Solutions Manual Trigonometry 4th Edition Mark Dugopolski

Instant download and all chappters Solutions Manual Trigonometry 4th Edition Mark Dugopolski https://testbankdata.com/download/solutions-manual-trigonometry-4th-edition-mark -dugopolski/ 3.1 Basic Identities

131

For Thought

12.

1. False, for example sin 0 = 0 but cos 0 = 1.

13. 1 − sin2 α = cos2 α

2. True, since 3. True, since

sin(x) cos(x) · = 1. cos(x) sin(x) 2 sin(x) cos(x) = 2 cos(x). sin(x)

4. False, since (sin x + cos x)2 = 1 + 2 sin(x) cos(x) 6= 1 = sin2 x + cos2 x. 5. False, since tan 1 =

p

7. False, since cos(−9) = cos(9) implies cos3 (−9) = cos3 (9). 8. True, since sin(x) csc(x) = 1. 9. False, since sin2 (x) + cos2 (x) = 1. 10. True, since (1 − sin x)(1 + sin x) = (1 − sin2 x).

3.1 Exercises

cot2 x

−

15. (sin β + 1)(sin β − 1) = sin2 β − 1 = − cos2 β 16. (1 + cos β)(1 − cos β) = 1 − cos2 β = sin2 β 17.

sin α 1 · cos α sin α = 1 + 1 = 2 sin α 1/ sin α 1/ sin α

1 + cos α ·

18.

(sin α + 1)(sin α − 1) sin2 α − 1 = = cos2 α cos2 α − cos2 α = −1 cos2 α

19. Since cot2 x = csc2 x − 1, we get √ cot x = ± csc2 x − 1. 20. Since sec2 (x) = tan2 (x) + 1, we have q

sec(x) = ± tan2 (x) + 1.

1. Pythagorean 2. even

21. sin(x) =

1 1 = q csc(x) ± 1 + cot2 (x)

22. cos(x) =

1 1 = q sec(x) ± tan2 (x) + 1

3. odd 4. identity 5. odd 6. even 7. reciprocal 8. sine, cosine 9.

1/ cos(x) 1 cos(x) = · = sin(x)/ cos(x) cos(x) sin(x) 1 = csc(x) sin(x)

=1

14. sec2 α − 1 = tan2 α

sec2 (1) − 1.

6. False, since sin2 (−6) = (− sin(6))2 = sin2 (6).

csc2 x

23. Since cot2 (x) = csc2 (x) − 1, we obtain 1 1 tan(x) = = p 2 cot(x) ± csc (x) − 1 24. cot(x) =

1 1 = p 2 tan(x) ± sec (x) − 1

cos(x)/ sin(x) cos(x) sin(x) = · = cos(x) 1/ sin(x) sin(x) 1

√ 5 25. Since sec α = 1 + = , we get 2 s √   2 2 5 2 2 cos α = √ = , sin α = 1 − √ = 5 √5 5 √ 1 5 √ or sin α = , csc α = 5, and cot α = 2. 5 5

sin(x) cos(x) + = sin2 (x) + cos2 (x) = 1 1/ sin(x) 1/ cos(x)

26. Since α is s in Quadrant II, we get √  2 3 7 cos α = − 1 − =− , 4 4

10.

11.

s

 2

1 2

132

Chapter 3

√ 4 4 7 sec α = − √ or sec α = − , 7 7 √ 3/4 3 3 7 tan α = √ = − √ or tan α = − , 7 − 7/4 7 √ 7 4 cot α = − , and csc α = . 3 3 v u u 27. Since sin α = −t1 −

√ !2 3 − = 5

√ 3 22 − 1− =− , we find 25 5 r

√ 5 5 22 csc α = − √ or csc α = − , 22 22 √ 5 5 3 sec α = − √ = − , 3 3 √ √ √ − 22/5 22 66 tan α = √ = √ = , and 3 − 3/5 3 √ √ 3 66 cot α = √ = . 22 22 √ √ √ 5 5 5 5 5 √ ·√ =− 28. cos α = =− , 20 4 −4 5 5 v √ !2 √ u u 5 11 t sin α = 1 − − = 4 4 √ √ √ 11/4 11 55 Also, tan α = √ =−√ =− , 5 − 5/4 5 √ 5 55 cot α = − √ = − , and 11 55 √ 4 4 11 csc α = √ = . 11 11 29. Since α is in Quadrant IV, we get s √   1 2 10 csc α = − 1 + − =− , 3 3 √ 3 3 10 sin α = − √ = − , 10 10 s

2

3 1 − −√ = 10 √ r 9 1 10 1− =√ = , 10 10 10 √ sec α = 10, and tan α = −3. 

Trigonometric Identities

√ 1 3 30. Since sin α = √ = , we obtain 3 3 s √ √   1 2 2 6 cos(α) = 1 − √ =√ = , 3 3 3 √ √ 3 6 sec(α) = √ = , 2 2 √ √ 1/ 3 1 2 tan(α) = √ √ = √ = , 2 2/ 3 2 √ and cot(α) = 2. 31. (− sin x) · (− cot x) = sin(x) · 32. sec x−sec x = 0

33.

cos x = cos(x) sin x

sin(y)+(− sin(y)) = 0

34. cos y + cos y = 2 cos y 35.

sin(x) − sin(x) + =0 cos(x) cos(x)

36.

cos(x) cos(x) − = −2 cot(x) − sin(x) sin(x)

37. (1 + sin α)(1 − sin α) = 1 − sin2 α = cos2 α 38. (1 − cos α)(1 + cos α) = 1 − cos2 α = sin2 α 39. (− sin β)(cos β)(1/ sin β) = − cos β 1 40. (− tan β) − sin β 



cos β = tan β cot β = 1

41. Odd, since sin(−y) = − sin(y) for any y, even if y = 2x. 42. Even, since cos(−y) = cos(y) for any y, even if y = 2x. 43. Neither, since f (−π/6) 6= f (π/6) and f (−π/6) 6= −f (π/6). 44. Odd, f (−x) = 2 sin(−x) cos(−x) = 2(− sin(x)) cos(x) = −f (x). 45. Even, since sec2 (−t) − 1 = sec2 (t) − 1. 46. Neither, since f (−π/6) 6= f (π/6) and f (−π/6) 6= −f (π/6).

cos α =

47. Even, f (−α) = 1 + sec(−α) = 1 + sec(α) = f (α) 48. Neither, since f (−π/6) 6= f (π/6) and f (−π/6) 6= −f (π/6).

3.1 Basic Identities

49. Even, f (−x) =

133

sin(−x) − sin(x) = = f (x) −x −x

50. Odd, f (−x) = (−x) cos(−x) = −x cos(x) = −f (x) 51. Odd, f (−x) = −x + sin(−x) = −x − sin(x) = −f (x) 52. Even, f (−x) = csc((−x)2 ) = csc(x2 ) = f (x) 53. If γ = π/3, then (sin(π/3) + cos(π/3))2 = !2 √ √ √ 3 1 ( 3 + 1)2 4+2 3 + = = 6= 1 2 2 4 4 while sin2 (π/3) + cos2 (π/3) = 1. Thus, it is not an identity. 2 54. If x = π/4, then tan −1 = 1−1 = 0 √ (π/4) 2 while sec (π/4) = ( 2)2 = 2. Thus, it is not an identity.

55. If β = π/6, then 1 (1 + sin(π/6)) = 1 + 2 2



2

 2

=

3 2

=

1 59. If y = π/6, then sin(π/6) = while 2 1 sin(−π/6) = − . Thus, it is not an identity. 2 1 60. If y = π/3, then cos(−π/3) = and 2 1 − cos(π/3) = − . Thus, it is not an identity. 2 61. If y = π/6, then cos2 (π/6) − sin2 (π/6) = √ !2  2 3 1 3 1 1 − = − = and 2 2 4 4 2 √ 3 sin(2 · π/6) = sin(π/3) = . 2 Thus, it is not an identity. 1 62. If x = π/6, then cos(2 · π/6) = cos(π/3) = 2 and √ !  √ 3 1 3 2 cos(π/6) sin(π/6) = 2 = . 2 2 2

9 4

Thus, it is not an identity. 63. 1 −

and 1 + sin2 (π/6) = 1 +

 2

1 2

5 = . 4

64.

56. If α = π/6, then √ sin(2π/6) = sin(π/3) =

3 2

65.

−(tan2 t + 1) − sec2 t = = −1 sec2 t sec2 t

66.

cos w(sin2 w + cos2 w) cos(w) · 1 = = cos2 w sec w sec w

67.

(1 − cos2 α) − cos2 α 1 − 2 cos2 α = =1 1 − 2 cos2 α 1 − 2 cos2 α

68.

− sin3 θ − sin2 θ = = tan2 θ (− cos2 θ) sin θ(sin2 θ − 1)

69.

tan x(tan2 x − sec2 x) tan x(−1) = = tan2 x − cot x − cot x

70.

sin2 x + cos2 x 1 = = csc x sin x sin x

and √

√

1 3 3 · = . 2 2 4

So, it is not an identity. 57. If α = 7π/6, then sin(7π/6) = − s

1−

cos2



7π 6

1 while 2



is a positive number.

Thus, it is not an identity. 58. If α = 3π/4, then tan(3π/4) = −1 while p

sin2 x(sin2 x − 1) = sin2 x(− cos2 x) · cos x = 1/ cos x − sin2 (x) cos3 (x)

Thus, it is not an identity.

sin(π/6) cos(π/6) =

1 = 1 − sec2 (x) = − tan2 (x) cos2 (x)

sec2 (3π/4) − 1 is a positive number. Then it is not an identity.

71.

1 cos2 (x)/ sin2 (x) 1 cos2 (x) − = − = sin x sin3 x sin3 x sin3 x 1 − cos2 x sin2 x 1 = = = csc x 3 3 sin x sin x sin x

134

72.

Chapter 3

1/ cos2 x 1 cos2 x 1− = 1 − · = cos2 x sin2 x sin2 x/ cos2 x 1 1− = 1 − csc2 x = − cot2 x sin2 x

73. (sin2 x − cos2 x)(sin2 x + cos2 x) = (sin2 x − cos2 x)(1) = sin2 x − cos2 x 74. (csc2 x + cot2 x)(csc2 x − cot2 x) = (csc2 x + cot2 x)(1) = csc2 x + cot2 x 75. Let b be the hypotenuse of the smaller right triangle shown below. Let a be the side opposite the head angle θ. Then a + b = r, the radius of the wheel.

R 900

80. a) 225◦ ·

π 5π = ◦ 180 4

b) −210◦ · c) 270◦ ·

π 7π =− ◦ 180 6

π 3π = 180◦ 2

81. A = θr2 /2 = 15◦ ·

82. ω =

Trigonometric Identities

π 62 3π 2 · = ft 180◦ 2 2

180 radians/hr = 1/4

180 radians/min = 12 radians/min 60/4 √

b 83. a) 0 d) 0

a

2 2 √ 2 3 e) 3 b)

c) Undefined f) 1

√ g)

Θ T

3 3

h) − 1

84. Hypotenuse

The two right triangles are similar triangles. Applying right triangle trigonometry, b=

R . cos θ

Consequently,

√

52 + 8 2 =

√ 89 ≈ 9.4 m, and

5 acute angles are arcsin √ 89   8 ◦ arcsin √ ≈ 58.0 89 



≈ 32.0◦ and

85. Let x be the number of ants. Then

R a=r−b=r− . cos θ

x = 10a + 6 = 7b + 2 = 11c + 2 = 13d + 2

Thus, R r − cos a θ = . tan θ tan θ r cos θ − R Simplifying, we obtain T = . sin θ

T =

for some positive integers a, b, c, d. The smallest positive x satisfying the system of equations is x = 4006 ants.

76. In Exercise 75, let R = 3, r = 13.5, and T = 4.9. Then 4.9 =

13.5 cos θ − 3 sin θ

Using a calculator, we find θ ≈ 58.0◦ . 79. 360◦ − 35◦ = 325◦

86. We study a general case of three tangent circles with radii r < x < y that are tangent to the right most line, as shown below. The angle α is formed from the middle line to the right most line.

3.1 Basic Identities

135

y 4 = 542 (16y) y 3 = 542 (16) y = 36.

y

√

p

Thus, x = 16y = 16(36) = 24. Hence, the radii are x = 24 in. and y = 36 in. x

3.1 Pop Quiz r

1.

Α

cos x 1 1 · = = csc x sin x cos x sin x

2. Since α is in Quadrant I, we get x r

=

x + r + r csc α r csc α

x r

=

x+r +1 r csc α

(1)

x(csc α − 1) = r(1 + csc α) csc α − 1 r = 1 + csc α x y + 2x + r + r csc α . r csc α

(2)

=

(3)

3.1 Linking Concepts

y+x r csc α

(y − x) csc α = y + x y·

(csc α − 1) 1 + csc α

= x

r x

= x

y·

By (2)

√ yr = x

(4)

In particular, assume r = 16 in. and there is another circle of radius 54 in. that is tangent to the circle of radius y and tangent to the left and right most lines. Applying (4), we find √ p y = 54x and x = 16y Solve for y as follows: y =

q

p

54 16y

4. 2 cos2 α + 2 sin2 α = 2(cos2 α + sin2 α) = 2 5. No, for if x = π/2 then sin(2x) = sin π = 0 and 2 sin x = 2 sin(π/2) = 2.

If we subtract (1) from (3), we obtain y−x r

r

f (−x) = cos(3(−x)) = cos(−(3x)) = cos(3x).

Likewise, by similar triangles we have =

 2 1

3. Even, since

x csc α = x + r + r csc α

y r

√ 8 2 2 cos α = 1 − = = , and 3 9 3 √ √ cos α 2 2/3 cot α = = = 2 2. sin α 1/3 s

Using similar triangles, we find

a) Since x = 100(4) cos 60◦ = 200 and √ y = −16(4)2 + 100(4) sin 60◦ = 200 3 − 256, after t = √ 4 the coordinates are (200, 200 3 − 256) ≈ (200, 90.4). √ b) Note, y = −16t2 +100t sin 60◦ = −16t2 +50 3t. Set y = 0. √ −t(16t − 50 3) = 0 √ 50 3 t = 0, 16 √ 25 3 t = 0, 8 √ 25 3 The projectile is in the air for seconds. 8 c) Using the answer from part b), we get √ ! 25 3 x = 100t cos 60◦ = 50t = 50 ≈ 270.6. 8 The projectile lands 270.6 feet from the gun.

136

Chapter 3

√ d) The vertex of the function y = −16t2 + 50 3t

For Thought

(given by the height) can be shown 1. True,

to be

! √ 25 3 1875 , ≈ (2.7, 117.2). 16 16

sin x sin x = sin x · = sin2 x. 1/ sin x 1

2. False, if x = π/3 then

e) If A is the time in the air which is the same as the number of seconds before the projectile lands, then

3. True,

1/ cos x 1 sin x sin x = · = = tan x. 1/ sin x cos x 1 cos x

−16A2 + vo A sin θ = 0 4. True, sin x ·

−A(16A − vo sin θ) = 0. vo sin θ . 16

5. True,

ymax

√ √ 1 + sin(π/4) cos(π/4) 1 + ( 2/2)( 2/2) √ = = cos(π/4) 2/2

vo sin θ + vo sin(θ) 32 

v 2 sin2 θ vo2 sin2 θ = − o + 64 32 ymax =

vo2 sin2 θ 64

h) Since x = vo t cos θ, we find t =

x . Then vo cos θ

y = −16t2 + vo t sin θ  2   x x = −16 + vo sin θ vo cos θ vo cos θ y = −

16 sec2 θ 2 x + x tan θ. vo2

sin(π/4) √ = 2+1 cos(π/4)

and

1 + (1/2) 3 2 3 √ = ·√ =√ . 2 2/2 2 2

g) The t-coordinate of the vertex of y = −16t2 + vo sin(θ)t is −b −vo sin θ vo sin θ = = . The maximum 2a −32 32 height, ymax , is given by 2

cos x sin x + = 1 + tan x. cos x cos x

sec(π/4) +

d = x = vo t cos θ = vo A cos θ =   vo sin θ v2 vo cos θ = o sin θ cos θ, 16 16 2 v i.e., d = o sin θ cos θ. 16

vo sin θ = −16 32

1 sin x = = tan x . cos x cos x

6. False, if x = π/4 then

f ) The distance d from the gun to the point where the projectile lands is given by



cot(π/3) = tan(π/3)

√ √ 3/3 1 √ = and tan2 (π/3) = ( 3)2 = 3. 3 3

The maximum height is 117.2 feet.

Then A =

Trigonometric Identities

1 + sin x 1 + sin x = = 2 (1 − sin x)(1 + sin x) 1 − sin x 1 . 1 − sin x

7. True, 

8. True, since tan x · cot x = tan x ·

1 = 1. tan x

9. False, if x = π/3 then (1 − cos(π/3))2 = 2 = 1/4 and (1 − 1/2)2 = (1/2) √ 2 sin (π/3) = ( 3/2)2 = 3/4. 10. False, if x = π/6 then (1 − csc(π/6))(1 + csc(π/6)) √ 2 = (1 − 2)(1 + 2) = 2 −3 and cot (π/6) = ( 3) = 3.

3.2 Exercises 1. complicated 2. sines, cosines

3.2 Verifying Identities

137

3. numerator, denominator

29. (tan α − 4) (tan α − 2)

4. identity

30. (2 cot α + 3) (cot α − 1)

5. D, since cos x tan x = cos x ·

sin x = sin x. cos x

6. I, since

31. (2 sec β + 1)2

32.

(3 csc θ − 2)2

33. (tan α − sec β) (tan α + sec β) 34. sin2 y − cos2 x sin2 y + cos2 x =
(sin y − cos x) (sin y + cos x) sin2 y + cos2 x


sec x cot x =

1 cos x 1 · = = csc x. cos x sin x sin x




7. A, since csc2 x − cot2 x = 1.

35. cos β sin2 β + sin β − 2 = cos β (sin β + 2) (sin β − 1)

sin x cos x 8. J, since + = 1 + cot x. sin x sin x

36. tan θ cos2 θ − 2 cos θ − 3 = tan θ (cos θ − 3) (cos θ + 1)

9. B, for 1 − sec2 x = − tan2 x.

37. 2 sec2 x − 1

10. C, for csc2 x − 1 = cot2 x.

38. cos2 x − 1 = [(cos x − 1)(cos x + 1)]2 = (cos x − 1)2 (cos x + 1)2





2


2

csc x sin x − = 1 − sin2 x = cos2 x. csc x csc x cos x sec x 12. E, for − = cos2 x − 1 = − sin2 x. sec x sec x 11. H, since

13. G, for csc2 x = 1 + cot2 x. 14. F, since




sin2 x + cos2 x 1 = . cos x sin x cos x sin x

39. cos α(sin α + 1) + (sin α + 1) = (cos α + 1) (sin α + 1) 40. sin θ (2 sin θ + 1) − cos θ (2 sin θ + 1) = (sin θ − cos θ) (2 sin θ + 1) 41. Rewrite the left side of the equation.

15. sin2 (α) − 1 = − cos2 α

sin(x) cot(x) =

16. tan2 (α) − 4 17.

2 cos2 β

sin(x) ·

− cos β − 1

cos(x) sin(x)

=

cos x

18. 2 csc2 β − 7 csc β + 3 42. Rewrite the left side of the equation. 19. csc2 x+2 csc x sin x+sin2 x = csc2 x+2+sin2 x 20.

4 cos2 x 4 cos2 x

− 4 cos x sec x + − 4 + sec2 x

21. 4 sin2 θ − 1

22.

sec2 x

cos2 (x) tan2 (x) =

= cos2 (x) ·

9 sec2 θ − 4

2

23. 9 sin θ+12 sin θ+4 24.

sin2 (x) cos2 (x)

=

sin2 (x)

9 cos2 θ−12 cos θ+4

25. 4 sin4 y − 4 sin2 y csc2 y + csc4 y = 4 sin4 y − 4 + csc4 y 26. tan4 y + 2 tan2 y cot2 y + cot4 y = tan4 y + 2 + cot4 y

43. Rewrite the left side of the equation. 1 − sec(x) cos3 (x) = 1−

1 · cos3 (x) = cos(x)

27. (2 sin γ + 1) (sin γ − 3)

1 − cos2 (x) =

28. (cos γ − 3) (cos γ + 2)

sin2 (x)

138

Chapter 3

44. Rewrite the left side of the equation.

48.

3

1 − csc(x) sin (x) = 1−

Trigonometric Identities

1 · sin3 (x) = sin(x)

cos(x) sin2 (x) + cos3 (x) sin(x)

=

cos(x)[sin2 (x) + cos2 (x)] sin(x)

=

cos(x)[1] sin(x)

=

1 − sin2 (x) = 2

cos (x)

cot(x) 45. 49. 1 + sec2 (x) sin2 (x) = 1 sin2 (x) = 1+ cos2 (x) sin2 (x) 1+ cos2 (x)

sin(x) cos(x) + csc(x) sec(x)

=

sin(x) cos(x) + 1/ sin(x) 1/ cos(x)

=

= sin2 (x) + cos2 (x) =

1 + tan2 (x) =

1

2

sec (x)

50. sin3 (x) csc(x) + cos3 (x) sec(x) =

46.

sin3 (x)

1 + csc2 (x) cos2 (x) = 1+

1 1 + cos3 (x) sin(x) cos(x)

1 cos2 (x) = sin (x)

sin2 (x) + cos2 (x) =

2

cos2 (x) 1+ sin2 (x)

=

1 = 51. Rewrite the left side of the equation.

1 + cot2 (x) =

tan(x) cos(x) + csc(x) sin2 (x) = sin x + sin x =

csc2 (x)

2 sin x 47.

52. sin3 (x) + sin(x) cos2 (x) cos(x)

=

sin(x)[sin2 (x) + cos2 (x)] cos(x)

=

sin(x)[1] cos(x)

=

cot(x) sin(x) − cos2 (x) sec(x) = cos x − cos x = 0 53. (1 + sin α)2 + cos2 α = 1 + 2 sin α + sin2 α + cos2 α =

tan(x) Copyright 2015 Pearson Education, Inc.

2 + 2 sin α

3.2 Verifying Identities

139

54.

59. (1 + cot α)2 − 2 cot α = 

2



1 + 2 cot α + cot α − 2 cot α = 1 + cot2 α = csc2 α = 1 = sin2 α 1 = 1 − cos2 α 1 (1 − cos α)(1 + cos α)

sec(x) tan(x) − tan(x) sec(x)

=

sec2 (x) − tan2 (x) tan(x) sec(x)

=

1 tan(x) sec(x)

=

cot(x) cos(x) 60. 1 − sin2 x 1 − sin x

=

(1 − sin x)(1 + sin x) 1 − sin x

=

55. 2 − csc(β) sin(β) = 2−1 =

1 + sin(x) =

1 =

csc x 1 + csc x csc x csc x + 1 csc x

sin2 (β) + cos2 (β) 56. 



1 + sin2 β



=



cos2 (β) 2 − cos2 (β)



=

1 − sin2 β

61. Rewrite the right side of the equation.

2 cos2 β − cos4 β 57.

= tan x + cot x = sin x cos x + = cos x sin x sin2 x + cos2 x sin(x) cos(x)

=

1 sin(x) cos(x)

=

= =

csc x csc x − sin x csc x sin x · csc x − sin x sin x 1 1 − sin2 x 1 cos2 x

= sec2 x 62. Rewrite the right side of the equation.

sec(x) csc(x) 58. csc x cot x − cot x csc x

=

=

csc2 x − cot2 x cot(x) csc(x)

=

=

1 cot(x) csc(x)

=

=

tan x csc x

=

=

csc x − 1 cot2 x csc x − 1 csc2 x − 1 csc x − 1 (csc x − 1)(csc x + 1) 1 sin x · csc x + 1 sin x sin x sin x + 1

140

63.

Chapter 3

64.

(− cos θ)(sin θ + 1) − cos2 θ

Rewrite the right side of the equation. cos(−x) − csc(−x) cos x cos x + csc x = cos x cos x csc x = + cos x cos x = 1 + csc x sec x

sin(−x) sin x − sin x (− tan(x))2 − sin x

67.

Rewrite the right side of the equation.

= =

68.

sec x

sin θ(1 + cos θ) sin2 θ 1 + cos θ sin θ

1 + sin(y) csc(y) · 1 − sin(y) csc(y)

=

csc(y) + 1 csc(y) − 1

sin y + cos y sin y − cos y · sin y − cos y sin y − cos y

=

sin2 y − cos2 y sin2 y − 2 sin y cos y + cos2 y

=

=

(1 − cos2 y) − cos2 y 1 − 2 sin y cos y

=

=

1 − 2 cos2 y 1 − 2 cos y sin y

=

=

69.

Rewrite the left side of the equation.

=

cot x + tan x csc x

=

=

cos x sin x + sin x cos x 1 sin x

=

cos2 x + sin2 x sin x cos x 1 sin x

=

=

66. −1 tan θ − sec θ −1 cos θ · tan θ − sec θ cos θ

=

= Rewrite the left side of the equation.

sin θ(1 + cos θ) 1 − cos2 θ

1 + sin(y) 1 − sin(y)

sin y + cos y sin y − cos y

2

1 csc θ − cot θ 1 sin θ · csc θ − cot θ sin θ sin θ 1 − cos θ sin θ 1 + cos θ · 1 − cos θ 1 + cos θ

=

Rewrite the right side of the equation.

tan2 x + 1 =

65.

=

1 + sin(θ) cos θ

Rewrite the left side of the equation. tan2 (−x) −

Trigonometric Identities

=

− cos(θ) sin θ + 1 · sin(θ) − 1 sin θ + 1

=

1 sin x cos x 1 sin x

(− cos θ)(sin θ + 1) sin2 θ − 1

=

1 cos x

=