6
sin 2( 5π )=− 6
√
3 ; 2
sin π3 =
√
2 2 1 2
x = π: sin 2π = 0; sin π = 0 x = 2π: sin 2(2π) = 0; sin 2π = 0 Thus, among the numbers in the given set, the solutions of sin 2x = sin x are 0, π3 , π, and 2π. 223 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
(12) We employ the same technique used in the previous item. x=0:
2 tan 0 + 4 sin 0 = 0 2 + sec 0 = 3
π 6
:
2 tan π6 + 4 sin π6 = 2 + sec π6 =
x=
π 4
:
x=
π 3
:
√ 2 3+6 3
√ 2 3+6 3
√ 2 tan π4 + 4 sin π4 = 2 2 + 2 √ 2 + sec π4 = 2 + 2 √ 2 tan π3 + 4 sin π3 = 4 3 2 + sec π3 = 4
π 2
x=
2π 3
:
Both tan π2 and sec π2 are undefined. : 2 tan 2π + 4 sin 2π =0 3 3 2 + sec 2π =0 3
3π 4
x=
5π 6
√ : 2 tan 3π + 4 sin 3π =2 2−2 4 4 √ 2 + sec 3π = 2 − 2 4 : 2 tan 5π + 4 sin 5π = 6 6 2 + sec 5π = 6
2 tan π + 4 sin π = 0
EP E
x=π:
√ 6−2 3 3
√ 6−2 3 3
D
x=
C O
x=
PY
x=
2 + sec π = 1
x = 2π : 2 tan 2π + 4 sin 2π = 0 2 + sec 2π = 3
D
After checking the equal values, the solutions of 2 tan x + 4 sin x = 2 + sec x among the given choices are π6 , 2π , and 5π . 2 3 6
3.8.2. Equations with One Term From the preceding discussion, you may observe that there may be more solutions of a given equation outside the given set. We now find all solutions of a given equation. We will start with a group of equations having straightforward techniques in finding their solutions. These simple techniques involve at least one of the following ideas:
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(1) equivalent equations (that is, equations that have the same solutions as the original equation); (2) periodicity of the trigonometric function involved; (3) inverse trigonometric function; (4) values of the trigonometric function involved on the interval [0, π] or [0, 2π] (depending on the periodicity of the function); and (5) Zero-Factor Law: ab = 0 if and only if a = 0 or b = 0.
PY
To “solve an equation” means to find all solutions of the equation. Here, unless stated as angles measured in degrees, we mean solutions of the equation that are real numbers (or equivalently, angles measured in radians).
C O
Example 3.8.2. Solve the equation 2 cos x − 1 = 0. Solution. The given equation is equivalent to 1 cos x = . 2
D
On the interval [0, 2π], there are only two solutions of the last equation, and these are x = π3 (this is in QI) and x = 5π (in QIV). 3
EP E
Because the period of cosine function is 2π, the complete solutions of the equation are x = π3 + k(2π) and x = 5π + k(2π) for all integers k. 2 3 In the preceding example, by saying that the “complete solutions are x = + k(2π) and x = 5π + k(2π) for all integers k,” we mean that any integral 3 value of k will produce a solution to the given equation. For example, when is a solution of the equation. When k = −2, k = 3, x = π3 + 3(2π) = 19π 3 5π 7π x = 3 + (−2)(2π) = − 3 is another solution of 2 cos x − 1 = 0. The family of solutions x = π3 + k(2π) can be equivalently enumerated as x = 19π + 2kπ, while 3 7π the family x = 5π + k(2π) can also be stated as x = − + 2kπ. 3 3
D
π 3
Example 3.8.3. Solve: (1 + cos θ)(tan θ − 1) = 0. Solution. By the Zero-Factor Law, the given equation is equivalent to 1 + cos θ = 0
or
tan θ − 1 = 0
cos θ = −1
tan θ = 1
θ = π + 2kπ, k ∈ Z
θ=
π 4
+ kπ, k ∈ Z.
Therefore, the solutions of the equation are θ = π + 2kπ and θ = k ∈ Z.
π 4
+ kπ for all 2
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Example 3.8.4. Find all values of x in the interval [−2π, 2π] that satisfy the equation (sin x − 1)(sin x + 1) = 0. Solution. sin x − 1 = 0
or
sin x + 1 = 0 sin x = −1
sin x = 1 x=
π 2
or −
3π 2
Solutions:
x= π , 2
− 3π , 2
3π , 2
or −
π 2
− π2
2
PY
Example 3.8.5. Solve: cos x = 0.1.
3π 2
C O
Solution. There is no special number whose cosine is 0.1. However, because 0.1 ∈ [−1, 1], there is a number whose cosine is 0.1. In fact, in any one-period interval, with cos x = 0.1 > 0, we expect two solutions: one in QI and another in QIV. We use the inverse cosine function. From Lesson 3.7, one particular solution of cos x = 0.1 in QI is x = cos−1 0.1. We can use this solution to get a particular solution in QIV, and this is x = 2π − cos−1 0.1, which is equivalent to x = − cos−1 0.1.
D
From the above particular solutions, we can produce all solutions of cos x = 0.1, and these are x = cos−1 0.1+2kπ and x = − cos−1 0.1+2kπ for all k ∈ Z. 2
EP E
Example 3.8.6. Solve: 3 tan θ + 5 = 0. Solution.
3 tan θ + 5 = 0
=⇒
tan θ = − 35
We expect only one solution in any one-period interval. tan θ = − 35 =⇒ θ = tan−1 − 53 + kπ, k ∈ Z
2
D
? Example 3.8.7. The voltage V (in volts) coming from an electricity distributing company is fluctuating according to the function V (t) = 200 + 170 sin(120πt) at time t in seconds. (1) Determine the first time it takes to reach 300 volts. (2) For what values of t does the voltage reach its maximum value?
Solution. (1) We solve for the least positive value of t such that V (t) = 300. 200 + 170 sin(120πt) = 300 100 sin(120πt) = 170 226 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
120πt = sin−1 t=
100 170
sin−1 100 170 ≈ 0.00167 seconds 120π
(2) The maximum value of V (t) happens when and only when the maximum value of sin(120πt) is reached. We know that the maximum value of sin(120πt) is 1, and it follows that the maximum value of V (t) is 370 volts. Thus, we need to solve for all values of t such that sin(120πt) = 1.
C O
PY
sin(120πt) = 1 π 120πt = + 2kπ, k nonnegative integer 2 π + 2kπ t= 2 120π 1 + 2k ≈ 0.00417 + 0.017k t= 2 120
This means that the voltage is maximum when t ≈ 0.00417 + 0.017k for each nonnegative integer k. 2
D
3.8.3. Equations with Two or More Terms
EP E
We will now consider a group of equations having multi-step techniques of finding their solutions. Coupled with the straightforward techniques we learned in the preceding discussion, these more advanced techniques involve factoring of expressions and trigonometric identities. The primary goal is to reduce a given equation into equivalent one-term equations. Example 3.8.8. Solve: 2 cos x tan x = 2 cos x. Solution.
D
2 cos x tan x = 2 cos x 2 cos x tan x − 2 cos x = 0 (2 cos x)(tan x − 1) = 0
2 cos x = 0
or
tan x − 1 = 0
cos x = 0
tan x = 1
x = π2 + 2kπ or x = 3π + 2kπ, 2 k∈Z
x = π4 + kπ, k∈Z
Solutions:
π 2
+ 2kπ,
3π 2
+ 2kπ,
π 4
+ kπ, k ∈ Z
2
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Example 3.8.9. Solve for x ∈ [0, 2π): sin 2x = sin x. Solution. sin 2x = sin x sin 2x − sin x = 0 2 sin x cos x − sin x = 0
Sine Double-Angle Identity
(sin x)(2 cos x − 1) = 0 2 cos x − 1 = 0
or
x = 0 or x = π
PY
sin x = 0
cos x =
x=
or x =
5π 3
π 5π , 3 3
C O
Solutions: 0, π,
π 3
1 2
2
Tips in Solving Trigonometric Equations (1) If the equation contains only one trigonometric term, isolate that term, and solve for the variable.
EP E
D
(2) If the equation is quadratic in form, we may use factoring, finding square roots, or the quadratic formula. (3) Rewrite the equation to have 0 on one side, and then factor (if appropriate) the expression on the other side. (4) If the equation contains more than one trigonometric function, try to express everything in terms of one trigonometric function. Here, identities are useful.
D
(5) If half or multiple angles are present, express them in terms of a trigonometric expression of a single angle, except when all angles involved have the same multiplicity wherein, in this case, retain the angle. Half-angle and double-angle identities are useful in simplification.
Example 3.8.10. Solve for x ∈ [0, 2π): 2 cos2 x = 1 + sin x. Solution. 2 cos2 x = 1 + sin x 2(1 − sin2 x) = 1 + sin x
Pythagorean Identity
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2 sin2 x + sin x − 1 = 0 (2 sin x − 1)(sin x + 1) = 0 2 sin x − 1 = 0 sin x = x=
π 6
or
sin x + 1 = 0
1 2
or x =
Factoring
sin x = −1 5π 6
x=
3π 2
2
π 5π 3π , 6, 2 6
Solutions:
PY
Example 3.8.11. Solve for x ∈ [0, 2π) in the equation 3 cos2 x + 2 sin x = 2. Solution. 3 cos2 x + 2 sin x = 2 3(1 − sin2 x) + 2 sin x = 2
C O
Pythagorean Identity
(3 sin x + 1)(sin x − 1) = 0 3 sin x + 1 = 0 sin x = − 31
Factoring
sin x − 1 = 0
or
sin x = 1
EP E
D
x = sin−1 (− 31 ) + 2π or x = π − sin−1 (− 13 )
x=
π 2
Solutions: 2π − sin−1 ( 13 )+, π + sin−1 ( 13 ),
π 2
2
D
One part of the last solution needs further explanation. In the equation sin x = − 31 , we expect two solutions in the interval [0, 2π): one in (π, 3π ) (which 2 is QIII), and another in ( 3π , 2π) (which is QIV). Since no special number satisfies 2 −1 1 π π sin x = − 3 , we use inverse sine function. Because the range of sin is [− 2 , 2 ], we know that − π2 < sin−1 (− 13 ) < 0. From this value, to get the solution in ( 3π , 2π), 2 −1 1 we simply add 2π to this value, resulting to x = sin (− 3 ) + 2π. On the other hand, to get the solution in (π, 3π ), we simply add − sin−1 (− 31 ) to π, resulting to 2 x = π − sin−1 (− 13 ). Example 3.8.12. Solve: sin2 x + 5 cos2
x 2
= 2.
Solution. sin2 x + 5 cos2
x 2
=2 x sin2 x + 5 1+cos =2 2
Cosine Half-Angle Identity
2 sin2 x + 5 cos x + 1 = 0 229 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
2(1 − cos2 x) + 5 cos x + 1 = 0 Pythagorean Identity 2 cos2 x − 5 cos x − 3 = 0 (2 cos x + 1)(cos x − 3) = 0 2 cos x + 1 = 0
cos x − 3 = 0
or
cos x = 3
x = 2π + 2kπ or 3 4π x = 3 + 2kπ, k∈Z
no solution
Solutions:
2π 3
4π 3
+ 2kπ,
PY
cos x = − 12
+ 2kπ, k ∈ Z
2
Example 3.8.13. Solve for x ∈ [0, 2π) in the equation tan 2x − 2 cos x = 0.
C O
Solution.
tan 2x − 2 cos x = 0 sin 2x − 2 cos x = 0 cos 2x sin 2x − 2 cos x cos 2x = 0
D
Apply the Double-Angle Identities for Sine and Cosine, and then factor.
EP E
2 sin x cos x − 2(cos x)(1 − 2 sin2 x) = 0 (2 cos x)(2 sin2 x + sin x − 1) = 0 (2 cos x)(2 sin x − 1)(sin x + 1) = 0
2 cos x = 0
or
2 sin x − 1 = 0 1 2
sin x =
x = π2 or x = 3π 2
x = π6 or x = 5π 6
D
cos x = 0
or
sin x + 1 = 0 sin x = −1 x=
3π 2
These values of x should be checked in the original equation because tan 2x may not be defined. Upon checking, this is not the case for each value of x obtained. The solutions are π2 , 3π , π6 , 5π , and 3π . 2 2 6 2 ? Example 3.8.14. A weight is suspended from a spring and vibrating vertically according to the equation f (t) = 20 cos 45 π t − 56 , where f (t) centimeters is the directed distance of the weight from its central position at t seconds, and the positive distance means above its central position. 230
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(1) At what time is the displacement of the weight 5 cm below its central position for the first time? (2) For what values of t does the weight reach its farthest point below its central position? Solution. (1) We find the least positive value of t such that f (t) = −5. 20 cos 54 π t − 56 = −5 cos 45 π t − 56 = − 41
5
PY
There are two families of solutions for this equation. • 45 π t − 65 = cos−1 − 14 + 2kπ, k ∈ Z cos−1 (− 14 )+2kπ t = 65 + 4 π
D
C O
In this family of solutions, the least positive value of t happens when k = 0, and this is 5 cos−1 − 41 + 2(0)π ≈ 1.5589. t= + 4 6 π 5 • 54 π t − 65 = 2π − cos−1 − 14 + 2kπ, k ∈ Z 2π−cos−1 (− 14 )+2kπ t = 65 + 4 5
π
EP E
Here, the least positive value of t happens when k = −1, and this is 5 2π − cos−1 − 41 + 2(−1)π t= + ≈ 0.1078. 4 6 π 5
Therefore, the first time that the displacement of the weight is 5 cm below its central position is at about 0.1078 seconds.
D
(2) The minimum value of f (t) happens when and only when the minimum value of cos 45 π t − 56 is reached. The minimum value of cos 54 π t − 56 is −1, which implies that the farthest point the weight can reach below its central position is 20 cm. Thus, we need to solve for all values of t such that cos 54 π t − 65 = −1. cos 45 π t − 56 = −1 4 5 π t − = cos−1 (−1) + 2kπ, k ≥ 0 5 6 4 π t − 56 = π + 2kπ 5 t=
5 6
+
π+2kπ 4 π 5
=
25 12
+ 52 k
Therefore, the weight reaches its farthest point (which is 20 cm) below its 25 central position at t = 12 + 52 k for every integer k ≥ 0. 2 231 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
More Solved Examples 1. Give a particular solution of the following equation. (a) sin2 x − 1 = 0 √ (b) cot x = 3
(d) csc2 x − csc x − 2 = 0
(c) sec 3x = −1
(f) 2 cos x − 3 = 0
(e) cos2 2x = sin2 x
Solution:
PY
(a) x = − π2 is a solution because sin2 (− π2 ) − 1 = (−1)2 − 1 = 0. √ √ (b) Note that cos π6 = 23 and sin π6 = 21 . Thus, cot π6 = 3. So, x = solution.
π 6
is a
(c) Since sec θ = −1 if and only if cos θ = −1, a particular solution of the equation in 3x is π, that is, 3x = π. Hence, x = π3 is a solution.
(e) x =
π 2
3π 2
= 1. As a consequence, csc2
C O
(d) Note that csc 3π = −1. So, csc2 2 − 2 = 1 − (−1) − 2 = 0. csc 3π 2 is a solution.
3π 2
−
(f) Because cos x must not be more than 1, then the equation has no solution.
D
2. What is the solution set of the following trigonometric equation sin2 x + cos2 x = 1?
EP E
Solution: The equation is the Pythagorean Identity, meaning any element of the domain of sin x and cos x satisfies the equation. The domain of both sin x and cos x is R. Therefore, the solution set of this trigonometric equation is R. One may try the numbers − π6 , 0, and π4 for illustration. (a) x = − π6
π π π 2 π 2 2 sin x + cos x = sin − + cos − = sin − + cos − 6 6 6 6 ! √ 2 2 1 3 1 3 = − + = + = 1. 2 2 4 4 2
2
D
2
(b) x = 0 sin2 x + cos2 x = sin2 0 + cos2 0 = 02 + 12 = 0 + 1 = 1. (c) x =
π 4
π π sin2 x + cos2 x = sin2 + cos2 4 4 √ !2 √ !2 2 2 1 1 = + = + = 1. 2 2 4 4 232 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
3. Find the solution set of the trigonometric equation tan2 x + 1 = sec2 x. Solution: Notice that this is a fundamental identity. Thus, the solution of this equation is any number common to the domain of the tangent and secant function. That is, the solution set is , ± 5π , ...} {x ∈ R | cos x 6= 0} = {x ∈ R | x 6= ± π2 , ± 3π 2 2 = {x ∈ R | x 6= 4. Find all solutions of
√
(2k+1)π ,k 2
∈ Z}.
3 tan x + 1 = 0.
5π 6
and x =
11π . 6
C O
Solution: The solutions are x =
PY
Solution: The equation is equivalent to tan x = − √13 . This is true only if x = 5π + kπ where k ∈ Z. 6 √ 5. What are the solutions of 3 tan x + 1 = 0, where x ∈ [0, 2π].
6. Determine all solutions of 4 cos2 x − 1 = 0.
EP E
D
Solution: Note that the equation is quadratic in form, so we can apply techniques in solving quadratic equations. For this case, we factor the expression on the left and obtain, (2 cos x − 1)(2 cos x + 1) = 0. Consequently, we have cos x = 1/2 or cos x = −1/2. The first equation have solutions of the form (π/3 + 2kπ) or (5π/3 + 2kπ) where k ∈ Z, while the second equation have solutions of the form (2π/3 + 2kπ) or (4π/3 + 2kπ). Combining the two solutions, one observes that the solution set of the original equation may be given by π 2π 4π 5π 7π , , , , , ... . 3 3 3 3 3
D
We can write this in a more compact form as kπ : k 6= 3j, where j ∈ Z . 3
7. Find the solutions of 4 cos2 x − 1 = 0 within the closed interval [0, 2π]. Solution: Similar to Example 6, the solution of the above equation is kπ : k 6= 3j, where j ∈ Z . 3 Since we are to find solutions in [0, 2π], we take k = 1, 2, 4, and 5 to obtain the solutions π/3, 2π/3, 4π/3, and 5π/3.
233 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
8. If x ∈ [0, 2π), solve the equation 2 sin2 x =
√
3 sin x.
√ Solution: First, we write the equation as 2 sin2 x − 3 sin x = 0. Then, we factor out sin x and get √ √ sin x(2 sin x − 3) = 0 =⇒ sin x = 0 or sin x = 3/2. The first of these equations has solutions x = 0 and x = π, while the second has solutions x = π/3 and 2π/3. The solutions of the original equation is the union of the two, i.e., the solution set is {0, π, π3 , 2π }. 3
9. Solve 2 cos2 x + 5 cos x − 3 = 0, where x ∈ [0, 2π).
1 and cos x = −3. 2
C O
cos x =
PY
Solution: By factoring the left hand side of the given equation, we get (2 cos x− 1)(cos x + 3) = 0. This gives us two equations, namely
First, we remark that the second equation does not have a solution because cos x should be more than or equal to -1. Hence, the solution of the first equation is the solution of the original equation. Thus, the solution set is { π3 , 5π }. 3 10. Determine the solution set of the equation cos 2x = sin x on [0, 2π).
D
Solution: Combining the equation cos 2x = sin x with the cosine double-angle identity cos 2x = 1 − 2 sin2 x, we get
EP E
sin x = 1 − 2 sin2 x.
This is equivalent to
2 sin2 x + sin x − 1 = 0 =⇒ (2 sin x − 1)(sin x + 1) = 0
D
=⇒ sin x = 1/2 and sin x = −1.
The solutions of the first equation is x = π/6 and x = 5π/6. The number x = 3π/2 is the solution of the second equation. Therefore, the solution set of , 3π }. the original equation is { π6 , 5π 6 2
11. Solve cos x = cos 2x, for x ∈ [0, 2π). Solution: cos x = cos 2x ⇒ cos x = 2 cos2 x − 1 ⇒
0 = 2 cos2 x − cos x − 1
⇒
0 = (2 cos x + 1)(cos x − 1). 234
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The given trigonometric equation is equivalent to solving 2 cos x + 1 = 0 and cos x = 1. For 2 cos x + 1 = 0 which is the same as cos x = −1/2, the solutions in the given interval are x = 2π/3, 4π/3. For cos x = 1, the solution is x = 0. , 4π }. Therefore, the solution set of the original equation is {0, 2π 3 3 12. A lighthouse at sea level is 34 mi from a boat. It is known that the top of the lighthouse is 42.5 mi from the boat and that x = r cos θ, where x is the horizontal distance, r is the distance of the top of the lighthouse from the boat, and θ is the angle of depression from the top of lighthouse. Find θ. Solution: 34 4 x = = r 42.5 5 4 =⇒ θ = cos−1 ≈ 0.6435 (or 36.87◦ ). 5 For this case, we used a calculator to find the value of the unknown variable θ since 54 is not a special value for cosine.
C O
PY
x = r cos θ =⇒ cos θ =
EP E
D
13. Three cities, A, B, and C, are positioned in a triangle as seen in the figure below.
It is known that City A is 140√mi from City C, while City B is 210 mi from City C. Cities A and B are 70 7 mi apart. Also, by the Cosine Law, we have
D
z 2 = x2 + y 2 − 2xy cos γ
where x, y, and z are the respective distances of BC, AC, AB, and γ = m∠ACB. Find γ. Solution: Substituting the corresponding values of x, y, and z, the problem is now equivalent to solving the equation 34300 = 44100 + 19600 − 58800 cos γ ⇒ −29400 = −58800 cos γ ⇒
1 2
= cos γ
⇒
π 3
= γ. 235
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Supplementary Problems 3.8 1. Find all solutions of the equation 2 cos x − cos x sin x = 0. 2. Determine the solution set of the equation csc2 x + 1 = 0. 3. What are the solutions of sec2 x + sec x − 2 = 0. 4. Find the solutions of the equation 4 sin2 x − 1 = 0 on [0, 2π). √ 5. Find the values of x ∈ [0, 2π) for which csc 2x = 2.
PY
6. What is the solution set of the equation sin θ = csc θ? 7. Solve t = sin−1 (cos 2t).
8. Let x ∈ [0, 2π). Find the solutions of the equation cos2 4x + sin2 2x = 1.
C O
9. If a projectile, such as a bullet, is fired into the air with an initial velocity v at an angle of elevation θ, then the height h of the projectile at time t is given by h(t) = −16t2 + vt cos θ meters. If the initial velocity is 109 meters per second, at what angle should the bullet be fired so that its height is 45 meters above the floor in 2 seconds.
EP E
D
10. In a baseball field, a pitcher throws the ball at a speed of 60 km/h to the catcher who is 100 m away. When the ball leaves a starting point at an angle of elevation of θ , the horizontal distance the ball travels is determined by 2 d = v32 sin θ, where d is measured in meters and velocity in kilometers per hour. At what angle of elevation (in degrees) is the ball thrown?
4
D
Lesson 3.9. Polar Coordinate System Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) locate points in polar coordinate system; (2) convert the coordinates of a point from rectangular to polar system and vice versa; and (3) solve situational problems involving polar coordinate system.
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Lesson Outline (1) Polar coordinate system: pole and polar axis (2) Polar coordinates of a point and its location (3) Conversion from polar to rectangular coordinates, and vice versa (4) Simple graphs and applications Introduction
D
EP E
D
C O
PY
Two-dimensional coordinate systems are used to describe a point in a plane. We previously used the Cartesian or rectangular coordinate system to locate a point in the plane. That point is denoted by (x, y), where x is the signed distance of the point from the y-axis, and y is the signed distance of the point from the x-axis. We sketched the graphs of equations (lines, circles, parabolas, ellipses, and hyperbolas) and functions (polynomial, rational, exponential, logarithmic, trigonometric, and inverse trigonometric) in the Cartesian coordinate plane. However, it is often convenient to locate a point based on its distance from a fixed point and its angle with respect to a fixed ray. Not all equations can be graphed easily using Cartesian coordinates. In this lesson, we also use another coordinate system, which can be presented in dartboard-like plane as shown below.
3.9.1. Polar Coordinates of a Point We now introduce the polar coordinate system. It is composed of a fixed point called the pole (which is the origin in the Cartesian coordinate system) and a 237 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
fixed ray called the polar axis (which is the nonnegative x-axis).
C O
PY
In the polar coordinate system, a point is described by the ordered pair (r, θ). The radial coordinate r refers to the directed distance of the point from the pole. The angular coordinate θ refers to a directed angle (usually in radians) from the polar axis to the segment joining the point and the pole.
D
EP E
D
Because a point in polar coordinate system is described by an order pair of radial coordinate and angular coordinate, it will be more convenient to geometrically present the system in a polar plane, which serves just like the Cartesian plane. In the polar plane shown below, instead of rectangular grids in the Cartesian plane, we have concentric circles with common center at the pole to identify easily the distance from the pole (radial coordinate) and angular rays emanating from the pole to show the angles from the polar axis (angular coordinate).
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Example 3.9.1. Plot the following points in one polar plane: A(3, π3 ), B(1, 5π ), 6 7π 19π 7π 17π 17π 5π C(2, 6 ), D(4, 12 ), E(3, −π), F (4, − 6 ), G(2.5, 4 ), H(4, 6 ), and I(3, − 3 ).
EP E
D
C O
PY
Solution.
D
As seen in the last example, unlike in Cartesian plane where a point has a unique Cartesian coordinate representation, a point in polar plane have infinitely many polar coordinate representations. For example, the coordinates (3, 4) in the Cartesian plane refer to exactly one point in the plane, and this particular point has no rectangular coordinate representations other than (3, 4). However, the coordinates (3, π3 ) in the polar plane also refer to exactly one point, but this point has other polar coordinate representations. For example, the polar coordinates (3, − 5π ), (3, 7π ), (3, 13π ), and (3, 19π ) all refer to the same point as 3 3 3 3 π that of (3, 3 ). The polar coordinates (r, θ + 2kπ), where k ∈ Z, represent the same point as that of (r, θ). In polar coordinate system, it is possible for the coordinates (r, θ) to have a negative value of r. In this case, the point is |r| units from the pole in the opposite direction of the terminal side of θ, as shown in Figure 3.41. 239
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PY
Figure 3.41
C O
), B(−4, 11π ), Example 3.9.2. Plot the following points in one polar plane: A(−3, 4π 3 6 C(−2, −π), and D(−3.5, − 7π ). 4
D
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D
Solution. As described above, a polar point with negative radial coordinate lies on the opposite ray of the terminal side of θ.
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Points in Polar Coordinates 1. For any θ, the polar coordinates (0, θ) represent the pole. 2. A point with polar coordinates (r, θ) can also be represented by (r, θ + 2kπ) or (−r, θ + π + 2kπ) for any integer k.
3.9.2. From Polar to Rectangular, and Vice Versa
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We now have two ways to describe points on a plane – whether to use the Cartesian coordinates (x, y) or the polar coordinates (r, θ). We now derive the conversion from one of these coordinate systems to the other.
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D
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We superimpose the Cartesian and polar planes, as shown in the following diagram.
Figure 3.42
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Suppose a point P is represented by the polar coordinates (r, θ). From Lesson 3.2 (in particular, the boxed definition on page 139), we know that x = r cos θ
and
y = r sin θ.
Conversion from Polar to Rectangular Coordinates x = r cos θ (r, θ) −→ −→ (x, y) y = r sin θ Given one polar coordinate representation (r, θ), there is only one rectangular coordinate representation (x, y) corresponding to it.
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Example 3.9.3. Convert the polar coordinates (5, π) and (−3, π6 ) to Cartesian coordinates. Solution.
x = 5 cos π = −5 (5, π) −→ y = 5 sin π = 0 √ x = −3 cos π = − 3 3 6 2 (−3, π6 ) −→ y = −3 sin π = − 3
√
2
−→ (− 3 2 3 , − 32 )
2
PY
6
−→ (−5, 0)
As explained on page 239 (right after Example 3.9.1), we expect that there are infinitely many polar coordinate representations that correspond to just one given rectangular coordinate representation. Although we can actually determine all of them, we only need to know one of them and we can choose r ≥ 0.
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Suppose a point P is represented by the rectangular coordinates (x, y). Referring back to Figure 3.42, the equation of the circle is p x2 + y 2 = r 2 =⇒ r = x2 + y 2 .
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We now determine θ. If x = y = 0, then r = 0 and the point is the pole. The pole has coordinates (0, θ), where θ is any real number. 3π 2
(or their
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If x = 0 and y 6= 0, then we may choose θ to be either π2 or equivalents) depending on whether y > 0 or y < 0, respectively.
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Now, suppose x 6= 0. From the boxed definition again on page 139, we know that y tan θ = , x where θ is an angle in standard position whose terminal side passes through the point (x, y). Conversion from Rectangular to Polar Coordinates (x, y) = (0, 0) −→ (r, θ) = (0, θ), θ ∈ R (y, π ) if y > 0 2 (0, y) −→ (r, θ) = (|y|, 3π ) if y < 0 y6=0 2
(x, 0) if x > 0 (x, 0) −→ (r, θ) = (|x|, π) if x < 0 x6=0
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(x, y)
−→ (r, θ)
x6=0, y6=0
r=
p
x2 + y 2
tan θ =
y x
θ same quadrant as (x, y) Given one rectangular coordinate representation (x, y), there are many polar coordinate representations (r, θ) corresponding to it. The above computations just give one of them.
PY
Example 3.9.4. Convert each Cartesian coordinates to polar coordinates (r, θ), where r ≥ 0. (4) (6, −2) (1) (−4, 0) (5) (−3, 6)
√ (3) (−3, − 3)
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(2) (4, 4)
(6) (−12, −8)
Solution. (1) (−4, 0) −→ (4, π)
tan θ =
y x
=
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(2) The point (4, 4) is in QI. p √ √ r = x2 + y 2 = 42 + 42 = 4 2 4 4
= 1 =⇒ √ π 4 2, 4
θ=
π 4
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(4, 4) −→ √ (3) (−3, − 3) in QIII q √ √ r = (−3)2 + (− 3)2 = 2 3 √
√
7π 6
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tan θ = −−33 = 33 =⇒ θ = √ √ (−3, − 3) −→ 2 3, 7π 6
(4) (6, −2) in QIV p √ r = 62 + (−2)2 = 2 10 −2 6
= − 13 =⇒ θ = tan−1 − 13 √ (6, −2) −→ 2 10, tan−1 − 13 tan θ =
(5) (−3, 6) in QII p √ r = (−3)2 + 62 = 3 5 = −2 =⇒ θ = π + tan−1 (−2) √ (−3, 6) −→ 3 5, π + tan−1 (−2) tan θ =
6 −3
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(6) (−12, −8) in QIII p √ r = (−12)2 + (−8)2 = 4 13 tan θ =
−8 −12
=
2 3
(−12, −8) −→
=⇒ θ = π + tan−1 √ 4 13, π + tan−1 23
2 3
2
3.9.3. Basic Polar Graphs and Applications
PY
From the preceding session, we learned how to convert polar coordinates of a point to rectangular and vice versa using the following conversion formulas: y r2 = x2 + y 2 , tan θ = , x = r cos θ, and y = r sin θ. x Because a graph is composed of points, we can identify the graphs of some equations in terms of r and θ.
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Graph of a Polar Equation
The polar graph of an equation involving r and θ is the set of all points with polar coordinates (r, θ) that satisfy the equation.
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As a quick illustration, the polar graph of the equation r = 2 − 2 sin θ consists of all points (r, θ) that satisfy the equation. Some of these points are (2, 0), (1, π6 ), ). (0, π2 ), (2, π), and (4, 3π 2
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Example 3.9.5. Identify the polar graph of r = 2, and sketch its graph in the polar plane.
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Solution. Squaring the equation, we get r2 = 4. Because r2 = x2 + y 2 , we have x2 + y 2 = 4, which is a circle of radius 2 and with center at the origin. Therefore, the graph of r = 2 is a circle of radius 2 with center at the pole, as shown below.
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In the previous example, instead of using the conversion formula r2 = x2 + y 2 , we may also identify the graph of r = 2 by observing that its graph consists of points (2, θ) for all θ. In other words, the graph consists of all points with radial distance 2 from the pole as θ rotates around the polar plane. Therefore, the graph of r = 2 is indeed a circle of radius 2 as shown. . Example 3.9.6. Identify and sketch the polar graph of θ = − 5π 4
D
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D
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Solution. The graph of θ = − 5π consists of all points (r, − 5π ) for r ∈ R. If 4 4 r > 0, then points (r, − 5π ) determine a ray from the pole with angle − 5π from 4 4 5π the polar axis. If r = 0, then (0, − 4 ) is the pole. If r < 0, then the points (r, − 5π ) determine a ray in opposite direction to that of r > 0. Therefore, the 4 is a line passing through the pole and with angle − 5π with graph of θ = − 5π 4 4 respect to the polar axis, as shown below.
Example 3.9.7. Identify (and describe) the graph of the equation r = 4 sin θ. Solution.
r r2 x2 + y 2 x2 + y 2 − 4y x2 + (y − 2)2
= 4 sin θ = 4r sin θ = 4y =0 =4
Therefore, the graph of r = 4 sin θ is a circle of radius 2 and with center at (2, π2 ). 245 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
PY
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? Example 3.9.8. Sketch the graph of r = 2 − 2 sin θ. Solution. We construct a table of values. x
0
π 6
r
2
1
x
7π 6
r
3
π 4
π 3
π 2
0.59 0.27 4π 3
3π 2
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5π 4
0
4
3π 4
5π 6
π
1
2
0.27 0.59
5π 3
7π 4
3.73 3.41
11π 6
2π
3
2
D
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3.41 3.73
2π 3
This heart-shaped curve is called a cardioid.
2
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? Example 3.9.9. The sound-pickup capability of a certain brand of microphone is described by the polar equation r = −4 cos θ, where |r| gives the sensitivity of the microphone to a sound coming from an angle θ (in radians). (1) Identify and sketch the graph of the polar equation. (2) Sound coming from what angle θ ∈ [0, π] is the microphone most sensitive to? Least sensitive? r r2 x2 + y 2 x2 + 4x + y 2 (x + 2)2 + y 2
= −4 cos θ = −4r cos θ = −4x =0 =4
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Solution. (1)
D
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D
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This is a circle of radius 2 and with center at (2, π).
(2) We construct a table of values. x r
0
π 6
−4 −3.46
π 4
π 3
π 2
2π 3
−2.83
−2
0
2
3π 4
5π 6
2.83 3.46
π 4
From the table, the microphone is most sensitive to sounds coming from angles θ = 0 and θ = π, and least sensitive to sound coming from an angle θ = π2 . 2 247 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
More Solved Examples 1. Locate in the polar plane the following polar points: M (1, π/3), A(0, π), T (π, 0), and H(4, 5π/3).
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Solution:
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2. Locate in the polar plane the following polar points: W (−1, 7π/4), X(2, −π/6), Y (4, −5π/6) and Z(−3, −11π/3).
D
Solution:
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3. Convert the following polar points to Cartesian coordinates. (a) (5, 5π/4) to Cartesian coordinates (b) (−2, 3π/4) to Cartesian coordinates (c) (π, π) (d) (0, 10)
PY
Solution: (a) Using the conversion formulas with r = 5 and θ = 5π/4, we get √ x = r cos θ = 5 cos(5π/4) = −5 2/2 and
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√ y = r sin θ = 5 sin(5π/4) = −5 2/2. √ √ Therefore, (5, 5π/4) in Cartesian coordinate is(−5 2/2, −5 2/2).
(b) Using the conversion formulas with r = −2 and θ = 5 = 3π/4, we get √ x = r cos θ = −2 cos(3π/4) = 2 2/2 and
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√ y = r sin θ = −2 sin(3π/4) = −2 2/2. √ √ Therefore, (−2, 3π/4) in Cartesian coordinate is (2 2/2, −2 2/2).
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(c) Notice here that π is used in two different ways. First is π, with numerical value approximately equal to −3.14, is used as a radius and second, as an angle equivalent to 180◦ . That is, the point is in the negative x-axis π units away from the origin. Hence, the Cartesian coordinate of (π, π) is (−π, 0). (d) Since the radius is 0, then the polar point (0, 10) is the origin with Cartesian coordinate (0, 0).
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4. Convert the following Cartesian points to polar coordinates. √ (a) (5, −5) (c) (−5 3, −15) √ (d) (8, 0) (b) (−3, 3) Solution: (a) The point (x, y) = (5, −5) is in the fourth quadrant. Using the conversion formulas, we get p p √ r = x2 + y 2 = 52 + (−5)2 = 5 2 and θ = tan−1 (y/x) = tan−1 (−1) = −π/4. √ Therefore, (5, −5) in polar coordinate is (5 2, −π/4). 249
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(b) Similarly, we use the conversion formulas to get q √ √ √ r = (−3)2 + ( 3)2 = 12 = 2 3 and
√ θ = tan−1 [ 3/(−3)] = −π/6.
Note that the point is in the second quadrant so we must use − pi6 + π. There√ √ fore, (−3, 3) in polar coordinate is (2 3, 5π/6). (c) The point is in the third quadrant. q √ √ √ r = (−5 3)2 + (−15)2 = 300 = 10 3
PY
and
√ tan θ = −15/(−5 3) ⇒ θ = 4π/3. √ √ Therefore, (−5 3, −15) in polar coordinate is (10 3, 4π/3).
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(d) Using the conversion formula, one can show that the point (8, 0) in polar coordinate is also (8, 0). 5. Identify (and describe) the graph of the equation r = 4 sin θ. Using a graphing software, graph the following equations. (a) r = 2 sin θ
(c) θ = 2r
(d) r = 2 − 2 cos θ
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(b) r = −5
Solution: (a) r = 2 sin θ is a circle with radius 1 centered at (1, π2 ).
EP E
(b) r = −5 is a standard circle with radius 5. (c) Notice that as θ increases, the r also increases. The graph of θ = 2r is a spiral rotating counter-clockwise from the pole.
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(d) The graph of r = 2 − 2 cos θ is a cardioid.
(a)
(b) 250
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(c)
(d)
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6. A boy is flying a kite with an angle of elevation of 60◦ from where he stands. What is the direct distance of the kite from him, if the the kite is 6 ft above the ground?
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Solution: The problem can be illustrated as follows:
Here, r (in ft) is the distance of the kite from the boy and θ is the angle of depression. To solve for r, we apply the formula y = r sin θ. Thus, √ √ √ r = y/ sin θ = 6/ sin(60◦ ) = 6/( 3/2) = 12/ 3 = 4 3. √ Therefore, the kite is 4 3 ft away from the boy. 251 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Supplementary Problems 3.9 1. Give two more pairs of coordinates that describe the same point. (a) (13, π/3)
(b) (0, 0)
(c) (15, 15π/4)
2. Locate the following points in the polar coordinate plane: (a) P (3, −π) (b) Q(−3, 7π/4)
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(c) R(5/2, 5π/2) (d) S(−8, −23π/6)
3. Transform the following to Cartesian coordinates:
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(a) (3, −π) (b) (−3, 7π/4) (c) (5/2, 5π/2) (d) (−8, −23π/6)
(a) (−9, 40) (b) (15, 20)
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(c) (5/2, 5π/2)
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4. Transform the following to polar coordinates:
(d) (14, −14)
5. Consider the equation in polar form r = 4 cos 2θ. (a) Complete the table
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θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 r
(b) Plot the points obtained in part (a) in a polar coordinate system. 6. A helicopter is hovering 800 feet above a road. A truck driver observes the helicopter at a horizontal distance of 600 feet. Find the angle of elevation of the helicopter from the truck driver.
4
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Topic Test 3 for Unit 3 12 1. Let θ be an angle in QIII such that cos θ = − . Find the values of the six 13 trigonometric functions of 2θ. cot2 x − 1 2. Prove that cot(2x) = . 2 cot x 3. Using half-angle identities to find the exact values of the following. (b) tan 7.5◦
−1 40 (b) cos sin 41
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4. Find the exact value of the following. 103π −1 (a) tan cot 6
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(a) tan 15◦
5. Let y ∈ [0, 2π). Find the solutions of the equation
sin−1 (cos2 y − cos y − 1) = −π/2. 6. Let θ ∈ [0, 2π]. Find all the solutions of the equation
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4 cos2 θ sin θ = 3 sin θ.
EP E
7. Let r = −2 − 2 sin θ. Complete the table and plot the points (r, θ) in the same polar coordinates. θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 r
D
8. Transform the following points from Cartesian to polar coordinates. (a) (−42, −56)
(c) (0, 7)
(e) (2π, 2π)
(b) (100, 100)
(d) (7, 0)
(f) (5, 12)
9. Transform the following points from polar to Cartesian coordinates. (a) (3, π/3)
(c) (−1, −π)
(e) (2π, 2π)
(b) (45, 7π/4)
(d) (5, 0)
(f) (9, 17π/6)
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Topic Test 4 for Unit 3 7 1. Let θ be an angle in the 2nd quadrant such that cos θ = − . Find the 25 following. (a) cos(2θ)
(b) sin(2θ)
(c) tan(2θ)
2. Given that cos 48◦ ≈ 0.6691. Find the approximate value of the following. (b) sin2 24◦
(c) tan2 24◦
PY
(a) cos2 24◦
3. Using half-angle identities to find the exact values of the following. (a) tan(π/12)
(b) tan(π/24) −1
1 −1 3 + cos . 7 5
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4. Find the exact value of cos cos
5. Let x ∈ [0, 2π). Find the solutions of the equation √ √ √ 4 sin2 x + (2 3 − 2 2) sin x − 6 = 0.
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6. Let θ ∈ [0, 2π]. Find all the solutions of the equation
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2 sin2 (2θ) − sin(2θ) − 1 = 0.
7. Let r = 2 + 2 cos θ. Complete the table and plot the points (r, θ) in the same polar coordinates. θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2
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r
8. Transform the following points from Cartesian to polar coordinates. (a) (21, −28)
(c) (0, −5)
(e) (π, π)
(b) (−100, −100)
(d) (−5, 0)
(f) (15, 8)
9. Transform the following points from polar to Cartesian coordinates. (a) (4, π/6)
(c) (1, π)
(e) (π, π)
(b) (100, 5π/4)
(d) (−5, 0)
(f) (15, 8π/3)
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PY
C O
4
Answers to
D
Odd-Numbered Exercises in Supplementary Problems and
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All Exercises in Topic Tests
D
4
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Supplementary Problems 1.1 (page 17) 1 2
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3 3. center −4, ,r=1 4
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PY
1. center (0, 0), r =
D
5. center (7, −6), r = 11
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5 3
D
EP E
D
9. center
5 7 7 − , ,r= 2 2 4
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PY
7. center (−2, −4), r =
11. (x − 17)2 + (y − 5)2 = 144
19. (x + 10)2 + (y − 7)2 = 36
13. (x − 15)2 + (y + 7)2 = 49
21. (x + 2)2 + (y − 3)2 = 12
15. (x − 15)2 + (y + 7)2 = 9
23. (x − 2.5)2 + (y − 0.5)2 = 14.5
17. (x + 2)2 + (y − 3.5)2 = 31.25
25. (x + 5)2 + (y + 1)2 = 8
27. Set up a Cartesian coordinate system by assigning C as the origin. Then the circle on the left end has radius 100 and has equation x2 + y 2 = 10000. A radius of the circle on the right end can be drawn from C to the upper right corner of the figure; this radius has length (by the Pythagorean theorem) 257 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
√ 3002 + 1002 = 100000. Then the circle on the right end has equation x2 + y 2 = 100000. We want the at y = 50. In this case, the √ √ length of the segment 2 left endpoint has x = − √ 10000 − 50 = − 7500 and the √ right endpoint √ has √ x = √100000√− 502 = 97500. Then the total length is 97500 − (− 7500) = 50 3 + 50 39 m ≈ 398.85 m. √
Supplementary Problems 1.2 (page 31)
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D
C O
PY
1. vertex (0, 0), focus (−9, 0), directrix x = 9, axis y = 0
D
3. vertex (−1, 7), focus (−2, 7), directrix x = 0, axis y = 7
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7. (y − 11)2 = 36(x − 7) 9. (x + 10)2 = 34(y − 3)
13. (y − 8)2 = −8(x + 3) 15. ≈ 4.17 cm 17. 3.75 cm
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11. (y − 9)2 = −80(x − 4)
C O
PY
5. vertex (3, 2), focus
3 11 3, − , directrix y = , axis x = 3 2 2
1.
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Supplementary Problems 1.3 (page 45) center:
(0, 0)
foci:
F1 (−2, 0), F2 (2, 0) √ √ vertices: V1 (−2 2, 0), V2 (2 2, 0) W1 (0, −2), W2 (0, 2)
D
covertices:
3.
center: foci:
(1, 1) F1 (1 −
√
3, 1), √ F2 (1 + 3, 1)
vertices: V1 (−1, 1), V2 (3, 1) covertices:
W1 (1, 0), W2 (1, 2)
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5.
center: foci:
(7, −5) F1 (1, −5), F2 (13, −5)
vertices: V1 (−3, −5), V2 (17, −5) W1 (7, −13), W2 (7, 3)
(x − 2)2 (y − 8)2 + =1 49 16
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7.
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covertices:
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9. The center is (−9, 10) and c = 12. √ We see that the given point (−9, 15) is a covertex, so b = 5. Then a = 52 + 122 = 13. Therefore, the equation is (x + 9)2 (y − 10)2 + = 1. 169 25
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11. Since the major axis is vertical, the center has the same x coordinate as the focus and the same y coordinate as the covertex; that is, the center is (−9, 10). (x + 9)2 + Then c = 5, b = 10, and a2 = 125. Therefore, the equation is 100 (y − 10)2 = 1. 125
13. Recall that the unit is 100 km. The vertices of the ellipse are at (3633, 0) and (−4055, 0). Then the center of the ellipse is at (−211, 0). Then a = 3844 (x + 211)2 and c = 211. It follows that b2 = 14731815. The equation is + 14776336 2 y = 1. 14731815 15. Set up a coordinate system with the center of the ellipse at the origin. Then a = 60 and b = 20. We want the length of the segment with endpoints (on the 2 y2 ellipse) having x = 45 (or −45). The y coordinates are given by 45 + 20 2 = 1, 602 q 2 or y = ± 202 1 − 45 ≈ ±13.23. Hence, the desired width is 26.46 ft. 602 260
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Supplementary Problems 1.4 (page 59) 1.
center: foci:
(0, 0) √ √ F1 (− 181, 0), F2 ( 181, 0)
3.
center:
(0, 5)
√ √ F1 (− 19, 5), F2 ( 19, 5) √ √ vertices: V1 (− 15, 5), V2 ( 15, 5) 2 asymptotes: y − 5 = ± √ x 15
D
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D
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foci:
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vertices: V1 (−10, 0), V2 (10, 0) 9 asymptotes: y = ± x 10
5.
center: foci:
(−3, −3) F1 (−3, −3 − F2 (−3, −3 +
vertices: V1 (−3, −3 −
√ √
15),
15)
√
6), √ V2 (−3, −3 + 6) √ 6 asymptotes: y + 3 = ± (x + 3) 3
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7.
y2 (x + 7)2 − =1 144 145
9.
(x + 10)2 (y + 4)2 − =1 81 256
11. The intersection (−4, 8) of the two asymptotes is the center of the hyperbola. Then the hyperbola is vertical and c = 13. Since the slopes of the asymptotes 5 5 , we have ab = 12 . are ± 12 √ Since c = 13, we have a2 + b2 = 169 and b = 169 − a2 . It follows that
PY
(y − 8)2 (x + 4)2 a 5 √ =⇒ a = 5 and b = 12 =⇒ − = 1. = 12 25 144 169 − a2
C O
13. The midpoint (9, 1) of the two given corners is the center of the hyperbola. Since the transverse axis is horizontal, a = 7 and b = 2. Therefore, the (x − 9)2 (y − 1)2 equation is − = 1. 49 4 Supplementary Problems 1.5 (page 66) 1. pair of intersecting lines
5. parabola
3. parabola
7. empty set
EP E
D
(x − 5)2 (y − 2)2 + = 1; so its foci 9. The standard equation of the ellipse is 36 100 are (5, 10) and (5, −6) while its vertices are (5, 12) and (5, −8). The equations of the circles are (x − 5)2 + (y − 10)2 = 4, (x − 5)2 + (y − 10)2 = 324, (x − 5)2 + (y + 6)2 = 4, and (x − 5)2 + (y + 6)2 = 324. (y + 5)2 (x + 9)2 − = 1. Its auxil25 25 iary rectangle has corners (−14, 0), (−4, 0), (−4, −10), (−14, −10). The equation of the circle is (x + 9)2 + (y + 5)2 = 50.
D
11. The standard equation of the hyperbola is
13. The equation simplifies to (x + 7)2 + (y − 3)2 =
r+2 . r−1
Its graph r+2 > 0; that is, when r ∈ (−∞, −2) ∪ (1, +∞). r−1 r+2 = 0; that is, when r = −2. (b) is a point if r−1 r+2 (c) is the empty set if < 0; that is, when r ∈ (−2, 1). r−1 (a) is a circle if
262 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Supplementary Problems 1.6 (page 77)
C O
EP E
D
9 4 4 89 (b) − , − , − , 3 20 3 20
PY
1. (a) (1, 6)
D
(c) (1, 6), (1, 2)
263 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
D
EP E
D
(e) No solution
C O
PY
√ √ 3 (d) 1, , (1 − 15, −1), (1 + 15, −1) 2
3. Let (x, y) be the ordered pair that satisfies the conditions. The resulting system of equations is 1 x2 = 2y 2 + 8 5 2 2 x +y = 16 1 1 1 1 1 1 1 1 Solving yields , , − , , , − , and − , − . 2 4 2 4 2 4 2 4
264 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
5. We have the system x2 + (y − 6)2 = 36 x2 = 4ky, where the first equation is a circle above the x-axis, tangent to the x-axis at x = 0, and the second equation is a parabola facing up/down, depending on k. Substituting the second equation in the first equation yields y 2 +(4k−12)y = 0. Note that y = 0 is already a root. We now consider two cases.
PY
If k > 0, the system might have one or two solutions. To ensure that the solution is unique, we set the discriminant to be nonpositive: 4k − 12 ≤ 0 ⇒ k ≥ 3. Thus k ∈ (−∞, 0] ∪ (3, +∞).
C O
If k ≤ 0, the system will always have a unique solution.
Topic Test 1 for Unit 1 (page 78)
EP E
D
1. (a) Since the coefficients of x2 and y 2 are equal, the graph is a circle, a point, or the empty set. Completing the squares, we see that the equation is equivalent to 2 2 3 1 + y+ = 4. x− 2 2 Hence, the graph is a circle with center (0.5, −1.5) and radius 2.
(b) By inspection, the graph is a parabola. Completing the squares, we see that the equation is equivalent to (x + 2)2 = 14(y + 4). Hence, the graph has vertex at (−2, −4) and is opening upward.
D
(c) Since the coefficients of x2 and y 2 are of opposite signs, the graph is a hyperbola or a pair of intersecting lines. Completing the squares, we see that the equation is equivalent to (x − 7)2 (y + 3)2 − = 1. 4 3
Hence, the graph is a horizontal hyperbola with center at (7, −3). (d) Since the coefficients of x2 and y 2 have the same sign and are unequal, the graph is an ellipse, a point, or the empty set. Completing the squares, we see that the equation is equivalent to (x − 8)2 y 2 + = 0. 2 7 Hence, the graph is the point (8, 0). 265 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
2. (a) The equation is equivalent to
center: foci:
x2 y2 + = 1. This is a vertical ellipse. 7 25
(0, 0) √ √ F1 (0, −3 2), F2 (0, 3 2)
center: C(1, −4)
F1 (1, −14), F2 (1, 6)
D
foci:
(y + 4)2 (x − 1)2 − = 1. This is a vertical 64 36
C O
(b) The equation is equivalent to hyperbola.
PY
vertices: V1 (0, −5), V2 (0, 5) √ √ covertices: W1 (− 7, 0), W2 ( 7, 0)
vertices: F1 (1, −12), F2 (1, 6)
EP E
asymptotes: y + 4 = ± 43 (x − 1)
3. (a) The parabola opens to the right and has focal distance c = 6. Its equation is (y − 3)2 = 24(x + 1).
D
(b) The intersection (−2, −5) of the two asymptotes is the center of the hyperbola. Then the hyperbola is horizontal and a = 5. Using the slopes of the asymptotes, we have ab = 12 . It follows that b = 12 and the 4 2 2 (y + 5) (x + 2) − = 1. equation is 25 144
4. Multiplying the first equation by 2, we get 2(x − 1)2 + 2(y + 1)2 = 10. By subtracting the second equation from this new equation, we get the equation 2(y + 1)2 + 8 = 10 − y. This has solutions y = 0 and y = −5/2. When y = 0, the corresponding √x values are 3 and −1. When y = −5/2, the corresponding x values are ± 211 + 1. Therefore, the solutions are (−1, 0), √
(3, 0), ±
11 2
+ 1, − 52 .
266 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
5. Set up a coordinate system by making the center of the door’s base the origin. Then the ellipse has center (0, 2) with a = 1/2 and b = 0.3; then its is equation 2 x2 + (y−2) = 1. 0.52 0.32 To determine if the cabinet can be pushed through the doorway, we determine the height of the doorway when x = 0.25 (or −0.25). We solve for y from the 2 2 equation 0.25 + (y−2) = 1. Solving for the y coordinate, we see that the height 0.52 0.32 is ≈ 2.2598 m. Hence, the cabinet cannot be pushed through the doorway.
Manipulating this equation gives us
= 4(x2 − 6y + 9) = 36 − 48 = −12
C O
x2 + (y + 1)2 −3(x2 − 8x + 16) + (y + 1)2 −3(x − 4)2 + (y + 1)2 (x − 4)2 (y + 1)2 − 4 12
PY
6. Let (x, y) be the coordinates of the point. This point satisfies p x2 + (y + 1)2 = 2|x − 3|.
= 1.
D
Therefore, the point traces a horizontal hyperbola with center at (4, −1).
EP E
Topic Test 2 for Unit 1 (page 79)
1. (a) By inspection, the graph is a parabola. Completing the squares, we see that the equation is equivalent to (y − 5)2 = −8(x − 5). Hence, the graph has vertex at (5, 5) and is opening to the left.
D
(b) Since the coefficients of x2 and y 2 are equal, the graph is a circle, a point, or the empty set. Completing the squares, we see that the equation is equivalent to (x + 5)2 + (y + 9)2 = −4. Hence, the graph is the empty set. (c) Since the coefficients of x2 and y 2 have the same sign and are unequal, the graph is an ellipse, a point, or the empty set. Completing the squares, (x + 2)2 (y − 1)2 + = 1. Hence, we see that the equation is equivalent to 4 9 the graph is a vertical ellipse with center (−2, 1).
(d) Since the coefficients of x2 and y 2 are of opposite signs, the graph is a hyperbola or a pair of intersecting lines. Completing the squares, we (y − 4)2 (x − 6)2 see that the equation is equivalent to − = 0. Hence, 11 17 the graph is a pair of intersecting lines given by the equations y − 4 = ± 11 (x − 6). 17 267 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
2. (a) The equation is equivalent to
center: foci:
x2 y 2 − = 1. This is a horizontal hyperbola. 64 64
(0, 0) √ F1 (−8 2, 0), √ F2 (8 2, 0)
vertices: V1 (−8, 0), V2 (8, 0)
(x + 3)2 (y − 2)2 + = 1. This is a horizontal 49 4
C O
(b) The equation is equivalent to ellipse.
PY
asymptotes: y = ±x
center: C(−3, 2) foci:
√ √ F1 (−3 − 3 5, 2), F2 (−3 + 3 5, 2)
vertices: F1 (−10, 2), F2 (4, 2)
W1 (−3, 0), W2 (−3, 4)
D
EP E
D
covertices:
3. (a) The parabola opens downward and has focal distance c = 5. Its equation is (x − 7)2 = −20(y + 7). (b) Since the ellipse has vertical or horizontal major axis, the center is at either (−1, 12) or (−5, 3). Since the major axis is longer than the minor axis, the center must be at (−5, 3). Then the ellipse is vertical with a = 9 and b = 4. Its equation is (x + 5)2 (y − 3)2 + = 1. 16 81 268 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
4. Completing the squares, we see that the first equation is equivalent to 9(x + 3)2 = 4y 2 + 36. On the other hand, the second equation is equivalent to 9(x + 3)2 = 36y + 36. Subtracting the second equation from the first, we get 4y 2 − 36y = 0, which has solutions y = 0 and y = 9. When y = 0, the corresponding x √ values are −5 and −1. When y = 9, the corresponding x √ valuesare −3 ± 2 10. Therefore, the solutions are (−5, 0), (−1, 0), −3 ± 2 10, 9 .
PY
5. Set up a coordinate system so that the opening of the hose (the parabola’s vertex) is at (0, 3) and that the water flows towards the positive x-axis. Then the x-axis (y = 0) corresponds to the ground; it follows the parabola passes through the point (2, 0). Hence, the equation of the parabola is x2 = − 43 (y−3).
C O
If Nikko stands on a 1.5-ft stool and the vertex remains at (0, 3), the line y = −1.5 will correspond to the ground. will strike the q Hence, the water √ 4 ground when y = −1.5. This gives x = − 3 (−1.5 − 3) = 6. Therefore, the √ water will travel 6 − 2 ft further before striking the ground. 6. Let (x, y) be the coordinates of the point. This point satisfies p 2 (x − 2)2 + y 2 = |y − 5|. 3
D
Manipulating this equation gives us
EP E
4 (x − 2)2 + y 2 = (y 2 − 10y + 25) 9 2 2 9(x − 2) + 5(y + 8y) = 100 9(x − 2)2 + 5(y + 4)2 = 100 + 80 (x − 2)2 (y + 4)2 + = 1. 20 36
D
Therefore, the point traces a vertical ellipse with center at (2, −4). Supplementary Problems 2.1 (page 85) 1. a3 = a1 + (3 − 1)d = 35; a10 = a1 +(10 − 1)d = 77 ⇒ d = 6, a1 = 23 ⇒ a5 = 47. n (2(17) + (n − 1)3) = 30705 ⇒ 3n2 + 31n − 61410 = 0. Using the 3. sn = 2 quadratic formula and noting that n must be a whole number, we have n = 138. a1 a1 (1 − r3 ) 5. We have s = 108 = and s3 = 112 = = 108 (1 − r3 ) ⇒ r = 1−r 1−r 1 − ⇒ a1 = 144. 27 269 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
7. Note that 0.123123 . . . = 0.123 + 0.000123 + 0.000000123 + . . . = 0.123 + 0.123(0.001) + 0.123(0.001)2 + . . ., which is an infinite geometric series with 41 0.123 = . r = 0.001. Thus, 0.123 = 1 − 0.001 333 4 (2a1 + (4 − 1)d) 9. We have s4 = = 80 ⇒ 2a1 + 3d = 40. Since the sum of the 2 first two numbers are one-third of the sum of the last two numbers, we have 1 (a1 + a2 ) = a3 + a4 ⇒ 4a1 + 14d = 0. Combining yields d = 10, and thus 3 a1 = 5, a2 = 15, a3 = 25, a4 = 35.
PY
11. Note that this is a geometric series with common ratio 2n − 1. Thus, the sum will have a finite value if |2n − 1| < 1 ⇒ −1 < 2n − 1 < 1 ⇒ 0 < n < 1. Thus, n ∈ (0, 1).
1. (a)
10 X √ i=3
(b)
2i
5 X
=
x2 x4 x6 x8 x10 + + + + 2 4 8 16 32
D
(c)
√ √ i 10 3 3 3· = + ... + = 26 3 2 2 2 2
5 X x2i i=1
C O
Supplementary Problems 2.2 (page 95)
(−1)i xi−1 = x − x2 + x3 − x4
EP E
i=2
3. (a)
150 X
(4i + 2) = 4
150 X
i=1
(b)
120 X
i+
150 X
i=1
i(i−5) =
i=3
120 X
i=1
2=4
150(151) + 2(150) = 45, 600 2
(i2 −5i)−1(1−5)−2(2−5) =
i=1
120(121)(2(120) + 1) +10 = 6
D
583, 230
(c)
130 X
(2i−3)(2i+3) =
i=1
130 X
2
(4i −9) = 4
i=1
130 X
2
i−
i=1
130 X
9=
i=1
130(131)(2(130) + 1) + 6
9(130) = 741, 975 5. s =
200 X i=1
2
(i − 1) − i
2
=
200 X
(1 − 2i) ⇒
i=1
200 X i=1
i=
200 − s 2
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Supplementary Problems 2.3 (page 108) 1. Part 1.
1 1+2 =2− 1 2 2
The formula is true for n = 1. Part 2. k X i k+2 Assume: =2− k . i 2 2 i=1
PY
k+1 X k+3 i = 2 − k+1 . To show: i 2 2 i=1
C O
k+1 k X X i i k+1 = + k+1 i i 2 2 2 i=1 i=1
k+2 k+3 + k+1 2k 2 k+3 = 2 − k+1 . 2
D
=2−
3. Part 1.
EP E
1(1!) = (1 + 1)! − 1
The formula is true for n = 1. Part 2.
Assume:
k X
i(i!) = (k + 1)! − 1.
D
i=1
To show:
k+1 X
i(i!) = (k + 2)! − 1.
i=1 k+1 X i=1
i(i!) =
k X
i(i!) + (k + 1)[(k + 1)!]
i=1
= (k + 1)! − 1 + (k + 1)[(k + 1)!] = (k + 2)(k + 1)! − 1 = (k + 2)! − 1.
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5. Part 1. 1−
1 1 1 = = 2 2 2(1)
The formula is true for n = 2. Part 2.
PY
1 1 1 1 1 · 1− ··· 1 − · 1− = . Assume: P = 1 − 2 3 k−1 k 2k 1 1 1 1 To show: 1 − ··· 1 − · 1− = . 2 k k+1 2(k + 1)
7. Part 1.
C O
1 1 1 1− ··· 1 − =P · 1− 2 k+1 k+1 k 1 · = 2k k + 1 1 . = 2(k + 1)
D
43(1)+1 + 23(1)+1 + 1 = 273 = 7(39) The number is divisible by 7 for n = 1.
EP E
Part 2.
Assume: 43k+1 + 23k+1 + 1 is divisible by 21. Prove: 43(k+1)+1 + 23(k+1)+1 + 1 is divisible by 21. 43(k+1)+1 +23(k+1)+1 +1 = 64·43k+1 +8·23k+1 +1 = 56·43k+1 +8 43k+1 + 23k+1 + 1 − 7
D
9. Part 1.
52(1)+1 · 21+2 + 31+2 · 22(1)+1 = 1216 = 19(64)
The number is divisible by 19 for n = 1.
Part 2. Assume: 52k+1 · 2k+2 + 3k+2 · 22k+1 is divisible by 19. Prove: 52(k+1)+1 · 2(k+1)+2 + 3(k+1)+2 · 22(k+1)+1 is divisible by 19. 52(k+1)+1 · 2(k+1)+2 + 3(k+1)+2 · 22(k+1)+1 = 50 · 52k+1 · 2k+2 + 12 · 3k+2 · 22k+1 = 12 (52n+1 · 2n+2 + 3n+2 · 22n+1 ) + 38 · 52n+1 · 2n+2
272 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
11. Part 1.
101 5 + + 41+2 = 1029 = 3(343) 3 3 The number is divisible by 3 for n = 1.
Part 2. 10k 5 + + 4k+2 is divisible by 3. 3 3 10k+1 5 + + 4k+3 is divisible by 3. Prove: 3 3 k k+1 5 5 k+3 10k 5 10 5 10 k+2 k+2 −6·4n+2 −9· + +4 = 10· + +4·4 = 10 + +4 3 3 3 3 3 3 3
13. Part 1. 1 ≤ 2 −
PY
Assume:
1 =1 1
C O
Part 2 k X 1 1 ≤2− Assume: 3 i k i=1 k+1 X 1 1 Prove: ≤2− 3 i k+1 i=1
EP E
D
k+1 X 1 1 (k + 1)3 − k 1 ≤ 2 − + = 2 − . Note that 0 < (k + 1)2 ⇒ 3 3 3 i k (k + 1) (k + 1) i=1 (k + 1)3 − k (k + 1)2 1 (k + 1)2 < (k + 1)3 − k, thus 2 − < 2 − =2− . 3 3 (k + 1) (k + 1) k+1
Supplementary Problems 2.4 (page 119)
D
1. (a) (2x − 3y)5 = 32x5 − 240x4 y + 720x3 y 2 − 1080x2 y 3 + 810xy 4 − 243y 5 √ 4 x 2 32 8 1 8 (b) − 2 = − x−11/2 + 16x−8 + x−3 + x2 − x−1/2 3 x 3 3 81 27 √ √ 4 (c) (1 + x) = 4x3/2 + x2 + 6x + 4 x + 1 10
3. Approximating yields (2.1)
≈
4 X 10 k=0
k
210−k (0.1)k = 1667.904, which has
an approximate error of −0.08. 19 X 19 5. In sigma notation we have (−3)k = (1 − 3)19 = (−2)19 . k k=0
273 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Topic Test 1 for Unit 2 (page 121) 1. (a) G, r = 3/2
(b) O
(c) O
2. First, a3 = a2 + 5 = a1 + 10. Also, a2 = 10, and a3 = 15.
i2
i=1
Thus,
+ 3i + 2
50 X 2i3 + 9i2 + 13i + 6
=
i2 + 3i + 2
i=1
=
(2i + 3)(i + 1)(i + 2) = 2i + 3. (i + 1)(i + 2)
50 X
(2i + 3) = 2700.
PY
3. We have
50 X 2i3 + 9i2 + 13i + 6
a3 + 1 a2 + 2 = . Thus, a1 = 5, a2 + 2 a1 + 4
i=1
C O
k 8 16−2k 1 =⇒ 16 − 2k = 8 =⇒ k = 4 4. (a) x − 2 k 4 35 8 8 1 = x8 =⇒ x − 2 8 4 28 28 19 27 19 28−19 (b) k = 19 =⇒ (n3 ) (−3m)19 = − 3 n m 19 19
D
5. For n = k + 1:
EP E
1 1 1 + + ··· + 1·3 3·5 (2(k + 1) − 1)(2(k + 1) + 1) 1 2k 2 + 3k + 1 k+1 k + = = = 2k + 1 (2k + 1)(2k + 3) (2k + 1) · (2k + 3) 2k + 3
D
6. a1 = 10, 000, r = 1.04, 60 − 20 = 40 1 − (1.04)40 s40 = 10, 000 · ≈ 499, 675.83 pesos 1 − (1.04) Topic Test 2 for Unit 2 (page 122) 1. (a) G, r = 4/5
(b) O
(c) A, d = 5/2
2. We have a1 + a2 = 2a1 + d = 9 and a1 + a2 + a3 = 3a1 + 2d = 9 yielding a1 = 9, d = −9. Using sn = −126, we get n = 7. 3. (a)
50 X
(2i + 1)(i − 3) =
i=1
(b)
30 X i=1
50 X
2i2 − 5i − 3 = 79, 472
i=1
r
i2 − 2i + 1 = 4
30 X i=1
i−1 435 = 2 2 274
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
k 1 8 8 24−4k 8 3 8−k 4. (x ) = x =⇒ 24 − 4k = 0 =⇒ k = 6 =⇒ = 28 k x k 6 5. (a) For n = k + 1: 1 + 4 + 7 + · · · + (3(k + 1) − 2) k(3k − 1) 3k 2 + 5k + 2 (k + 1)(3k + 2) = + (3k + 1) = = 2 2 2
Supplementary Problems 3.1 (page 133)
PY
(b) For n = k + 1: 3(n+1) + 7(n+1)−1 + 8 = 7 (3n + 7n−1 + 8) − 4 · 3n − 6 · 8, where 4 · 3n is divisible by 12 for n ≥ 1, and 6 · 8 = 48 = 12(4).
5. 2110◦ − 5(360◦ ) = 310◦
C O
360 6 6 rev = rev 1. = 432◦ 5 5 1 rev π 6π 6π 24π ◦ 3. 216 = 216 = rad; s = 4 = cm 180 5 5 5
2 189π 9 7π = cm2 2 6 16 s √ 5π 2(15) 6 π 6 2 ◦ 9. θ = 150 = ; A = 15 in ; r = in =√ = 5π 6 π π 6 s 6 180◦ 11. r = 6 in; s = 6 in; θ = = = 1 rad; 1 rad = 1 ≈ 57.30◦ r 6 π 13.
EP E
D
7π 9 1 7. θ = ; r = cm; A = 6 2 2
8π cm 3
D
π 15. θ = 20◦ = rad; A = 800 cm2 9 v √ √ √ u 2(800) 120 120 π 120 π π 40 π u r = t π = √ = cm; s = = cm π π 9 3 π 9 17. r = 6 cm; θ = 54◦ =
3π 10
1 2 3π 54π Area of shaded region = 2×area of sector AOE = 2 (6) = cm2 2 10 5 √ √ 1 2π 1 19. Asegment = Asector − Atriangle = (6)2 − (3)(6 3) = (12π − 9 3) cm2 2 3 2 275 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Supplementary Problems 3.2 (page 143) 1.
33π 33π π 33π is coterminal with − 8π = , and terminates in QI. 4 4 4 4
=⇒
√ 13 sec θ = 3
2 √ 17 − 4 13 3 = 9 2 3
C O
2 7. tan θ = − , cos θ > 0 3 √ 13 − sec θ + tan θ = √3 sec θ − tan θ 13 + 3
PY
3. The secant function is positive in QI and QIV. The cotangent function is negative in QII and QIV. Therefore, the angle θ is in QIV. ! √ π 5π 3 1 5π is in QII. The reference angle is , and therefore P = − , . 5. 6 6 6 2 2
√ 2 r 2 3 9. csc θ = 2, cos θ < 0; r = 2, y = 1, x = − 3; sec θ = = √ = − x 3 − 3 √
D
EP E
D
11. csc θ = −4 and θ not in QIII =⇒ θ in QIV 4 csc θ = =⇒ r = 4, y = −1 −1 p √ √ x = (4)2 − (−1)2 = ± 15, θ is in Quadrant IV, x = 15 √ √ x 15 r 4 15 y 1 cos θ = = sec θ = = sin θ = = − r 4 x 15 r 4 √ √ y 15 x r tan θ = = − cot θ = = − 15 csc θ = = −4 y x 15 y p √ 13. x = −2, y = 4 =⇒ r = (−2)2 + (4)2 = 2 5 √ √ √ x 5 r y 2 5 cos θ = = − sec θ = = − 5 sin θ = = r 5 x r 5 r √ y x 1 tan θ = = −2 csc θ = = 52 cot θ = = − y x y 2 √ p √ √ 10 2 2 15. x = 2, y = −6, r = (2) + (−6) = 2 10; sec θ = 10, csc θ = − 3 10 80 sec2 θ − csc2 θ = 10 − = 9 9 17. cos θ = sin 2π = 3
√
3 2
and
3π 2
< θ < 2π
=⇒
θ=
11π 6
276 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
19. f (x) = sin 2x + cos 2x + sec 2x + csc 2x + tan 2x + cot 2x √ √ √ 7π 2 2 √ =− + + 2 − 2 − 1 − 1 = −2 f 8 2 2 Supplementary Problems 3.3 (page 170)
π 4π k
= 2 =⇒ k = 8
9. y = 3 sec 2(x − π) − 3
1 4
5. y = 3 sin 34
8π 9
+
2π 3
− 5 = − 13 2
4
,8 3 3
PY
= 8π
7. domain = R; range =
1. P = 3.
2π
11. Asymptotes: x =
3π 2
+ 2kπ, k ∈ Z
C O
13. (a) P = 8π, phase shift = − π4 , domain = R, range = [−3, 1]
π 3
+ kπ, k ∈ Z , range = R
D
EP E
D
(b) P = π, phase shift = − π6 , domain = x|x 6=
277 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
π , 2
domain =
x|x 6=
π 2
+
2kπ , 3
k ∈ Z , range =
C O
PY
(c) P = 4π , phase 3 shift = 3 1 −∞, − 2 ∪ − 2 , ∞
π 12
+
kπ , 2
k ∈ Z , range =
D
EP E
D
(d) P = π, phase shift = − π6 , domain = x|x 6= (−∞, 1] ∪ [3, ∞)
1 15. y = 8 cos 10 (t − 10π); at t = 10, y ≈ −4.32 (that is, the mass is located about 4.32 cm below the resting position)
278 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Supplementary Problems 3.4 (page 179) sin x tan x − sin x sin x sin x 1 1. = cos x − = · − 1 = sec x − 1 sin x sin x sin x cos x sin x sin A + sin2 A cos2 A sin A + 1 cos2 A = = =1 3. sin A + 1 + sin A 1 + sin A 1 + sin A
PY
1 1 cos x + sin x + csc x + sec x x cos x = cos x + sin x 5. = sin x cos x = sin cos x sin x cot x + tan x cos2 x + sin2 x + sin x cos x sin x cos x
C O
sin x sin x + sin x cos x − sin x tan x + sin x cos x 7. = cos x = 1 1 + cos x cos x csc x + cot x + sin x sin x sin x sin x(1 + cos x) sin2 x 1 − cos2 x cos x = = = 1 + cos x cos x cos x sin x cot θ cot θ a cos θ = = = 2 2 sin θ csc θ 1 + cot θ 1 + a2
D
9. sin θ cos θ = sin2 θ ·
cos a = sec a + tan a
15.
cos2 a cos a 1 − sin2 a = = = 1 − sin a 1 sin a 1 + sin a 1 + sin a + cos a cos a
1 1 1 + cos a + 1 − cos a 2 2 + = = = = 2 csc2 a 1 − cos a 1 + cos a (1 − cos a)(1 + cos a) 1 − cos2 a sin2 a
D
13.
EP E
1 1 + sin a +1 csc a + 1 1 + sin a 11. = sin a = sin a = 1 1 − sin a csc a − 1 1 − sin a −1 sin a sin a
tan α = 17. 1 − tan2 α
sin α sin α cos α cos α = sin2 α cos2 α − sin2 α 1− 2 cos2 α sin α sin α cos2 α = cos α cos2 α − sin2 α sin α cos α sin α cos α = = 2 2 cos α − (1 − cos α) 2 cos2 α − 1
279 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
cos α sin α − cot α − sin α sec α 19. = sin α cos α 1 1 sec α csc α · cos α sin α cos2 α − sin2 α · cos α sin α = cos2 α − sin2 α = sin α cos α Supplementary Problems 3.5 (page 188)
=
π 2 π 2 π 2 π 2 π 2 π 2
θ=
11π 6
+ kπ − B + kπ − B + kπ cos B − cos + kπ cos B + sin + kπ cos B + kπ sin B
π 2 π 2
C O
sin = cos
=⇒
PY
√ 2π 3 1. cos θ = sin = and θ in QIV 3 2 π sin 3. tan A = tan + kπ − B = 2 cos
sin sin
+ kπ sin B + kπ sin B
= cot B
EP E
D
5. sin 105◦ − cos 15◦ = sin(90◦ + 15◦ ) − cos 15◦ = cos 15◦ − cos 15◦ = 0 √ 7. cot α = 7, csc β = 10, and α and β are acute √ √ √ √ 3 2 7 2 2 10 , sin α = , sin β = , cos β = =⇒ cos α = 10 10 10 10 cos(α + β) = cos α cos β − sin α sin β √ ! √ ! √ ! √ ! √ 7 2 2 10 3 2 2 5 = − = 10 10 10 10 5 2 3 sin(x − π) + sin(x + π) = sin x cos π − cos x sin π + sin x cos π + cos x sin π 2 4 = 2 sin x cos π = 2 (−1) = − 3 3
D
9. 3 sin x = 2 =⇒ sin x =
4 3 and A in QII =⇒ cos A = − 5 5 4 3 cos B = and B in QIV =⇒ sin B = − 5 5 4 4 −3 −3 7 (a) sin(A − B) = − = 5 5 5 5 25
11. sin A =
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−3 4 4 −3 24 (b) cos(A − B) = + =− 5 5 5 5 25 7 7 (c) tan(A − B) = 25 = − 24 24 − 25 cos(A − B) < 0 and sin(A − B) > 0 =⇒ A − B in QII 4 5 and cos β = 5 13 sin(α + β) + sin(α − β) = sin α cos β + cos α sin β + sin α cos β − cos α sin β 5 8 4 = = 2 sin α cos β = 2 5 13 13 √
17, A in QI
=⇒
√
csc B =
34 , 3
B in QI
=⇒
=⇒
A + B = 45◦
D
1 3 + 4 5 = 1 tan(A + B) = 1 3 1− 4 5
1 4 3 tan B = 5 tan A =
C O
15. csc A =
PY
13. Given: sin α =
π 23π + tan π 23π 3π 9 36 17. = tan + = tan = −1 π 23π 9 36 4 1 − tan tan 9 36
EP E
tan
19. sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ Topic Test 1 for Unit 3 (page 190)
D
π 5π 1. r = 6 cm, θ = 37.5 = 37.5 = rad 180 24 5π 5π (a) s = 6 = cm 24 4 1 2 5π 15π (b) A = (6) = cm2 2 24 4 p √ 2. x = −1, y = −2, r = (−1)2 + (−2)2 = 5 √ √ √ −2 5 − 5 10 − 3 5 + +2= sin θ + cos θ + tan θ = 5 5 5 ◦
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12 5 12 , A is in QII =⇒ cos A = − , tan A = − 13 13 5 5 3 4 3 cos B = − , B is in QIV =⇒ sin B = − , cos B = , tan A = − 3 5 5 4 (a) cos(A − B) = cos A cos B + sin A sin B 5 4 12 3 56 = − + − =− 13 5 13 5 65 3 12 − + − 33 tan A + tan B 5 4 =− = (b) tan(A − B) = 12 3 1 + tan A tan B 56 1+− − 5 4
PY
3. sin A =
tan 57◦ + tan 78◦ = tan(57 + 78)◦ = tan 135◦ = −1 1 − tan 57◦ tan 78◦ p √ cos x tan x + sin x sin x 5. = cos x + = 2 cos x = 2 1 − sin2 x = 2 1 − a2 tan x tan x
C O
4.
EP E
D
6. cos6 x + sin6 x = (cos2 x)3 + (sin2 x)3 = (cos2 x + sin2 x)(cos4 x − cos2 x sin2 x + sin4 x) = cos4 x − cos2 x sin2 x + sin4 x = cos4 x − cos2 x(1 − cos2 x) + (1 − cos2 x)2 = cos4 x − cos2 x + cos4 x + 1 − 2 cos2 x + cos4 x = 3 cos4 x − 3 cos2 x − 1
D
7. Connect the three diagonals of the hexagon. In doing!this, the hexagon is √ 1 3 , . Same coordinates divided into 6 equilateral triangles. Hence, B 2 2 for C, E and F , except that they will just vary in signs depending on the quadrant. x π 1 2π 8. y = 2 sin + − 1 =⇒ y = 2 sin x+ −1 2 3 2 3 2π P = 4π, Phase Shift = , Amplitude = 2, Range = [−3, 1] 3
282 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
Topic Test 2 for Unit 3 (page 191) π π π 1. Asector = cm2 , θ = 30◦ = 30 = rad 3 180 6 π π π 1 π 2 = r =⇒ r = 2 cm =⇒ 2 = cm 3 2 6 6 3 p 2. x = 8, y = −6, r = (8)2 + (−6)2 = 10
PY
(sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ 8 1 −6 = = 1 + 2 sin θ cos θ = 1 + 2 10 10 25
4. sin 160 cos 35 − sin 70 cos 55 = sin 20 cos 35 − cos 20 sin 35
C O
8 3. sin A = − 17 π π 15 −8 7 sin − A + cos − A = cos A + sin A = + = 2 2 17 17 17
√
EP E
D
√ 2− 6 = sin(20 − 35) = − sin(45 − 30) = 4 π π √ π π tan + tan √ 7π 1+ 3 4 3 √ = −2 − 3 5. tan = = tan + = π π 12 4 3 1− 3 1 − tan tan 4 3
D
3 4 4 6. cos A = − , A is in QIII =⇒ sin A = − , tan A = 5 5 3 24 24 7 tan B = , B is in QIII =⇒ sin B = − , cos B = − , 7 25 25 (a) sin(A + B) = sin A cos B + cos A sin B 4 7 3 24 4 = − − + − − = 5 25 5 25 5 4 24 1− 1 − tan A tan B 3 3 7 =− (b) cot(A + B) = = 4 24 tan A + tan B 4 + 3 7 7.
tan2 x tan2 x tan x = = = sin x cos x 3 2 tan x + tan x tan x(1 + tan x) sec2 x
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8.
sin x = sin x cos x sec x 1 sin x − cos x = 3
2 1 1 =⇒ sin2 x − 2 sin x cos x + cos2 x = (sin x − cos x) = 3 9 1 8 4 =⇒ 1 − 2 sin x cos x = =⇒ −2 sin x cos x = − =⇒ sin x cos x = 9 9 9 π x 1 π − + 2 = − tan x− +2 9. y = tan 18 3 3 6 π P = 3π, phase shift = 6
EP E
D
C O
PY
2
Supplementary Problems 3.6 (page 200) 1. (a) sin 2θ =
(b) cos 2θ = 1 8 √ 1− 5 2
√ 4 2 9
(c) tan 2θ =
√ 4 2 7
(e) csc 2θ =
7 9
(d) sec 2θ =
9 7
(f) cot 2θ =
√ 9 2 8 √ 7 2 8
7. cot 4θ = 1/(tan 4θ) = −7/24
5. tan x =
9. sin2
D
3. cos(2t) =
11.
tan tan
1 y 2 1 y 2
5π 8
=
√ 2+ 2 4
and cos2
5π 8
=
√ 2− 2 4
1 − cos y 1 − cos y − sin y −1 −1 sin y sin y = = sin y sin y + 1 + cos y +1 +1 1 + cos y 1 + cos y 1 − cos2 y − sin y − sin y cos y = sin2 y + sin y + sin y cos y sin2 y − sin y − sin y cos y = sin2 y + sin y + sin y cos y sin y − 1 − cos y = . sin y + 1 + cos y 284
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√ ◦
13. (a) cos 105 =
√ 2+ 3 2
(b) tan 22.5◦ =
√
2−1
Supplementary Problems 3.7 (page 219)
PY
1. (a) sin[sin−1 (1/2)] = 1/2 √ √ (b) cos[cos−1 (− 2/2)] = − 2/2 √ √ (c) tan[tan−1 (− 3)] = − 3 √ √ (d) sin[arctan( 3)] = − 3/2 √ (e) cos[arccos( 2)] does not exist √ (f) tan[arcsin(1/4)] = 15/15 √ (g) cos(sin−1 3/2) = 1/2 3. (a) sin[2 cos−1 (−4/5)] = −24/25
C O
(b) cos[2 sin−1 (5/13)] = 119/169
(c) sin[sin−1 (3/5) + cos−1 (−5/13)] = 33/65 √ (d) cos[sin−1 (1/2) − cos−1 (8/17)] = (15 + 8 3)/34 √ 5. (a) arcsec(− 2) = 3π/4
D
(b) arccsc(−2) = −π/6 √ (c) arccot 3 = π/6
EP E
(d) [sec−1 (−1)] · [cos−1 (−1)] = π · π = π 2 √ (e) 2 cot−1 3 + 3 csc−1 2 = 2(π/6) + 3(π/6) = 5π/6 (f) csc−1 0 does not exist
7. Vertex angle θ should be π/3.
D
Supplementary Problems 3.8 (page 236)
1. Solution set: {π/2, 3π/2, 5π/2, 7π/2, ...} = {(2k + 1)π/2 | k ∈ Z} 3. Solution set: {2kπ/3 | k ∈ Z} 5. Solution set: {π/8, 3π/8, 9π/8, 11π/8} 7. Solution set: {−π/2, π/6} 9. The bullet should be fired with an angle of θ = 60◦ .
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Supplementary Problems 3.9 (page 252) 1. (a) (13, 7π/3), (13, 13π/3)
(c) (15, 7π/4), (15, 23π/4)
(b) (0, 2π), (0, π/4) 3. (a) (−3, 0) √ √ (b) (−3 2/2, 3 2/2)
(c) (0, 5/2) √ (d) (−4 3, −4)
5. (a) r = 4 cos 2θ
r 4
2
−2
0
−4
PY
θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 −2
0
4
2
0
−2
−4
EP E
D
C O
(b)
2
Topic Test 3 for Unit 3 (page 253)
D
1. (a) cos(2θ) = (b) sin(2θ) =
120 (c) tan(2θ) = − 119
119 169 − 120 169
(d) sec(2θ) =
169 119
169 (e) csc(2θ) = − 120
(f) cot(2θ) = − 119 120
2 tan x . 1 − tan2 x √ √ 4− 6− 2 √ (b) tan 7.5◦ = √ 6− 2 9 −1 40 (b) cos sin = 41 41
2. Hint: Use the double-angle identity for tangent tan(2x) = 3. (a) tan 15◦ = 2 −
√
3
103π π 4. (a) tan cot = 6 3 π 3π 5. Solution Set = 0, , 2 2 −1
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6. Solution Set =
π 5π 7π 11π 0, π, 2π, , , , 6 6 6 6
7. r = −2 − 2 sin θ 0
π/6
π/4 π/3 √ √ −2 − 2 −2 − 3
r
−2
−3
θ
5π/6
π
7π/6
r
−3
−2
−1
π/2 −4
5π/4 4π/3 √ √ −2 + 2 −2 + 3
2π/3 3π/4 √ √ −2 − 3 −2 − 2 3π/2 0
D
EP E
D
C O
PY
θ
8. (a) (r, θ) = (−70, tan−1 34 ) √ (b) (r, θ) = (100 2, π4 )
(d) (r, θ) = (7, 0) √ (e) (r, θ) = (2π 2, π4 ) (f) (r, θ) = (13, tan−1
(c) (r, θ) = (7, π2 )
12 ) 5
√
9. (a) (x, y) = ( 32 , 3 2 3 )
(d) (x, y) = (5, 0)
√
(b) (x, y) =
√ ( 452 2 , − 452 2 )
(e) (x, y) = (2π, 0) √
(f) (x, y) = (− 9 2 3 , 92 )
(c) (x, y) = (1, 0) 287
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Topic Test 4 for Unit 3 (page 254) 1. (a) cos(2θ) = − 527 625
336 (b) sin(2θ) = − 625
2. (a) cos2 24◦ ≈ 0.8346
(b) sin2 24◦ ≈ 0.1655
3. (a) tan(π/12) = 2 −
336 527
(c) tan2 24◦ ≈ 0.1983 √ √ 4− 6− 2 √ (b) tan(π/24) = √ 6− 2
√ 3
√ 3 − 16 3 1 −1 3 + cos = 4. cos cos 7 5 35 π 3π 4π 5π 5. Solution set = , , , 4 4 3 3 5π 7π 13π 15π π 5π 6. Solution set = , , , , , 8 8 8 8 4 4 7. r = 2 + 2 cos θ 0
r
4
θ
5π/6 √ 2− 3
π 0
π/3
π/2
2π/3
3
2
1
3π/4 √ 2− 2
7π/6 5π/4 4π/3 3π/2 √ √ 2− 3 2− 2 1 2
D
EP E
r
π/6 π/4 √ √ 2+ 3 2+ 2
D
θ
PY
−1
C O
(c) tan(2θ) =
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8. (a) (r, θ) = (35, tan−1 (− 43 )) √ ) (b) (r, θ) = (100 2, 5π 4
(d) (r, θ) = (5, π) √ (e) (r, θ) = (π 2, π4 )
(c) (r, θ) = (5, − π2 ) √ 9. (a) (x, y) = (2 3, 2) √ √ (b) (x, y) = (−50 2, −50 2)
8 )) (f) (r, θ) = (17, tan−1 ( 15
(d) (x, y) = (−5, 0) (e) (x, y) = (−π, 0) √
, 152 3 ) (f) (x, y) = (− 15 2
D
EP E
D
C O
PY
(c) (x, y) = (−1, 0)
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References
[1] R.N. Aufmann, V.C. Barker, and R.D. Nation, College Trigonometry, Houghton Mifflin Company, 2008. [2] E.A. Cabral, M.L.A.N. De Las Pe˜ nas, E.P. De Lara-Tuprio, F.F. Francisco, I.J.L. Garces, R.M. Marcelo, and J.F. Sarmiento, Precalculus, Ateneo de Manila University Press, 2010.
PY
[3] R. Larson, Precalculus with Limits, Brooks/Cole, Cengage Learning, 2014. [4] L. Leithold, College Algebra and Trigonometry, Addison Wesley Longman Inc., 1989, reprinted by Pearson Education Asia Pte. Ltd., 2002.
C O
[5] M.L. Lial, J. Hornsby, and D.I. Schneider, College Algebra and Trigonometry and Precalculus, Addison-Wesley Educational Publisher, Inc., 2001. [6] J. Stewart, L. Redlin, and S. Watson, Precalculus: Mathematics for Calculus, Brooks/Cole, Cengage Learning, 2012. [7] M. Sullivan, Algebra & Trigonometry, Pearson Education, Inc., 2012.
D
EP E
D
[8] C. Young, Algebra and Trigonometry, John Wiley & Sons, Inc., 2013.
290 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2016.
