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CHAPTER
9
FORM
4
Trigonometry II SPM Topical Analysis
Year
2003
2004
2005
2006
2007
2008
2009
2010
2011
Paper
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
Number of questions
2
0
3
0
3
0
3
0
3
0
3
0
2
0
2
0
3
0
ONCEPT MAP
TRIGONOMETRY II TRIGONOMETRI II Unit Circle (Bulatan Unit) sin θ = y-coordinate (sin θ = koordinat-y) cos θ = x-coordinate (kos θ = koordinat-x) y-coordinate koordinat-y tan θ = ——————− (tan θ = ——————) x-coordinate koordinat-x Signs of sin θ , cos θ and tan θ for 0º θ 360º Tanda sin θ , kos θ dan tan θ bagi 0º θ 360º
Basic angle in quadrant I Sudut asas dalam sukuan I
Values of sin θ , cos θ and tan θ for 0º θ 360º Nilai-nilai sin θ , kos θ dan tan θ bagi 0 º θ 360º
Values of sine, cosine and tangent of special angles Nilai sinus, kosinus dan tangen bagi sudut-sudut khusus
Graphs of sin θ , cos θ and tan θ for 0º θ 360º Graf sin θ , kos θ dan tan θ bagi 0º θ 360º
COMPANION WEBSITE
Learning Objectives
9.1
Values of Sine, Cosine and Tangent of an Angle Facts
9.1a Unit Circle
0°, 90°, 180°, 270° and 360° are not in any quadrant.
1 A unit circle is a circle of radius 1 unit with its centre at the origin. 2 The diagram shows the unit circle. The x-axis and the y-axis divide the unit circle into four quadrants. y 1
4
9
II I Second First quadrant quadrant –1 Third O Fourth 1 quadrant quadrant
CHAPTER
F O R M
III
x
IV
–1
3 In the diagram, the angle θ is positive if point P moves in an anticlockwise direction around the unit circle. If OP is at the (a) first quadrant, 0° θ 90°, (b) second quadrant, 90° θ 180°, (c) third quadrant, 180° θ 270°, (d) fourth quadrant, 270° θ 360°. y (0, 1) 90°
P 0°/360° x (1, 0)
θ
180° (–1, 0)
O 270° (0, –1)
9.1b Coordinates and Ratios of
1
Coordinates for Points on the Unit Circle
y y (a) (b) P –1
1
1 35° O
1
x
–1
O –1
70° P
1
y
x
1 P(0.8, 0.6)
–1
Each of the above diagrams shows a unit circle. Determine (i) the quadrant in which the angle between the radius OP and the positive x-axis lies, (ii) the angle between the radius OP and the positive x-axis.
–1
1
x
–1
For any point on the circumference of a unit circle, (a) the value of the y-coordinate can be determined, i.e. 0.6, (b) the value of the x-coordinate can be determined, i.e. 0.8, (c) the ratio of the y-coordinate to the 0.6 3 x-coordinate can be obtained, i.e. —— =— 0.8 4
Solution (a) (i) Quadrant II (ii) 90° + 35° = 125° (b) (i) Quadrant III (ii) 180° + 70° = 250° Try: Question 1, Self Assess 9.1.
Trigonometry II
O
152
2
x
O N(–0.7, –0.7)
(b) For point N(–0.7, –0.7): (i) y-coordinate = –0.7 (ii) x-coordinate = –0.7 (iii) ratio of y-coordinate to x-coordinate –0.7 = —— –0.7 = 1
The diagram shows the unit circle. Determine (i) the y-coordinate, (ii) the x-coordinate, (iii) the ratio of the y-coordinate to the x-coordinate, of points: (a) M (b) N Try: Question 2, Self Assess 9.1.
9.1c Values of sin θ, cos θ and tan θ in
3
Quadrant I of the Unit Circle
y
y 1 1
–1
P(x , y)
1 y θ O x Q
1
0.5
x
30° 0.86 1
O
–1
x
In quadrant I, P is a point on the circumference of the unit circle with centre O. θ is the angle between the radius OP and the positive x-axis. From the diagram, PQ y (a) sin θ = —— = — = y OP 1 OQ x (b) cos θ = —— = — = x OP 1 PQ y (c) tan θ = —— = — OQ x
The diagram shows quadrant I of the unit circle. Use the diagram to determine the values of (a) sin 0°, (d) sin 30°, (b) cos 0°, (e) cos 30°, (c) tan 0°, (f) tan 30°. Solution
Therefore,
(d) sin 30° = 0.5
(a) sin 0° = 0 (b) cos 0° = 1 (c) tan 0° = 0
y-coordinate x-coordinate y-coordinate
———————— x-coordinate
(e) cos 30° = 0.86
sin θ = y-coordinate cos θ = x-coordinate y-coordinate tan θ = ——————— x-coordinate
0.5 = 0.58 (f) tan 30° = —— 0.86 Try: Question 3, Self Assess 9.1.
153
Trigonometry II
CHAPTER
M(0.5, 0.86)
9
Solution (a) For point M(0.5, 0.86): (i) y-coordinate = 0.86 (ii) x-coordinate = 0.5 (iii) ratio of y-coordinate to x-coordinate 0.86 = —— 0.5 = 1.72
y
F O R M 4
9.1d Signs of Trigonometric Ratios
5 In summary, the signs of the values of the trigonometric ratios of an angle in all four quadrants are shown in the following diagram.
1 The values of the trigonometric ratios of an angle in quadrant I (0° θ 90°) are as follows:
4
CHAPTER
F O R M
9
y sin θ=— r
y
Positive
sin (+) Quadrant II sin θ = (+) cos θ = (–) tan θ = (–)
P(x, y )
x θ =— cos r
Positive
y tan θ = — x
Positive
r O
y
θ x
x
Q
180°
2 The values of the trigonometric ratios of an angle in quadrant II (90° θ 180°) are as follows: y
0°/360° tan (+) Quadrant III sin θ = (–) cos θ = (–) tan θ = (+)
P(–x, y) y
θ
x
4
Positive
Determine whether the value of each of the following is positive or negative. (a) tan 50° (b) sin 110° (c) cos 220° (d) tan 345°
–x cos θ = —– Negative r y tan θ = —– –x
Negative
Solution (a) tan 50° is positive because 0° 50° 90° (in quadrant I). (b) sin 110° is positive because 90° 110° 180° (in quadrant II). (c) cos 220° is negative because 180° 220° 270° (in quadrant III). (d) tan 345° is negative because 270° 345° 360° (in quadrant IV).
3 The values of the trigonometric ratios of an angle in quadrant III (180° θ 270°) are as follows: –y sin θ = —– Negative r –x cos θ = —– Negative r –y tan θ = —– –x Positive y x =—
y
θ
–x
x
O
–y
r Try: Question 4, Self Assess 9.1.
P(–x, –y )
4 The values of trigonometric ratios of an angle in quadrant IV (270° θ 360°) are as follows: Negative
x cos θ =— r
Positive
–y tan θ = —– Negative x Trigonometry II
9.1e Special Angles 1 Special angles are angles with the values of sine, cosine or tangent equal to 0, ±0.5, ±1 or undefined. Thus, in the range of 0° θ 360°, 0°, 30°, 45°, 60°, 90°, 180°, 270° and 360° are special angles. 2 The following table shows the values of sine, cosine and tangent of the special angles.
y
–y sin θ = —– r
cos (+) Quadrant IV sin θ = (–) cos θ = (+) tan θ = (–) 270°
r
Q –x O
y sin θ=— r
90° All (+) Quadrant I sin θ = (+) cos θ = (+) tan θ = (+)
θ
x
O r
x –y P(x, –y )
154
θ
sin θ
cos θ
tan θ
0°
0
1
0
1 — = 0.5 2 1 —– = 0.7071 √2
√3 —– = 0.866 2 1 —– = 0.7071 √2
1 —– = 0.5774 √3
60°
√3 —– = 0.866 2
1 — = 0.5 2
√3 = 1.732
90°
1
0
undefined or ∞
180°
0
–1
0
270°
–1
0
undefined or –∞
360°
0
1
0
1
3 The values of sine, cosine and tangent of special angles are shown in the following diagrams. (c) (a) 30° 30°
2
sin 90° = 1 cos 90° = 0 tan 90° = ∞
2
3 60°
60°
1
1
90° 1
√3 1 sin 30° = — sin 60° = —– 2 2 3 √ 1 cos 30° = —– cos 60° = — 2 2 1 tan 30° = —– tan 60° = √3 √3 (b)
45°
2
1
sin 180° = 0 cos 180° = –1 tan 180° = 0
–1 180°
1 O
0°/360°
270° –1
sin 270° = –1 cos 270° = 0 tan 270° = –∞
1 sin 45° = —– √2
sin 0° = 0 cos 0° = 1 tan 0° = 0 sin 360° = 0 cos 360° = 1 tan 360° = 0
1 cos 45° = —– √2
45° 1
tan 45° = 1
5 Find the values of the following. (a) 4 sin 90° – 2 cos 60° (b) 3 sin 270° + 4 cos 180° (c) 2 tan 45° – 3 sin 30° (d) 3 tan 60° + 2 cos 360° Solution (a) 4 sin 90° – 2 cos 60° 1 = 4(1) – 2— 2 =4–1 = 3
(b) 3 sin 270° + 4 cos 180° = 3(–1) + 4(–1) = –3 – 4 = –7 (c) 2 tan 45° – 3 sin 30° = 2(1) – 3(0.5) = 2 – 1.5 = 0.5 (d) 3 tan 60° + 2 cos 360° = 3(1.732) + 2(1) = 5.196 + 2 = 7.196
Try: Question 5, Self Assess 9.1.
155
Trigonometry II
9
45°
CHAPTER
30°
F O R M 4
6 State the values of θ in each of the following for 0° θ 360°. (a) sin θ = 0 (b) cos θ = –1 (c) tan θ = 0 Solution (a) sin θ = 0 ∴ θ = 0°, 180° and 360°
(b) cos θ = –1 ∴ θ = 180°
cos 180° = –1 tan 0° = 0
(c) tan θ = 0 ∴ θ = 0°, 180° and 360°
sin 0° = 0
tan 180° = 0 tan 360° = 0
sin 180° = 0 sin 360° = 0
4
CHAPTER
F O R M
9
Try: Question 6, Self Assess 9.1.
9.1f Corresponding Angles 1 Angles in quadrants II, III and IV have corresponding angles in quadrant I which are the acute angles corresponding to them. The corresponding angles are known as basic angles. 2 The table below shows the corresponding angle, α, in quadrant I which corresponds to the angle θ in quadrants II, III or IV. Value of angle
Quadrant
Value of corresponding angle in quadrant I
90° θ 180°
II
α = 180° – θ
Diagram y P
θ
α
x
O
180° θ 270°
α = θ – 180°
III
y
θ
x
O
α P
270° θ 360°
α = 360° – θ
IV
y
θ O
x
α P
7 Determine the value of the angle in quadrant I which corresponds to the value of each of the following angles. (a) 145° (b) 200° (c) 325° Solution (a) 145° lies in quadrant II. The corresponding angle in quadrant I = 180° – 145° = 35°
(b) 200° lies in quadrant III. The corresponding angle in quadrant I = 200° – 180° = 20°
200°
(c) 325° lies in quadrant IV. The corresponding angle in quadrant I = 360° – 325° = 35°
145° 35°
O
x
156
x
O
20°
y
Try: Question 7, Self Assess 9.1.
Trigonometry II
y
y
O 325°
35°
x
9.1g Relationship between
Trigonometric Ratios of Corresponding Angles y
2 In quadrant II, where 90° θ 180°,
90° All (+)
180° 180° – θ
0°
θ
θ – 180° 360° – θ 360° tangent (+)
sin θ = sin (180° – θ) cos θ = –cos (180° – θ) tan θ = –tan (180° – θ)
x
cosine (+)
In quadrant III, where 180° θ 270°,
270°
sin θ = –sin (θ – 180°) cos θ = –cos (θ – 180°) tan θ = tan (θ – 180°)
The chart shows the operations of θ for 0° θ 360°. For 90° θ 360°, the following steps are used to state the relationship between sin θ, cos θ and tan θ and their corresponding angles, and to determine their values. Step 1: Determine the quadrant where θ is in and determine the signs of the values of sin θ, cos θ and tan θ.
9
sine (+)
In quadrant IV, where 270° θ 360°, sin θ = –sin (360° – θ) cos θ = cos (360° – θ) tan θ = –tan (360° – θ)
8 State the relationship between (a) sin 136°, (b) cos 225°, (c) tan 320°, and their respective values of corresponding angles in quadrant I. Then find their values.
y
Solution (a) asic angle in B
Since 225° is in quadrant III, cos 225° is negative. θ = 225°, cos 225° = –cos (225° – 180°) = –cos 45° cos 225° = –0.7071 (c)
y
320°
sine (+)
quadrant II
O 136°
44°
40°
x
asic angle in B quadrant IV
tangent (–) x
O
Since 320° is in quadrant IV, tan 320° is negative. θ = 320°, tan 320° = –tan (360° – 320°) = –tan 40° tan 320° = –0.8391
Since 136° is in quadrant II, sin 136° is positive. θ = 136°, sin 136° = sin (180° – 136°) = sin 44° sin 136° = 0.6947 (b) y 225°
asic angle in B quadrant III
45°
O
x
cosine (–) Try: Question 8, Self Assess 9.1.
157
Trigonometry II
CHAPTER
1
Step 2: Determine the values of the corresponding angles to θ in quadrant I and find the values of sin θ, cos θ and tan θ.
F O R M 4
9 (b) 4 tan 320° + 5 sin 198° = 4(–0.8391) + 5(–0.3090) = –3.3564 – 1.5450 = –4.9014
Find the value of each of the following using a calculator. (a) 3 cos 142° – 4 sin 278° (b) 4 tan 320° + 5 sin 198° (c) 7 cos 223° – 8 tan 335°
Press cos 142 = Display: –0.788010753
4
CHAPTER
F O R M
9
Solution
(a) 3 cos 142° – 4 sin 278° = 3(–0.7880) – 4(–0.9903) P ress = –2.3640 + 3.9612 sin 278 = = 1.5972 Display: –0.990268068
(c) 7 cos 223° – 8 tan 335° = –1.3890 Students can also obtain the answer directly by pressing: 7 cos 223 – 8 tan 335 = Display: –1.389014646
Try: Question 9, Self Assess 9.1.
(b) Draw a chart of four quadrants to determine the possible quadrant in which θ can be found, based on the sign (+ or –) of the value of sin θ, cos θ or tan θ. S A
9.1h Finding the Angles between 0°
and 360° Given Values of sin θ, cos θ or tan θ
If the value of sin θ, cos θ or tan θ is given and 0° θ 360°, the value of θ can be found using the following steps. (a) Find the basic angle in quadrant I which corresponds to θ.
T
C
(c) Find the values of θ in the right quadrants found in (b).
10 cos θ = –0.6025 Basic ∠ = 52.95° ∴ θ = 180° + 52.95° = 232.95°
(a) Given that sin θ = 0.6025 and 90° θ 180°, find the value of θ. (b) Given that cos θ = –0.6025 and 180° θ 270°, find the value of θ. (c) Given that tan θ = –1.732 and 0° θ 360°, find the values of θ.
(c)
1 80° θ 270°
90° S
Solution (a)
90° S 180°
(b)
270°
C
sin θ = 0.6025 Basic ∠ = 37.05° ∴ θ = 180° – 37.05° = 142.95°
tan θ = –1.732 Basic ∠ = 60° θ1 = 180° – 60° = 120° θ2 = 360° – 60° = 300° ∴ θ = 120° and 300°
SHIFT sin–1
0.6025 = Display: 37.04915756 9 0° θ 180°
90° S 180°
A
θ
0°/360°
52.95° T
C
T
0°/360°
Press
0°/360°
60°
θ2
θ
37.05°
A
θ1
60°
180°
A
T
C 270°
Try: Questions 10 – 14, Self Assess 9.1.
Trigonometry II
When finding basic angle using a calculator, do not press the (–) sign.
158
θ is negative tan in quadrant II and quadrant IV.
0° θ 360°
1
SPM Clone
’08
Paper 1/ Compulsory question/ 1 mark
5 Given that tan θ = —–, where 180° < θ < 360°, 12 find the value of sin θ. 12 5 A – —– C —– 13 13 5 12 B – —– D —– 13 13
PQ 5 sin θ = —– = – —– OQ 13 θ is positive in the tan third quadrant. P
y –12
–5
Solution In ΔOPQ, OQ = √(–5)2 + (–12)2 = 13
θ O
x
13
9
Q(–12, –5)
CHAPTER
Answer: B
9.1i Solving Problems In examinations, students can give their answers of angles in only degrees (correct to at least 1 decimal place) or in degrees and minutes.
2
The knowledge of trigonometric ratios is used to solve problems in daily life.
SPM Clone
’07
Paper 1/ Compulsory question/ 1 mark
CG 16 sin x° = —– = —– = BG BG 4BG = ∴BG =
In the diagram, ABCD and EDF are straight lines. 4 It is given that sin x° = — , CG = 16 cm, 5 CD = 38 cm and DE = 48 cm.
BC = √ 202 – 162 = 12 cm
G
x° A
B
∴BD = BC + CD = 12 + 38 = 50 cm
F
y° D
C
BE = √ BD2 – ED2 = √ 502 – 482 = 14 cm
E
Find the value of tan y°. 24 7 A – —– C —– 25 24 7 24 B – —– D —– 24 25 Solution
BE tan y° = – —– DE 14 = – —– 48 7 = – —– 24
G 16 cm x° A
38 cm
B
C
y° D
4 — 5 80 20 cm
F
Answer: B
48 cm E
159
Since y° is an obtuse angle (in the second quadrant), tan y° is negative.
Trigonometry II
F O R M 4
3
SPM Clone
’06
Paper 1/ Compulsory question/ 1 mark
Solution QR = 7 cm In ΔQRS, RS 2 = 252 – 72 = 576 RS = 24 cm
In the diagram, Q is the midpoint of the straight line PQR. 14 cm Q
P
R
x° 25 cm
tan x° = –tan ∠RQS RS = – —— QR 24 = – —– 7
4
CHAPTER
F O R M
9
S
Find the value of tan x°. 24 24 A – —– C – —– 25 7 24 24 B —– D —– 25 7 SPM Clone
SPM Clone
’04
’11
SPM Clone
’09
In the diagram, ABCD is a parallelogram and CDE is a straight line.
25 cm
E
x°
D
C
C
In the diagram, ABC is a straight line. Given that 7 1 find the length of AC, sin x° = —– and tan y° = —, 25 2 in cm. A 14 C 38 B 24 D 52
A
B
3 Given tan ∠ADE = —, find the value of cos 4 ∠ABC. 3 3 C – — A — 5 5 4 4 B — D – — 5 5
Solution 7 Given I n ΔBCD, sin x° = —– 25 7 BD —– = —– 25 25 BD = 7 cm BC 2 = DC 2 – BD 2 = 252 – 72 = 576 ∴ BC = 24 cm
Solution 3 In ΔADE, tan ∠ADE = — 4 AE 3 —– = — DE 4
1 Given I n ΔABD, tan y° = — 2 7 1 —– = — AB 2 AB = 14 cm ∴ AC = AB + BC = 14 cm + 24 cm = 38 cm
∴ AE = 3, DE = 4 AD = √32 + 42 = 5
4 cos ∠ADC = – cos ∠ADE = – — 5 4 cos ∠ABC = cos ∠ADC = – — 5
Since ABCD is a parallelogram, ∠ABC = ∠ADC.
Answer: C Trigonometry II
which corresponds to the angle x°.
Paper 1/ Compulsory question/ 1 mark
D
B
∠RQS is the basic angle
5
Paper 1/ Compulsory question/ 1 mark
A
(in quadrant II), tan x° is negative.
Answer: C
4
y°
Since 90° x° 180°
Answer: D 160
E 3 A
4
D
11 Since TQ = 12 cm, TS = SQ = 6 cm In ΔSQR, using the Pythagoras’ Theorem, SR = √ 62 + 82 = √ 100 = 10 cm Since 270° x° 360°
T
S
R x°
8 cm
Q
(in quadrant IV), sin x° is
negative. ∴ sin x° = –sin ∠SRQ SQ = – —— SR ∠SRQ is the basic angle which corresponds to the angle x°. 6 = – —– 10 3 = – — 5
In the above diagram, PQR and TSQ are straight lines. Given that PQ = 9 cm, PT = 15 cm, QR = 8 cm and TS = SQ, find the value of sin x°. Solution In ΔPQT, using the Pythagoras’ Theorem, TQ = √152 – 92 = √ 225 – 81 = √ 144 = 12 cm Try: Questions 15 – 17, Self Assess 9.1.
9.1 1 (a) (c) y
The diagram shows a unit circle. Four points P, Q, R and S are marked on the circumference of the unit circle. Determine (a) the y-coordinate, (b) the x-coordinate, (c) the ratio of the y-coordinate to the x-coordinate of each point.
y
1 P
1
20° –1
x
1
O
–1
(d)
y
–1
3
y 1
1 P
1
x
–1
O –1
–1
58° 1
x 75° 48° 22°
P O
Each of the above diagram shows a unit circle. Determine (i) the quadrant in which the angle between the radius OP and the positive x-axis lies, (ii) the angle between the radius OP and the positive x-axis. 2
y
x
4 Determine whether the value of each of the following is positive or negative. (a) sin 80° (e) cos 165° (i) tan 312° (b) sin 150° (f) cos 72° (j) tan 292° (c) sin 230° (g) cos 240° (k) tan 162° (d) sin 335° (h) cos 290° (l) tan 76°
P (0.8, 0.6)
R (–0.4, –0.92)
1
The diagram shows quadrant I of the unit circle. Use the diagram to determine the values of the following. (a) sin 22° (d) sin 48° (g) sin 75° (b) cos 22° (e) cos 48° (h) cos 75° (c) tan 22° (f) tan 48° (i) tan 75°
Q (–0.3, 0.95)
O
y 1
65° O
x
P –1
–1
(b)
1
76° O
x S(0.65, –0.76)
161
Trigonometry II
9
9 cm
P
CHAPTER
15 cm
F O R M 4
4
9 CHAPTER
F O R M
5 Find the value of each of the following. (a) 6 sin 30° – 45 cos 90° + 2 tan 180° (b) 10 cos 0° × 5 sin 90° (c) √ 2 sin 45° + tan 45° (d) 6 cos 60° – 2 tan 45° (e) 5 tan 360° – 2 sin 270° + cos 30° (f) 4 sin 180° + 6 tan 90° (g) 4 sin 270° × sin 90° – 3 sin 180° (h) 5 sin 0° + 6 sin 90° (i) cos 90° + 3 cos 180° + 4 cos 270° (j) 4 cos 90° + 7 cos 360° (k) 2 tan 45° + 4 tan 180° + 7 tan 360° (l) 4 tan 45° – 7 tan 180°
(g) cos 182° + 2 tan 228° (h) 3 tan 346° – 4 cos 303° (i) 2 cos 114° – 3 tan 145° 10 Given that 0° x 360°, find the values of x for each of the following. (a) sin x = 0.5621 (b) sin x = –0.6481 (c) cos x = 0.5120 (d) cos x = –0.9650 (e) tan x = 0.4163 (f) tan x = –0.8421
6 State the value(s) of x in each of the following for 0° x 360°. (a) sin x = 0 (c) cos x = 1 (e) tan x = 0 (b) sin x = 1 (d) cos x = –1 (f) tan x = –1
11 (a) Given that sin y = 0.3272 and 90° y 180°, find the value of y. (b) Given that sin y = – 0.9532 and 180° y 270°, find the value of y.
7 Complete the table below for the corresponding angles.
12 (a) Given that cos y = 0.6198 and 0° y 180°, find the value of y. (b) Given that cos y = –0.7638 and 180° y 360°, find the value of y.
Quadrant Quadrant Quadrant Quadrant I II III IV 100°
260°
280°
13 (a) Given that tan y = 0.6346 and 0° y 90°, find the value of y. (b) Given that tan y = –0.6176 and 180° y 360°, find the value of y.
309°
14 Given that 0° θ 360°, find the values of θ for each of the following. (a) sin θ = sin 72° (b) sin θ = –sin 31° (c) cos θ = cos 34° (d) cos θ = –cos 67° (e) tan θ = tan 85° (f) tan θ = – tan 43°
108° 244°
223° 144° 209° 350° 8 State the relationship between the values of sine, cosine and tangent below and their respective values of corresponding angles in quadrant I. Then find their values. (a) sin 98° (d) sin 203° (g) sin 300° (b) cos 116° (e) cos 242° (h) cos 326° (c) tan 153° (f) tan 260° (i) tan 348°
15 P 17 cm 48° Q
9 Find the value of each of the following. (a) 3 sin 218° – 4 cos 136° (b) cos 315° – 2 sin 196° (c) sin 100° + 2 cos 280° (d) 5 sin 94° + 3 tan 284° (e) 4 tan 239° – 2 sin 341° (f) sin 226° – 3 tan 333° Trigonometry II
R
S
T
In the diagram, QRST is a straight line. Calculate (a) the length of PR, (b) the length of RS, (c) the length of PQ, (d) the value of cos ∠PST. 162
16
17
C B 11 cm
θ°
E F
60°
12 cm
10 cm
13 cm
A
5.74 cm
D A
In the diagram, ABC is a straight line. Calculate the values of (a) tan ∠BDC, (c) sin ∠CAD, (b) cos ∠ABD, (d) tan θ .
B 4 cm C
D
9.2
(d) The value of sin x = 0 when x = 0°, 180° and 360°.
Graphs of Sine, Cosine and Tangent
Try: Questions 1 – 2, Self Assess 9.2.
The graphs of sine, cosine and tangent can be drawn based on the information about special angles of the unit circle. sin 90° = 1 cos 90° = 0 tan 90° = ∞ 1 90° sin 180° = 0 cos 180° = –1 180° –1 tan 180° = 0
2 y = cos x, 0° x 360° y 1
0° 1 360° 270° –1
sin 270° = –1 cos 270° = 0 tan 270° = –∞
sin 0° = 0 cos 0° = 1 tan 0° = 0
0
90° 180° 270° 360°
1 90°
sin 360° = 0 cos 360° = 1 tan 360° = 0
270° –1 cos 270° = 0
(a) The graph of y = cos x is in the form of a wave which is continuous. (b) The maximum value of the cosine graph is 1 when x = 0° and 360°. (c) The minimum value of the cosine graph is –1 when x = 180°. (d) The value of cos x = 0 when x = 90° and 270°.
1 y = sin x, 0° x 360° y 1 x
Try: Questions 3 – 4, Self Assess 9.2. sin 90° = 1 1 90°
–1 sin 180° = 0 180° –1
0° cos 0° = 1 360° cos 360° = 1
cos 180° = –1 180° –1
of Sine, Cosine and Tangent
90° 180° 270° 360°
cos 90° = 0
–1
9.2a Drawing and Comparing Graphs
0
x
3 y = tan x, 0° x 360° y
0° sin 0° = 0 360° sin 360° = 0 270° –1 sin 270° = –1
0
(a) The graph of y = sin x is in the form of a wave which is continuous. (b) The maximum value of the sine graph is 1 when x = 90°. (c) The minimum value of the sine graph is –1 when x = 270°.
90° 180° 270° 360°
x
tan 90° = ∞ 1 90°
tan 180° = 0 180° –1
0° tan 0° = 0 360° tan 360° = 0 270° –1 tan 270° = ∞
163
Trigonometry II
CHAPTER
9
In the diagram, ABCD is a straight line. Calculate (a) the length of EF, (b) ∠CDF, (c) ∠ABF.
F O R M 4
(a) The graph of y = tan x is not continuous at x = 90° and 270°. (b) The graph of y = tan x has no maximum or minimum value. (c) The values of y = tan x range from –∞ to ∞. (d) The value of tan x = 0 when x = 0°, 180° and 360°.
(d) The value of sin 2x = 0 when x = 0°, 90°, 180°, 270° and 360°. 6 y = cos 2x, 0° x 360° y 1
Try: Questions 5 – 6, Self Assess 9.2.
4
0
y y = cos x 225° 0 45°90° 180° 270° 360°
9 CHAPTER
4
x
90° 135° 180° 225° 270° 315° 360°
–1
1 0.7071
F O R M
45°
–0.7071 –1
(a) The graph of y = cos 2x is in the form of a wave which is continuous. (b) The maximum value of the cosine graph is 1 when x = 0°, 180° and 360°. (c) The minimum value of the cosine graph is –1 when x = 90° and 270°. (d) The value of cos 2x = 0 when x = 45°, 135°, 225° and 315°.
x
y = sin x
When the graphs of y = sin x and y = cos x intersect at two points (45°, 0.7071) and (225°, –0.7071), sin x = cos x = 0.7071 and –0.7071, when x = 45° and 225° respectively.
7 y = tan 2x, 0° x 360°
Try: Questions 7 – 8, Self Assess 9.2.
y
5 y = sin 2x, 0° x 360° y 1
0
0
45°
90° 135° 180° 225° 270° 315° 360°
x
x
90° 135° 180° 225° 270° 315° 360°
(a) The graph of y = tan 2x is not continuous at 45°, 135°, 225° and 315°. (b) The graph of y = tan 2x has no maximum or minimum value. (c) The values of y = tan 2x range from –∞ to ∞. (d) The value of tan 2x = 0 when x = 0°, 90°, 180°, 270° and 360°.
–1
(a) The graph of y = sin 2x is in the form of a wave which is continuous. (b) The maximum value of the sine graph is 1 when x = 45° and 225°. (c) The minimum value of the sine graph is –1 when x = 135° and 315°.
6
45°
SPM Clone
’06
Paper 1/ Compulsory question/ 1 mark
Solution The information about special angles of the unit circle is used to draw the graph of y = cos x for 0° x 180°.
Which of the following graphs represents y = cos x for 0° x 180°? y A y C 1
1 0
90° 180°
0
x
x
cos 90° = 0 90°
–1
–1
B
90° 180°
y
D
1 0 –1
Trigonometry II
cos 180° = –1 180°
y 1
90° 180°
x
0
y cos 0° = 1 0°
1 0 –1
90° 180°
x
Answer: C
–1
164
90° 180°
x
7
SPM Clone
SPM Clone
’04 ’08
Paper 1/ Compulsory question/ 1 mark
Solution The information about special angles of the unit circle is used to draw the graph of y = tan x for 0° < x < 180°.
Which of the following represents the graph of y = tan x for 0° < x < 180°? A y C y
180°
x
0
B y
0
D
90°
180°
90° 180°
x
tan 180° = 0 180°
y
x
0
8
90° 180°
y
tan 90° = ∞ 90° tan 0° = 0 0°
0
90° 180°
x
9
90°
x
CHAPTER
0
Answer: C
SPM Clone
SPM Clone
’05
’09
Paper 1/ Compulsory question/ 1 mark
Which of the following graphs represents y = sin x for 0° < x < 270°? A C
Solution For the graph y = sin x, 0° < x < 270°: (a) The maximum value of sin x is 1 when x = 90°. (b) The minimum value of sin x is –1 when x = 270°. (c) When sin x = 0, x = 0° and 180°.
y
y 1
1
0
90° 180° 270°
x
0
90° 180° 270°
x
Answer: C
–1
–1
B
D
y
y
1
0
90° 180° 270°
x
0
90° 180° 270°
x
–1
9.2b Solving Problems The graphs of sine, cosine and tangent are very useful in solving problems involving periodic movements or changes.
12 (b) What trigonometric function does the graph represent? (c) At what hours are the tides at the mean sea level? (d) How many hours do the tides take to reach the maximum height?
The Bay of Fundy in Nova Scotia has the highest tides in the world. In one 12-hour period, the water starts at the mean sea level, rises to 7 metres above, drops to 7 metres below, then returns to the mean sea level. (a) Sketch a graph that shows the level of the tides over a 12-hour period. 165
Trigonometry II
F O R M 4
Solution (a)
(b) Graph of sine (c) At 0, 6 and 12 hours. (d) 3 hours
Height (metre) 7
0
3
6
9
12
Time (hour)
–7 Try: Questions 9 – 10, Self Assess 9.2.
9
4
9.2 1 Sketch the graph of y = sin x for 0° x 360°.
CHAPTER
F O R M
2
(b) the values of sin x when sin x = cos x. 8
y p
y 1
r 0
q
x
360°
y = sin x
0 –1
–1
y = cos x x
r 360° (p, q)
From the above graph of y = sin x, determine the value of p, of q and of r.
From the above diagram, state the value of p, of q and of r.
3 Sketch the graph of y = cos x for 0° x 180°.
9
y
4
Y
p 0
q
r
360°
5 Sketch the graph of y = tan x for 0° x 360°. y 1 135°
0
p
q
r
360°
x
–1
From the graph of y = tan x, determine the value of p of q and of r. y
7
y = cos x
1
0 90° –1
180°
Q
X
In the diagram, O is the centre of a circle of radius 10 cm. A particle P moves from X in an anticlockwise direction on the circumference of the circle at a constant speed. It takes twelve seconds to complete a circle. Meanwhile, a second particle Q moves along the diameter XY. PQ is always perpendicular to XY. The diagram shows the positions of P and Q two seconds after they have started to move from X. (a) Calculate ∠POQ. (b) Calculate the time taken, in seconds, when Q is back at the same position, as shown in the diagram. x 10 The straight line y = —— and the graphs of 360 y = sin x and y = cos x are drawn on the same axes for 0° x 360°. State the number of times (a) the graph of y = cos x cuts the x-axis, (b) the graph of y = sin x intersects the graph of y = cos x, (c) the graph of y = sin x intersects the straight x line y = ——. 360
From the above graph of y = cos x, determine the value of p, of q and of r. 6
O
x
–1
270° 360°
x
y = sin x
The above diagram shows the graphs of y = sin x and y = cos x for 0° x 360°. State (a) the values of x when cos x = 0, Trigonometry II
P
166
9 Paper 1 Multiple-choice Questions
1 Which of the following angles is not in the third quadrant? A 188° C 265° B 217° D 278°
5 Given that sin p° = –0.9397 and 180° p° 270°, find the value of p. A 190 C 240 B 220 D 250
2 The diagram shown is a circle of radius 1 unit with centre (0, 0). y A (0, 1) Q (– 0.866, 0.5)
θ
6 Given that cos q° = 0.6428 and 270° q° 360°, find the value of q. A 280 C 330 B 310 D 350
x
O
B (–1, 0)
Find the value of tan θ. A –0.866 C 0.5 B –0.577 D 1.732
7 Given that tan r° = –1.4281 and 0° r° 360°, the values of r are A 55 and 110 C 125 and 305 B 110 and 220 D 220 and 305
3 The diagram shown is a circle of radius 1 unit with centre O. y P(0.707, 0.707)
Q(–0.866, 0.5)
x
O R(– 0.6, – 0.8)
S(0.6, – 0.8)
Which of the following sides is the hypotenuse of an isosceles triangle? A OS C OQ B OR D OP
9 Given that sin y° = –sin 63° and 90° y° 270°, find the value of y. A 63 B 117 C 243 D 257
4 In the diagram, A, B and C are three points on the circumference of a quadrant.
10 3 cos 90° – 4 sin 180° + 5 tan 45° – 2 sin 270° = A 0 C 5 B 2 D 7
y A (– 0.5, 0.866) B (– 0.707, 0.707)
1 p
C (–0.866, 0.5)
q r O
8 Given that cos x° = cos 25° and 90° x° 360°, find the value of x. A 115 B 155 C 205 D 335
x
Which of the following statements is true about the diagram?
11 sin 250° + cos 250° + tan 250° = A –sin 70° + cos 70° + tan 70° B –sin 70° – cos 70° + tan 70° C sin 70° – cos 70° – tan 70° D sin 70° – cos 70° + tan 70°
167
12 In the diagram, ABC is a straight line. A
D B
9
I cos p = –sin r II tan q = 1 III cos p = –0.866 A III only B I and II only C I and III only D I, II and III
Values of Sine, Cosine and Tangent of an Angle
C
7 find Given that sin ∠DBA = —–, 25 the value of cos ∠DBA. 24 7 A – —– C —– 25 25 7 24 B – —– D —– 25 25 13 In the diagram, HIJ is a straight line and JI = 12 cm. K
60° J
I
H
Find the length, in cm, of KJ. A 24 C 8 B 12 D 6 4 14 In the diagram, cos ∠KJM = —. 5 M 7 cm L
J
32 cm
K
The length of KL, in cm, is A 8 C 25 B 24 D 40 15 In the diagram, PQR is a straight line. S 80° y° P
Q
R
Find the value of cos y°. A 0.1736 C –0.1736 B 0.6428 D –0.6428 Trigonometry II
CHAPTER
9.1
F O R M 4
16 In the diagram, PQRS is a rectangle. P
T
S
19 In the diagram, ADC is a straight line.
13 A —– 11 11 B —– 13 11 C – —– 13 13 D – —– 11
C
x° E
9 cm
D
9 cm
6 cm 80 cm
4
CHAPTER
F O R M
9
Q
R
A
Given that PT = TS, find the value of cos x°. 40 A —– 41 9 B —– 41 40 C – —– 41 9 D – —– 41
B
4 Given that tan ∠EAD = — and 3 9 —– , find the length sin ∠CAB = 41 of DC. A 40 cm C 30 cm B 32 cm D 26 cm 20 In the diagram, ABC is a straight line. C
5 cm
B
B
C
E 4 cm D
1 the Given that cos ∠ADE = —, 5 value of sin ∠ACB is 4 A — 5 3 B — 5 2 C — 5 1 D — 5 18 In the diagram, JKL is a straight line. L
K
J
15 cm
Trigonometry II
10 cm
6 cm
C
’04 D 6 cm A
K
J
The value of tan ∠MKL is 3 3 C – — A — 4 4 3 3 D – — B — 5 5 22 In the diagram, DEF is a straight line.
x° E
13 cm
D
Given that tan ∠DCF = 1, tan x° =
168
y° B
C
25 In the diagram, ABC and BDE are straight lines.
SPM Clone
’04 ’11
A
y°
B
F 2 cm
10 cm
The value of cos y° is 3 3 C – — A – — 4 5 4 4 D – — B – — 3 5
L
12 cm
E
24 In the diagram, ABC is a straight SPM Clone line.
21 In the diagram, JKL is a straight 4 line and sin ∠MJK = —. 5
C
D
x°
Find the value of tan x°. 3 3 C – — A — 4 4 4 4 D – — B — 3 3
Find the value of tan ∠ABD. 12 A —– 5 5 B —– 12 5 C – —– 12 12 D – —– 5
M
B 5 cm
13 cm
H 8 cm I
3 the Given that tan ∠LHI = —, 2 value of sin ∠LIK is 3 4 C — A — 4 5 2 3 D — B — 3 5
F
A
D
8 cm
’03
A
17 In the diagram, ACD is a straight line. A
23 In the diagram, ABCD is a straight SPM Clone line and B is the midpoint of AC.
D x°
C
E
10 cm
3 tan y° = — 2 Given that cos x° = —, 5 3 and AC = 18 cm, find the length, in cm, of DE. A 8 C 12 B 10 D 18
30 In the diagram below, O is the origin and POR is a straight line on ’07 a Cartesian plane. y O
x
θ
’05
R(12, –5) 24 cm
S
7 cm Q
x°
R
What is the value of cos x°? 7 7 A —– C – —– 24 24 7 7 B —– D – —– 25 25
12 cm S
Q
T
x° 10 cm
R Find the value of tan x°. 4 A – — 5 4 B — 5 4 C – — 3 4 D — 3
SPM Clone
y
R(0.8, 0.6)
31 The diagram shows a right-angled triangle ABC. O C
Find the value of sin x° + cos y°. A 1.11 B 1.31 C 1.46 D 1.66
p
θ B
q
’07
A
1 Given that cos θ = – —, find the 2 value of p. q A ——— tan 60°
9.2
G F E
Graphs of Sine, Cosine and Tangent
35 The diagram shows the graphs of y = cos x and y = sin x for 0° x 360°. y
33 In the diagram, KLMN is a SPM parallelogram and JKL is a straight Clone ’09 line.
36 The diagram shows the graph of y = cos x. The value of p is
q°
N
D
It is given that BF = 40 cm, GE = 3 13 cm, CE = 27 cm and sin p° = — . 5 Find the value of tan q°. 12 5 A – —– C —– 5 12 5 12 B – —– D —– 12 5
y = cos x
12 32 Given that sin x° = —–, where 13 SPM Clone 90° < x° < 270°. Find the value of ’08 cos x°. 5 5 C – —– A —– 12 12 5 5 D – —– B —– 13 13
A
C
x
1
1
29 In the diagram below, ABC and BFED are straight lines. p°
y° x°
’08
B q tan 60° q C ——— sin 30° D q sin 30°
SPM Clone
B
S(0.51, 0.86)
1
SPM Clone
28 In the diagram below, S is the SPM midpoint of the straight line QST. Clone ’06
34 In the diagram, point R and point S lie on the arc of a unit circle with ’10 centre O.
The value of cos θ is 12 5 A —– C – —– 13 13 5 12 B —– D – —– 13 13
9
P
27 In the diagram, PQR is a straight SPM line. Clone P
15 —– 17 8 —– 17 8 C – —– 17 15 D – —– 17 A B
SPM Clone
CHAPTER
26 Given that sin θ = –0.7071 and 90° θ 270°, find the value ’05 of θ. A 135° C 225° B 145° D 245° SPM Clone
x 360° y = sin x
0 –1
A(θ , –0.7071)
The value of θ is A 45° C 210° B 135° D 225°
y
M 1
0
J
K
8 Given sin ∠NKJ = —–, find the 17 value of cos ∠LMN.
169
L
p
360°
x
–1
A 30° B 45°
C 60° D 90° Trigonometry II
F O R M 4
y
0
x
q
C
A 90° B 180° C 270° D 360°
37 The diagram shows the graph of y = sin x.
y 1
D
y
y
4
1
1
0
0
360°
38 Which of the following graphs represents y = sin x°? ’03 A y
42 Which of the following graphs represents y = cos x for ’07 0° x 180°? A y
y
1 0
360°
1
x
180°
360°
C
y
y 0
1
1
360°
D
1
360°
y
1
360°
x
90°
x
–1
D
y 1
–1
D
180°
0
–1
x
y
C
0
x
–1
1
180°
90°
x
y
0
180°
0
x
–1
–1
C
x
y
B
1
360°
180°
–1
x
–1
180°
90°
0
–1
1
0
x
SPM Clone
SPM Clone
B
180°
–1
B
0
90°
x
9 CHAPTER
F O R M
x
90°
–1
40 Which of the following graphs SPM Clone represents y = cos 2x°? ’05 A
The value of q is A 360° B 270° C 180° D 90°
180°
0
41 Which of the following graphs represents y = sin x for ’06 0° x 180°? A
y 1 0
0
SPM Clone
180°
360°
x
y
–1 0
90°
180°
x
90°
180°
–1
43 Which of the following graphs SPM Clone represents y = tan x°? ’08 A y ’10
–1
39 The diagram shows the graph of SPM Clone y = tan x. Find the value of p. ’04 y
0
360°
x
B y
B
y
1 0
p
x
0
180° 90°
–1
Trigonometry II
170
x
x
0
360°
x
C
0
D
44 Which of the following graphs SPM represents the graph of y = sin θ? Clone ’09 A y
y
1 0
1 0
y
180°
360°
360°
180°
180°
360°
θ
D
y 1
B
x
90°
–1
θ
–1 0
y
’11
x
360°
C
y 1 90°
180°
θ
–1
θ
9
0
0
CHAPTER
–1
You may refer to COMPANION WEBSITE
Fully Worked Solutions Paper 1
1 Given that tan θ = –1.732 and 180° θ 360°, find the value of θ. A 240° C 300° B 260° D 340°
3 The diagram shows the graph of y = tan x for 0° x 360°.
2 In the diagram, PQR and QTS are straight lines. P
6 cm Q 12 cm y°
y
R
A(θ , 1) 0
T
x°
S
Online Tests
x
Find the value of θ. A 200° B 215° C 225° D 250°
Given that tan y° = 2 and QS = 3QT, find the value of cos x°.
COMPANION WEBSITE
4 C – — 5 4 D – — 3
3 A – — 5 3 B – — 4
Multiple-choice Questions
171
Trigonometry II
F O R M 4
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