Chapter 7 Analytic Trigonometry - bsdweb.bsdvt.orgshubbard/HPCanswers/PR_9e_ISM_07all.pdf · tan...

648

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Chapter 7

Analytic Trigonometry

Section 7.1

1. Domain: { } is any real numberx x ;

Range: { }1 1y y− ≤ ≤

2. { }| 1x x ≥ or { }| 1x x ≤

3. [ )3, ∞

4. True

5. 1; 3

2

6. 1

2− ; 1−

7. sinx y=

8. 0 x π≤ ≤

9. x−∞ ≤ ≤ ∞

10. False. The domain of 1siny x−= is 1 1x− ≤ ≤ .

11. True

12. True

13. 1sin 0−

We are finding the angle θ , 2 2

θπ π− ≤ ≤ , whose

sine equals 0.

1

sin 0,2 2

0

sin 0 0

θ θ

θ−

π π= − ≤ ≤

=

=

14. 1cos 1− We are finding the angle θ , 0 θ≤ ≤ π , whose cosine equals 1.

1

cos 1, 0

0

cos 1 0

θ θθ

−

= ≤ ≤ π=

=

15. ( )1sin 1− −


θπ π− ≤ ≤ ,

whose sine equals 1− .

( )1

sin 1,2 2

2

sin 12

θ θ

θ

−

π π= − − ≤ ≤

π= −

π− = −

16. ( )1cos 1− −

We are finding the angle θ , 0 θ≤ ≤ π , whose cosine equals 1− .

( )1

cos 1, 0

cos 1

θ θθ

−

= − ≤ ≤ π= π

− = π

17. 1tan 0−


θπ π− < < , whose

tangent equals 0.

1

tan 0,2 2

0

tan 0 0

θ θ

θ−

π π= − < <

=

=

18. ( )1tan 1− −



tangent equals 1− .

1

tan 1,2 2

4

tan ( 1)4

θ θ

θ

−

π π= − − < <

π= −

π− = −

Section 7.1: The Inverse Sine, Cosine, and Tangent Functions

649


19. 1 2sin

2−



sine equals 2

2.

1

2sin ,

2 2 2

4

2sin

2 4

θ θ

θ

−

π π= − ≤ ≤

π=

π=

20. 1 3tan

3−



tangent equals 3

3.

1

3tan ,

3 2 2

6

3tan

3 6

θ θ

θ

−

π π= − < <

π=

π=

21. 1tan 3−



tangent equals 3 .

1

tan 3,2 2

3

tan 33

θ θ

θ

−

π π= − < <

π=

π=

22. 1 3sin

2−

−



sine equals 3

2− .

1

3sin ,

2 2 2

3

3sin

2 3

θ θ

θ

−

π π= − − ≤ ≤

π= −

π− = −

23. 1 3cos

2−

−

We are finding the angle θ , 0 θ≤ ≤ π , whose

cosine equals 3

2− .

1

3cos , 0

25

6

3 5cos

2 6

θ θ

θ

−

= − ≤ ≤ π

π=

π− =

24. 1 2sin

2−

−



sine equals 2

2− .

1

2sin ,

2 2 2

4

2sin

2 4

θ θ

θ

−

π π= − − ≤ ≤

π= −

π− = −

Chapter 7: Analytic Trigonometry

650


25. 1sin 0.1 0.10− ≈

26. 1cos 0.6 0.93− ≈

27. 1tan 5 1.37− ≈

28. 1tan 0.2 0.20− ≈

29. 1 7cos 0.51

8− ≈

30. 1 1sin 0.13

8− ≈

31. 1tan ( 0.4) 0.38− − ≈ −

32. 1tan ( 3) 1.25− − ≈ −

33. 1sin ( 0.12) 0.12− − ≈ −

34. 1cos ( 0.44) 2.03− − ≈

35. 1 2cos 1.08

3− ≈

36. 1 3sin 0.35

5− ≈

37. 1 4cos cos

5

π−

follows the form of the equation

( )( ) ( )( )1 1cos cosf f x x x− −= = . Since 4

5

π is

in the interval 0,π , we can apply the equation

directly and get 1 4 4cos cos

5 5

π π− =

.

38. 1sin sin10

π− − follows the form of the

equation ( )( ) ( )( )1 1sin sinf f x x x− −= = . Since

10

π− is in the interval ,2 2

π π −

, we can apply

the equation directly and get

1sin sin10 10

π π− − = − .

39. 1 3tan tan

8


equation ( )( ) ( )( )1 1tan tanf f x x x− −= = . Since

3

8


π π −

, we can apply


1 3 3tan tan

8 8

π π− − = − .

40. 1 3sin sin

7



3

7


π π −

, we can apply


1 3 3sin sin

7 7

π π− − = − .

41. 1 9sin sin

8

π−

follows the form of the

equation ( )( ) ( )( )1 1sin sinf f x x x− −= = , but we

cannot use the formula directly since 9

8

π is not

in the interval ,2 2

π π −

. We need to find an

angle θ in the interval ,2 2

π π −

for which

9sin sin

8

π θ= . The angle 9

8

π is in quadrant III

so sine is negative. The reference angle of 9

8

π is

8

π and we want θ to be in quadrant IV so sine

will still be negative. Thus, we have

9sin sin

8 8

π π = −

. Since 8

π− is in the interval

,2 2

π π −

, we can apply the equation above and

get 1 19sin sin sin sin

8 8 8

π π π− − = − = − .


651


42. 1 5cos cos

3


equation ( )( ) ( )( )1 1cos cosf f x x x− −= = , but

we cannot use the formula directly since 5

3

π− is

not in the interval 0,π . We need to find an

angle θ in the interval 0,π for which

5cos cos

3

π θ − =

. The angle 5

3

π− is in

quadrant I so the reference angle of 5

3

π− is 3

π.

Thus, we have 5

cos cos3 3

π π − =

. Since 3

π is


above and get

1 15cos cos cos cos

3 3 3

π π π− − − = = .

43. 1 4tan tan

5

π−


equation ( )( ) ( )( )1 1tan tanf f x x x− −= = , but


5

π is

not in the interval ,2 2

π π −



π π −

for which

4tan tan

5

π θ =

. The angle 4

5

π is in quadrant

II so tangent is negative. The reference angle of 4

5

π is

5

π and we want θ to be in quadrant IV

so tangent will still be negative. Thus, we have

4tan tan

5 5

π π = −

. Since 5

π− is in the

interval ,2 2

π π −

, we can apply the equation

above and get

1 14tan tan tan tan

5 5 5

π π π− − = − = − .

44. 1 2tan tan

3


equation ( )( ) ( )( )1 1tan tanf f x x x− −= = . but we


3

π− is not


π π −

. We need to find an angle

θ in the interval ,2 2

π π −

for which

2tan tan

3

π θ − =

. The angle 2

3

π− is in

quadrant III so tangent is positive. The reference

angle of 2

3

π− is 3

π and we want θ to be in

quadrant I so tangent will still be positive. Thus,

we have 2

tan tan3 3

π π − =

. Since 3

π is in the

interval ,2 2

π π −


above and get 1 12tan tan tan tan

3 3 3

π π π− − − = = .

45. 1 1sin sin

4−


( )( ) ( )( )1 1sin sinf f x x x− −= = . Since 1

4 is in

the interval 1,1− , we can apply the equation

directly and get 1 1 1sin sin

4 4− =

.

46. 1 2cos cos

3− −


equation ( )( ) ( )( )1 1cos cosf f x x x− −= = .

Since 2

3− is in the interval 1,1− , we can

apply the equation directly and get

1 2 2cos cos

3 3− − = −

.


652


47. ( )1tan tan 4− follows the form of the equation

( )( ) ( )( )1 1tan tanf f x x x− −= = . Since 4 is a

real number, we can apply the equation directly

and get ( )1tan tan 4 4− = .

48. ( )( )1tan tan 2− − follows the form of the equation

( )( ) ( )( )1 1tan tanf f x x x− −= = . Since 2− is a


and get ( )( )1tan tan 2 2− − = − .

49. Since there is no angle θ such that cos 1.2θ = ,

the quantity 1cos 1.2− is not defined. Thus,

( )1cos cos 1.2− is not defined.

50. Since there is no angle θ such that sin 2θ = − ,

the quantity ( )1sin 2− − is not defined. Thus,

( )( )1sin sin 2− − is not defined.

51. ( )1tan tan π− follows the form of the equation

( )( ) ( )( )1 1tan tanf f x x x− −= = . Since π is a


and get ( )1tan tan π π− = .

52. Since there is no angle θ such that sin 1.5θ = − ,

the quantity ( )1sin 1.5− − is not defined. Thus,

( )( )1sin sin 1.5− − is not defined.

53. ( ) 5sin 2f x x= +

5sin 2y x= +

( )1 1

5sin 2

5sin 2

2sin

52

sin5

x y

y x

xy

xy f x− −

= += −

−=

−= =

The domain of ( )f x equals the range of

1( )f x− and is 2 2

xπ π− ≤ ≤ or ,

2 2

π π −

in

interval notation. To find the domain of ( )1f x−

we note that the argument of the inverse sine

function is 2

5

x − and that it must lie in the

interval 1,1− . That is,

21 1

55 2 5

3 7

x

x

x

−− ≤ ≤

− ≤ − ≤− ≤ ≤

The domain of ( )1f x− is { }| 3 7x x− ≤ ≤ , or

3,7− in interval notation. Recall that the

domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is

also 3,7− .

54. ( ) 2 tan 3

2 tan 3

f x x

y x

= −

= −

( )1 1

2 tan 3

2 tan 3

3tan

23

tan2

x y

y x

xy

xy f x− −

= −= +

+=

+= =

The domain of ( )f x equals the range of 1( )f x−

and is 2 2

xπ π− < < or ,

2 2

π π −

in interval

notation. To find the domain of ( )1f x− we note

that the argument of the inverse tangent function can be any real number. Thus, the domain of

( )1f x− is all real numbers, or ( ),−∞ ∞ in

interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its

inverse. Thus, the range of f is ( ),−∞ ∞ .

55. ( ) ( )2cos 3f x x= −

( )2cos 3y x= −


653


( )( )

( )

1

1 1

2cos 3

cos 32

3 cos2

1cos

3 2

x y

xy

xy

xy f x

−

− −

= −

= −

= − = − =


1( )f x− and is 03

xπ≤ ≤ , or 0,

3

π

in interval


that the argument of the inverse cosine function

is 2

x− and that it must lie in the interval 1,1− .

That is,

1 12

2 2

2 2

x

x

x

− ≤ − ≤

≥ ≥ −− ≤ ≤




2,2− .

56. ( ) ( )3sin 2f x x=

( )3sin 2y x=

( )( )

( )

1

1 1

3sin 2

sin 23

2 sin3

1sin

2 3

x y

xy

xy

xy f x

−

− −

=

=

=

= =


1( )f x− and is 4 4

xπ π− ≤ ≤ , or ,

4 4

π π −

in



function is 3

x and that it must lie in the interval

1,1− . That is,

1 13

3 3

x

x

− ≤ ≤

− ≤ ≤




3,3− .

57. ( ) ( )( )

tan 1 3

tan 1 3

f x x

y x

= − + −

= − + −

( )( )

( )( )( ) ( )

1

1

1 1

tan 1 3

tan 1 3

1 tan 3

1 tan 3

1 tan 3

x y

y x

y x

y x

x f x

−

−

− −

= − + −

+ = − −

+ = − −

= − + − −

= − − + =

(note here we used the fact that 1tany x−= is an

odd function).


1( )f x− and is 1 12 2

xπ π− − ≤ ≤ − , or

1 , 12 2

π π − − −

in interval notation. To find the

domain of ( )1f x− we note that the argument of

the inverse tangent function can be any real

number. Thus, the domain of ( )1f x− is all real

numbers, or ( ),−∞ ∞ in interval notation. Recall

that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is

( ),−∞ ∞ .

58. ( ) ( )cos 2 1f x x= + +

( )cos 2 1y x= + +

( )( )

( )( )

1

1

cos 2 1

cos 2 1

2 cos 1

cos 1 2

x y

y x

y x

y x

−

−

= + +

+ = −

+ = −

= − −


654



1( )f x− and is 2 2x π− ≤ ≤ − , or 2, 2π− − in


we note that the argument of the inverse cosine function is 1x − and that it must lie in the

interval 1,1− . That is, 1 1 1

0 2

x

x

− ≤ − ≤≤ ≤

The domain of ( )1f x− is { }| 0 2x x≤ ≤ , or

0,2 in interval notation. Recall that the


0,2 .

59. ( ) ( )3sin 2 1f x x= +

( )3sin 2 1y x= +

( )( )

( )

1

1

1 1

3sin 2 1

sin 2 13

2 1 sin3

2 sin 13

1 1sin

2 3 2

x y

xy

xy

xy

xy f x

−

−

− −

= +

+ =

+ =

= − = − =


1( )f x− and is 1 1

2 4 2 4x

π π− − ≤ ≤ − + , or

1 1,

2 4 2 4

π π − − − +

in interval notation. To find

the domain of ( )1f x− we note that the argument

of the inverse sine function is 3

x and that it must

lie in the interval 1,1− . That is,

1 13

3 3

x

x

− ≤ ≤

− ≤ ≤



domain of a function equals the range of its inverse and the range of a function equals the

domain of its inverse. Thus, the range of f is

3,3− .

60. ( ) ( )2cos 3 2f x x= +

( )2cos 3 2y x= +

( )( )

( )

1

1

1 1

2cos 3 2

cos 3 22

3 2 cos2

3 cos 22

1 2cos

3 2 3

x y

xy

xy

xy

xy f x

−

−

− −

= +

+ =

+ = = − = − =


1( )f x− and is 2 2

3 3 3x

π− ≤ ≤ − + , or

2 2,

3 3 3

π − − +

in interval notation. To find the

domain of ( )1f x− we note that the argument of

the inverse cosine function is 2

x and that it must

lie in the interval 1,1− . That is,

1 12

2 2

x

x

− ≤ ≤

− ≤ ≤




2,2− .

61. 1

1

4sin

sin4

2sin

4 2

x

x

x

ππ

π

−

−

=

=

= =

The solution set is 2

2

.


655


62. 1

1

2cos

cos2

cos 02

x

x

x

ππ

π

−

−

=

=

= =

The solution set is {0} .

63. ( )

( )

1

1

3cos 2 2

2cos 2

32

2 cos3

12

21

4

x

x

x

x

x

ππ

π

−

−

=

=

=

= −

= −


4 −

.

64. ( )

( )

1

1

6sin 3

sin 36

3 sin6

13

21

6

x

x

x

x

x

ππ

π

−

−

− =

= −

= −

= −

= −


6 −

.

65. 1

1

3 tan

tan3

tan 33

x

x

x

ππ

π

−

−

=

=

= =

The solution set is { }3 .

66. 1

1

4 tan

tan4

tan 14

x

x

x

ππ

π

−

−

− =

= −

= − = −

The solution set is { 1}− .

67. 1 1

1

1

1

4cos 2 2cos

2cos 2 0

2cos 2

cos

cos 1

x x

x

x

x

x

ππ

ππ

π

− −

−

−

−

− =

− =

=

== = −

The solution set is { 1}− .

68. 1 1

1

1

5sin 2 2sin 3

3sin

sin3

3sin

3 2

x x

x

x

x

π πππ

π

− −

−

−

− = −

= −

= −

= − = −


2

−

.

69. Note that 29 45 29.75θ ′= ° = ° .

a. ( ) ( )( )1

180 180cos tan 23.5 tan 29.7524 1

13.92 hours or 13 hours, 55 minutes

Dπ π− ⋅ ⋅

= ⋅ −π

≈

b. ( ) ( )( )1

180 180cos tan 0 tan 29.7524 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ −π

≈

c. ( ) ( )( )1

180 180cos tan 22.8 tan 29.7524 1


Dπ π− ⋅ ⋅

= ⋅ −π

≈

70. Note that 40 45 40.75θ ′= ° = ° .

a. ( ) ( )( )1

180 180cos tan 23.5 tan 40.7524 1


Dπ π− ⋅ ⋅

= ⋅ − π

≈

b. ( ) ( )( )1

180 180cos tan 0 tan 40.7524 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

c. ( ) ( )( )1

180 180cos tan 22.8 tan 40.7524 1


Dπ π− ⋅ ⋅

= ⋅ − π

≈


656


71. Note that 21 18 21.3θ ′= ° = ° .

a. ( ) ( )( )1

180 180cos tan 23.5 tan 21.324 1


Dπ π− ⋅ ⋅

= ⋅ − π

≈

b. ( ) ( )( )1

180 180cos tan 0 tan 21.324 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

c. ( ) ( )( )1

180 180cos tan 22.8 tan 21.324 1


Dπ π− ⋅ ⋅

= ⋅ − π

≈

72. Note that 61 10 61.167θ ′= ° ≈ ° .

a. ( ) ( )( )1

180 180cos tan 23.5 tan 61.16724 1


Dπ π− ⋅ ⋅

= ⋅ − π

≈

b. ( ) ( )( )1

180 180cos tan 0 tan 61.16724 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

c. ( ) ( )( )1

180 180cos tan 22.8 tan 61.16724 1


Dπ π− ⋅ ⋅

= ⋅ − π

≈

73. a. ( ) ( )( )1

180 180cos tan 23.5 tan 024 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

b. ( ) ( )( )1

180 180cos tan 0 tan 024 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

c. ( ) ( )( )1

180 180cos tan 22.8 tan 024 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

d. There are approximately 12 hours of daylight every day at the equator.

74. Note that 66 30 66.5θ ′= ° = ° .

a. ( ) ( )( )1

180 180cos tan 23.5 tan 66.524 1

24 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

b. ( ) ( )( )1

180 180cos tan 0 tan 66.524 1

12 hours

Dπ π− ⋅ ⋅

= ⋅ − π

≈

c. ( ) ( )( )1

180 180cos tan 22.8 tan 66.524 1

22.02 hours or 22 hours, 1 minute

Dπ π− ⋅ ⋅

= ⋅ − π

≈

d. The amount of daylight at this location on the winter solstice is 24 24 0− = hours. That is, on the winter solstice, there is no daylight. In general, for a location at 66 30 '° north latitude, it ranges from around-the-clock daylight to no daylight at all.

75. Let point C represent the point on the Earth’s axis at the same latitude as Cadillac Mountain, and arrange the figure so that segment CQ lies along the x-axis (see figure).

At the latitude of Cadillac Mountain, the effective

radius of the earth is 2710 miles. If point D(x, y) represents the peak of Cadillac Mountain, then the length of segment PD is

1 mile1530 ft 0.29 mile

5280 feet⋅ ≈ . Therefore, the

point ( , ) (2710, )D x y y= lies on a circle with

radius 2710.29r = miles. We now have

1

2710cos

2710.292710

cos 0.01463 radians2710.29

x

rθ

θ −

= =

= ≈

Finally, 2710(0.01463) 39.64 miless rθ= = ≈ ,

and 2 (2710) 39.64

24 t

π = , so

24(39.64)0.05587 hours 3.35 minutes

2 (2710)t

π= ≈ ≈

y

x θ

P

Q (271 0,0)

D ( )x , y

C

s

2710 mi 27102710


657


Therefore, a person atop Cadillac Mountain will see the first rays of sunlight about 3.35 minutes sooner than a person standing below at sea level.

76. ( ) 1 134 6tan tanx

x xθ − − = −

.

a. ( ) 1 134 610 tan tan 42.6

10 10θ − − = − ≈ °

If you sit 10 feet from the screen, then the viewing angle is about 42.6° .

( ) 1 134 615 tan tan 44.4

15 15θ − − = − ≈ °


( ) 1 134 620 tan tan 42.8

20 20θ − − = − ≈ °


b. Let r = the row that result in the largest viewing angle. Looking ahead to part (c), we see that the maximum viewing angle occurs when the distance from the screen is about 14.3 feet. Thus, 5 3( 1) 14.3

5 3 3 14.3

3 12.3

4.1

r

r

r

r

+ − =+ − =

==

Sitting in the 4th row should provide the largest viewing angle.

c. Set the graphing calculator in degree mode

and let 1 11

34 6tan tanY

x x− − = −

:

0 500°

90°

Use MAXIMUM:

0 500°

90°

The maximum viewing angle will occur when 14.3x ≈ feet.

77. a. 0a = ; 3b = ; The area is: 1 1 1 1tan tan tan 3 tan 0

03

square units3

b a

π

π

− − − −− = −

= −

=

b. 3

3a = − ; 1b = ; The area is:

1 1 1 1 3tan tan tan 1 tan

3

4 6

5 square units

12

b a

π π

π

− − − − − = − −

= − −

=

78. a. 0a = ; 3

2b = ; The area is:

1 1 1 13sin sin sin sin 0

2

03

square units3

b a

π

π

− − − − − = −

= −

=


658


b. 1

2a = − ;

1

2b = ; The area is:

1 1 1 11 1sin sin sin sin

2 2

6 6

square units3

b a

π π

π

− − − − − = − − = − −

=

79. Here we have 1 41 50 'α = ° , 1 87 37 'β = − ° ,

2 21 18'α = ° , and 2 157 50 'β = − ° .

Converting minutes to degrees gives

( )51 6

41α = ° , ( )371 60

87β = − ° , 2 21.3α = ° , and

( )52 6

157β = − ° . Substituting these values, and

3960r = , into our equation gives 4250d ≈ miles. The distance from Chicago to Honolulu is about 4250 miles. (remember that S and W angles are negative)

80. Here we have 1 21 18'α = ° , 1 157 50'β = − ° ,

2 37 47 'α = − ° , and 2 144 58'β = ° .

Converting minutes to degrees gives 1 21.3α = ° ,

( )51 6

157β = − ° , ( )472 60

37α = − ° , and

( )292 30

144β = ° . Substituting these values, and

3960r = , into our equation gives 5518d ≈ miles. The distance from Honolulu to Melbourne is about 5518 miles. (remember that S and W angles are negative)

Section 7.2

1. Domain: odd integer multiples of 2

x xπ ≠

,

Range: { } 1 or 1y y y≤ − ≥

2. True

3. 1 5

55=

4. secx y= , 1≥ , 0 , π

5. cosine

6. False

7. True

8. True

9. 1 2cos sin

2−

Find the angle , ,2 2

θ θπ π− ≤ ≤ whose sine

equals 2

2.

1

2sin ,

2 2 2

4

2 2cos sin cos

2 4 2

θ θ

θ

−

π π= − ≤ ≤

π=

π= =

10. 1 1sin cos

2−

Find the angle , 0 ,θ θ≤ ≤ π whose cosine

equals 1

2.

1

1cos , 0

2

3

1 3sin cos sin

2 3 2

θ θ

θ

−

= ≤ ≤ π

π=

π = =

11. 1 3tan cos

2−

−


equals 3

2− .

1

3cos , 0

25

6

3 5 3tan cos tan

2 6 3

θ θ

θ

−

= − ≤ ≤ π

π=

π− = = −

Section 7.2: The Inverse Trigonometric Functions [Continued]

659


12. 1 1tan sin

2− −



equals 1

2− .

1

1sin ,

2 2 2

6

1 3tan sin tan

2 6 3

θ θ

θ

−

π π= − − ≤ ≤

π= −

π − = − = −

13. 1 1sec cos

2−


equals 1

2.

1

1cos , 0

2

3

1sec cos sec 2

2 3

θ θ

θ

π−

= ≤ ≤ π

π=

= =

14. 1 1cot sin

2− −



equals 1

2− .

1

1sin ,

2 2 2

6

1cot sin cot 3

2 6

θ θ

θ

−

π π= − − ≤ ≤

π= −

π − = − = −

15. ( )1csc tan 1−


θ θπ π− < < whose tangent

equals 1.

( )1

tan 1,2 2

4

csc tan 1 csc 24

θ θ

θ

π−

π π= − < <

π=

= =

16. ( )1sec tan 3−



equals 3 .

( )1

tan 3,2 2

3

sec tan 3 sec 23

θ θ

θ

−

π π= − < <

π=

π= =

17. 1sin tan ( 1)− −



equals 1− .

1

tan 1,2 2

4

2sin tan ( 1) sin

4 2

θ θ

θ

−

π π= − − < <

π= −

π − = − = −

18. 1 3cos sin

2−

−



equals 3

2− .

1

3sin ,

2 2 2

3

3 1cos sin cos

2 3 2

θ θ

θ

−

π π= − − ≤ ≤

π= −

π − = − =


660


19. 1 1sec sin

2− −



equals 1

2− .

1

1sin ,

2 2 2

6

1 2 3sec sin sec

2 6 3

θ θ

θ

−

π π= − − ≤ ≤

π= −

π − = − =

20. 1 3csc cos

2−

−


equals 3

2− .

1

3cos 0

25

6

3 5csc cos csc 2

2 6

θ θ

θ

−

= − ≤ ≤ π

π=

π− = =

21. 1 15 2cos sin cos

4 2− − π = −


equals 2

2− .

1

2cos , 0

23

45 3

cos sin4 4

θ θ

θ

−

= − ≤ ≤ π

π=

π π =

22. 1 12 1tan cot tan

3 3− − π = −



equals 1

3− .

1

1tan ,

2 23

62

tan cot3 6

θ θ

θ

π−

π π= − − < <

π= −

π = −

23. 1 17 3sin cos sin

6 2− − π − = −



equals 3

2− .

1

3sin ,

2 2 2

3

7sin cos

6 3

θ θ

θ

π−

π π= − − ≤ ≤

π= −

π − = −

24. ( )1 1cos tan cos 13

− − π − = −


equals 1− .

1

cos 1, 0

3

cos tan3

θ θ

θ

π−

= − ≤ ≤ ππ=

π − =


661


25. 1 1tan sin

3−

Let 1 1sin

3θ −= . Since

1sin

3θ = and

2 2θπ π− ≤ ≤ , θ is in quadrant I, and we let

1y = and 3r = .

Solve for x: 2

2

1 9

8

8 2 2

x

x

x

+ =

=

= ± = ±

Since θ is in quadrant I, 2 2x = .

1 1 1 2 2tan sin tan

3 42 2 2

y

xθ− = = = ⋅ =

26. 1 1tan cos

3−

Let 1 1cos

3θ −= . Since

1cos

3θ = and 0 θ≤ ≤ π ,

θ is in quadrant I, and we let 1x = and 3r = . Solve for y:

2

2

1 9

8

8 2 2

y

y

y

+ =

=

= ± = ±

Since θ is in quadrant I, 2 2y = .

1 1 2 2tan cos tan 2 2

3 1

y

xθ− = = = =

27. 1 1sec tan

2−

Let 1 1tan

2θ −= . Since

1tan

2θ = and

2 2θπ π− < < , θ is in quadrant I, and we let

2x = and 1y = .

Solve for r: 2 2

2

2 1

5

5

r

r

r

+ =

=

=

θ is in quadrant I.

1 1 5sec tan sec

2 2

r

xθ− = = =

28. 1 2cos sin

3−

Let 1 2sin

3θ −= . Since

2sin

3θ = and


2y = and 3r = .

Solve for x: 2

2

2 9

7

7

x

x

x

+ =

=

= ±

Since θ is in quadrant I, 7x = .

1 2 7cos sin cos

3 3

x

rθ−

= = =

29. 1 2cot sin

3−

−

Let 1 2sin

3θ −

= −

. Since 2

sin3

θ = − and

2 2θπ π− ≤ ≤ , θ is in quadrant IV, and we let

2y = − and 3r = .

Solve for x: 2

2

2 9

7

7

x

x

x

+ =

=

= ±

Since θ is in quadrant IV, 7x = .

1 2 7 2 14cot sin cot

3 22 2

x

yθ−

− = = = ⋅ = − −

30. 1csc tan ( 2)− −

Let 1tan ( 2)θ −= − . Since tan 2θ = − and

2 2θπ π− < < , θ is in quadrant IV, and we let

1x = and 2y = − .

Solve for r: 2

2

1 4

5

5

r

r

r

+ =

=

= ±

Since θ is in quadrant IV, 5r = .

1 5 5csc tan ( 2) csc

2 2

r

yθ− − = = = = − −


662


31. 1sin tan ( 3)− −

Let 1tan ( 3)θ −= − . Since tan 3θ = − and

2 2θπ π− < < , θ is in quadrant IV, and we let

1x = and 3y = − .

Solve for r: 2

2

1 9

10

10

r

r

r

+ =

=

= ±

Since θ is in quadrant IV, 10r = .

1sin tan ( 3) sin

3 10 3 10

1010 10

y

rθ− − = =

−= ⋅ = −

32. 1 3cot cos

3−

−

Let 1 3cos

3θ −

= −

. Since 3

cos3

θ = − and

0 θ≤ ≤ π , θ is in quadrant II, and we let

3x = − and 3r = . Solve for y:

2

2

3 9

6

6

y

y

y

+ =

=

= ±

Since θ is in quadrant II, 6y = .

1 3cot cos cot

3

3 1 2 2

26 2 2

x

yθ−

− = =

− −= = ⋅ = −

33. 1 2 5sec sin

5−

Let 1 2 5sin

5θ −= . Since

2 5sin

5θ = and


2 5y = and 5r = .

Solve for x:

2

2

20 25

5

5

x

x

x

+ =

=

= ±


1 2 5 5sec sin sec 5

5 5

r

xθ−

= = = =

34. 1 1csc tan

2−

Let 1 1tan

2θ −= . Since

1tan

2θ = and

2 2θπ π− < < , θ is in quadrant I, and we let

2x = and 1y = .

Solve for r: 2 2

2

2 1

5

5

r

r

r

+ =

=

=


1 1 5csc tan csc 5

2 1

r

yθ− = = = =


4 2 4− − π π = − = −

36. 1 17 1 2cos sin cos

6 2 3− −π π = − =

37. 1cot 3− We are finding the angle , 0 ,θ θ< < π whose

cotangent equals 3 .

1

cot 3, 0

6

cot 36

θ θ

θ

−

= < < ππ=

π=

38. 1cot 1− We are finding the angle , 0 ,θ θ< < π whose

cotangent equals 1.

1

cot 1, 0

4

cot 14

θ θ

θ

−

= < < ππ=

π=


663


39. 1csc ( 1)− −

We are finding the angle ,2 2

θ θπ π− ≤ ≤ ,

0θ ≠ , whose cosecant equals 1− .

1

csc 1, , 02 2

2

csc ( 1)2

θ θ θ

θ

−

π π= − − ≤ ≤ ≠

π= −

π− = −

40. 1csc 2−


θ θπ π− ≤ ≤ ,

0θ ≠ , whose cosecant equals 2 .

1

csc 2, , 02 2

4

csc 24

θ θ θ

θ

−

π π= − ≤ ≤ ≠

π=

π=

41. 1 2 3sec

3−

We are finding the angle , 0θ θ≤ ≤ π , 2

θ π≠ ,

whose secant equals 2 3

3.

1

2 3sec , 0 ,

3 2

6

2 3sec

3 6

θ θ θ

θ

−

π= ≤ ≤ π ≠

π=

π=

42. ( )1sec 2− −

We are finding the angle , 0θ θ≤ ≤ π , 2

θ π≠ ,

whose secant equals 2− .

( )1

sec 2, 0 ,2

2

32

sec 23

θ θ θ

θ

−

π= − ≤ ≤ π ≠

π=

π− =

43. 1 3cot

3−

−

We are finding the angle , 0 ,θ θ< < π whose

cotangent equals 3

3− .

1

3cot , 0

32

3

3 2cot

3 3

θ θ

θ

−

= − < < π

π=

π− =

44. 1 2 3csc

3−

−


θ θπ π− ≤ ≤ ,

0θ ≠ , whose cosecant equals 2 3

3− .

1

2 3csc , , 0

3 2 2

3

2 3csc

3 3

θ θ θ

θ

−

π π= − − ≤ ≤ ≠

π= −

π− = −

45. 1 1 1sec 4 cos

4− −=

We seek the angle , 0 ,θ θ π≤ ≤ whose cosine

equals 1

4. Now

1cos

4θ = , so θ lies in quadrant

I. The calculator yields 1 1cos 1.32

4− ≈ , which is

an angle in quadrant I, so ( )1sec 4 1.32− ≈ .


664


46. 1 1 1csc 5 sin

5− −=

We seek the angle , ,2 2

π πθ θ− ≤ ≤ whose sine

equals 1

5. Now

1sin

5θ = , so θ lies in

quadrant I. The calculator yields 1 1sin 0.20

5− ≈ ,

which is an angle in quadrant I, so 1csc 5 0.20− ≈ .

47. 1 1 1cot 2 tan

2− −=

We seek the angle , 0 ,θ θ π≤ ≤ whose tangent

equals 1

2. Now

1tan


quadrant I. The calculator yields 1 1an 0.46

2− ≈ ,

which is an angle in quadrant I, so

( )1cot 2 0.46− ≈ .

48. 1 1 1sec ( 3) cos

3− − − = −


equals 1

3− . Now

1cos

3θ = − , θ lies in

quadrant II. The calculator yields

1 1cos 1.91

3− − ≈

, which is an angle in

quadrant II, so ( )1sec 3 1.91− − ≈ .

49. ( )1 1 1csc 3 sin

3− − − = −

We seek the angle , ,2 2

π πθ θ− ≤ ≤ whose sine

equals 1

3− . Now

1sin

3θ = − , so θ lies in

quadrant IV. The calculator yields

1 1sin 0.34

3− − ≈ −


quadrant IV, so ( )1csc 3 0.34− − ≈ − .

50. 1 11cot tan ( 2)

2− − − = −


equals 2− . Now tan 2θ = − , so θ lies in quadrant II. The calculator yields

( )1tan 2 1.11− − ≈ − , which is an angle in

quadrant IV. Since θ lies in quadrant II, 1.11 2.03θ π≈ − + ≈ . Therefore,

1 1cot 2.03

2− − ≈

.

51. ( )1 1 1cot 5 tan

5− − − = −


equals 1

5− . Now

1tan



1 1tan 0.42

5− − ≈ −


quadrant IV. Since θ is in quadrant II, 0.42 2.72θ π≈ − + ≈ . Therefore,

( )1cot 5 2.72− − ≈ .


665


52. ( )1 1 1cot 8.1 tan

8.1− − − = −


equals 1

8.1− . Now

1tan

8.1θ = − , so θ lies in


1 1tan 0.12

8.1− − ≈ −


quadrant IV. Since θ is in quadrant II,

0.12 3.02θ π≈ − + ≈ . Thus, ( )1cot 8.1 3.02− − ≈ .

53. 1 13 2csc sin

2 3− − − = −

We seek the angle , , 02 2

π πθ θ θ− ≤ ≤ ≠ ,

whose sine equals 2

3− . Now

2sin

3θ = − , so θ

lies in quadrant IV. The calculator yields

1 2sin 0.73

3− − ≈ −


quadrant IV, so 1 3csc 0.73

2− − ≈ −

.

54. 1 14 3sec cos

3 4− − − = −

We are finding the angle , 0 , 2

πθ θ π θ≤ ≤ ≠ ,

whose cosine equals 3

4− . Now

3cos

4θ = − , so

θ lies in quadrant II. The calculator yields

1 3cos 2.42

4− − ≈


quadrant II, so 1 4sec 2.42

3− − ≈

.

55. 1 13 2cot tan

2 3− − − = −

We are finding the angle , 0 ,θ θ π≤ ≤ whose

tangent equals 2

3− . Now

2tan

3θ = − , so θ

lies in quadrant II. The calculator yields

1 2tan 0.59

3− − ≈ −



0.59 2.55θ π≈ − + ≈ . Thus, 1 3cot 2.55

2− − ≈

.

56. ( )1 1 1cot 10 tan

10− − − = −

We are finding the angle , 0 ,θ θ π≤ ≤ whose

tangent equals 1

10− . Now

1tan

10θ = − , so θ

lies in quadrant II. The calculator yields

1 1tan 0.306

10− − ≈ −



0.306 2.84θ π≈ − + ≈ . So, ( )1cot 10 2.84− − ≈ .

57. Let 1tan uθ −= so that tan uθ = , 2 2

π πθ− < < ,

u−∞ < < ∞ . Then,

( )1

2

2 2

1 1cos tan cos

sec sec1 1

1 tan 1

u

u

θθ θ

θ

− = = =

= =+ +

58. Let 1cos uθ −= so that cos uθ = , 0 θ π≤ ≤ , 1 1u− ≤ ≤ . Then,

( )1 2

2 2

sin cos sin sin

1 cos 1

u

u

θ θ

θ

− = =

= − = −


666


59. Let 1sin uθ −= so that sin uθ = , 2 2

π πθ− ≤ ≤ ,

1 1u− ≤ ≤ . Then,

( )1

2 2

2

sintan sin tan

cossin sin

cos 1 sin

1

u

u

u

θθθ

θ θ

θ θ

− = =

= =−

=−

60. Let 1cos uθ −= so that cos uθ = , 0 θ π≤ ≤ , 1 1u− ≤ ≤ . Then,

( )1

2 2

2

sintan cos tan

cos

sin 1 cos

cos cos

1

u

u

u

θθθ

θ θθ θ

− = =

−= =

−=

61. Let 1sec uθ −= so that sec uθ = , 0 θ π≤ ≤ and

2

πθ ≠ , 1u ≥ . Then,

( )1 2 2

2

2 2

2

sin sec sin sin 1 cos

1 sec 11

sec sec

1

u

u

u

θ θ θ

θθ θ

− = = = −

−= − =

−=

62. Let 1cot uθ −= so that cot uθ = , 0 θ π< < , u−∞ < < ∞ . Then,

( )1 2

2

2 2

1sin cot sin sin

csc

1 1

1 cot 1

u

u

θ θθ

θ

− = = =

= =+ +

63. Let 1csc uθ −= so that csc uθ = , 2 2

π πθ− ≤ ≤ ,

1u ≥ . Then,

( )1

2 2

2

sincos csc cos cos cot sin

sin

cot cot csc 1

csc csc csc

1

u

u

u

θθ θ θ θθ

θ θ θθ θ θ

− = = ⋅ =

−= = =

−=


2

πθ ≠ , 1u ≥ . Then,

( )1 1 1cos sec cos

secu

uθ

θ− = = =

65. Let 1cot uθ −= so that cot uθ = , 0 θ π< < , u−∞ < < ∞ . Then,

( )1 1 1tan cot tan

cotu

uθ

θ− = = =


2

πθ ≠ , 1u ≥ . Then,

( )1 2

2 2

tan sec tan tan

sec 1 1

u

u

θ θ

θ

− = =

= − = −

67. 1 112 12cos sin

13 13g f − − =

Let 1 12sin

13θ −= . Since

12sin

13θ = and

2 2

π πθ− ≤ ≤ , θ is in quadrant I, and we let

12y = and 13r = . Solve for x: 2 2 2

2

2

12 13

144 169

25

25 5

x

x

x

x

+ =

+ =

=

= ± = ±


1 112 12 5cos sin cos

13 13 13

xg f

rθ− − = = = =


667


68. 1 15 5sin cos

13 13f g − − =

Let 1 5cos

13θ −= . Since

5cos

13θ = and

0 θ π≤ ≤ , θ is in quadrant I, and we let 5x = and 13r = . Solve for y:

2 2 2

2

2

5 13

25 169

144

144 12

y

y

y

y

+ =

+ =

=

= ± = ±

Since θ is in quadrant I, 12y = .

1 15 5 12sin cos sin

13 13 13

yf g

rθ− − = = = =

69. 1 1

1

7 7cos sin

4 4

2 3cos

2 4

g fπ π

π

− −

−

=

= − =

70. 1 1

1

5 5sin cos

6 6

3sin

2 3

f gπ π

π

− −

−

=

= − = −

71. 1 13 3tan sin

5 5h f − − − = −

Let 1 3sin

5θ − = −

. Since

3sin

5θ = − and

2 2

π πθ− ≤ ≤ , θ is in quadrant IV, and we let

3y = − and 5r = . Solve for x: 2 2 2

2

2

( 3) 5

9 25

16

16 4

x

x

x

x

+ − =

+ =

=

= ± = ±


1 13 3tan sin

5 5

3 3tan

4 4

h f

y

xθ

− − − = −

−= = = = −

72. 1 14 4tan cos

5 5h g − − − = −

Let 1 4cos

5θ − = −

. Since

4cos

5θ = − and

0 θ π≤ ≤ , θ is in quadrant II, and we let 4x = − and 5r = . Solve for y:

2 2 2

2

2

( 4) 5

16 25

9

9 3

y

y

y

y

− + =

+ =

=

= ± = ±


1 14 4tan cos

5 5

3 3tan

4 4

h g

y

xθ

− − − = −

−= = = = −

73. 1 112 12cos tan

5 5g h− − =

Let 1 12tan

5θ −= . Since

12tan

5θ = and

2 2


5x = and 12y = . Solve for r: 2 2 2

2

5 12

25 144 169

169 13

r

r

r

= +

= + =

= ± = ±

Now, r must be positive, so 13r = .

1 112 12 5cos tan cos

5 5 13

xg h

rθ− − = = = =

74. 1 15 5sin tan

12 12f h− − =

Let 1 5tan

12θ −= . Since

5tan

12θ = and

2 2


12x = and 5y = . Solve for r: 2 2 2

2

12 5

144 25 169

169 13

r

r

r

= +

= + =

= ± = ±

Now, r must be positive, so 13r = .

1 15 5 5sin tan sin

12 12 13

yf h

rθ− − = = = =


668


75. 1 1

1

4 4cos sin

3 3

3cos

2 6

g fπ π

π

− −

−

− = −

= =

76. 1 1

1

5 5cos sin

6 6

1 2cos

2 3

g fπ π

π

− −

−

− = −

= − =

77. 1 11 1tan cos

4 4h g − − − = −

Let 1 1cos

4θ − = −

. Since

1cos

4θ = − and

0 θ π≤ ≤ , θ is in quadrant II, and we let 1x = − and 4r = . Solve for y:

2 2 2

2

2

( 1) 4

1 16

15

15

y

y

y

y

− + =

+ =

=

= ±


1 11 1tan cos

4 4

15tan 15

1

h g

y

xθ

− − − = −

= = = = −−

78. 1 12 2tan sin

5 5h f − − − = −

Let 1 2sin

5θ − = −

. Since

2sin

5θ = − and

2 2

π πθ− ≤ ≤ , θ is in quadrant IV, and we let

2y = − and 5r = . Solve for x: 2 2 2

2

2

( 2) 5

4 25

21

21

x

x

x

x

+ − =

+ =

=

= ±


1 12 2tan sin

5 5

2 2 21tan

2121

h f

y

xθ

− − − = −

−= = = = −

79. a. Since the diameter of the base is 45 feet, we

have 45

22.52

r = = feet. Thus,

1 22.5cot 31.89

14θ − = = °

.

b. 1cot

cot cot

r

hr

r hh

θ

θ θ

−=

= → =

Here we have 31.89θ = ° and 17h = feet.

Thus, ( )17cot 31.89 27.32r = ° = feet and

the diameter is ( )2 27.32 54.64= feet.

c. From part (b), we get cot

rh

θ= .

The radius is 22

612

1 = feet.

6137.96

cot 22.5 /14

rh

θ= = ≈ feet.

Thus, the height is 37.96 feet.

80. a. Since the diameter of the base is 6.68 feet,

we have 6.68

3.342

r = = feet. Thus,

1 3.34cot 50.14

4θ − = = °

b. 1cot

cot cot

r

hr r

hh

θ

θθ

−=

= → =

Here we have 50.14θ = ° and 4r = feet.

Thus, ( )4

4.79cot 50.14

h = =°

feet. The

bunker will be 4.79 feet high.

c. 1 4.22cot 54.88

6TGθ − = = °

From part (a) we have 50.14USGAθ = ° . For

steep bunkers, a larger angle of repose is required. Therefore, the Tour Grade 50/50 sand is better suited since it has a larger angle of repose.

Section 7.3: Trigonometric Equations

669


81. a. 2

12

2cot

2

2cot

2

x

y gt

x

y gt

θ

θ −

=+

= +

The artillery shell begins at the origin and

lands at the coordinates ( )6175,2450 . Thus,

( )( )

12

1

2 6175cot

2 2450 32.2 2.27

cot 2.437858 22.3

θ −

−

⋅ = ⋅ +

≈ ≈ °

The artilleryman used an angle of elevation of 22.3° .

b.

( ) ( )

0

0

sec

6175 sec 22.3sec

2.272940.23 ft/sec

v t

x

xv

t

θ

θ

=

°= =

=

82. Let. 1 1

2cot cos

1

xy x

x

− −= =+

−10 10

32π

2π−

Note that the range of 1coty x−= is ( )0, π , so

1 1tan

x− will not work.

83. 1 1 1sec cosy x

x− −= =

0−5

π

5

84. 1 1 1csc siny x

x− −= =

−10 10

2

π

2

π−

_

_

85 – 86. Answers will vary.

Section 7.3

1. 3 5 1

4 6

6 3

4 2

x x

x

x

− = − +=

= =


2

.

2. 2 1

, 2 2

−

3.

( ) ( )24 5 0

4 5 1 0

x x

x x

− − =− + =

4 5 0 or 1 0

5 or 1

4

x x

x x

− = + =

= = −


1, 4

−

.

4. 2 1 0x x− − =

( ) ( ) ( )( )( )

21 1 4 1 1

2 1

1 1 4

2

1 5

2

x− − ± − − −

=

± +=

±=

The solution set is 1 5 1 5

, 2 2

− +

.


670


5.

[ ][ ]2(2 1) 3(2 1) 4 0

(2 1) 1 (2 1) 4 0

2 (2 5) 0

x x

x x

x x

− − − − =− + − − =

− =

2 0 or 2 5 0

50 or

2

x x

x x

= − =

= =


0, 2

.

6. 3 25 2x x x− = −

Let 31 5 2y x= − and 2

2y x x= − . Use

INTERSECT to find the solution(s):

−6

6

−6 6

In this case, the graphs only intersect in one

location, so the equation has only one solution. Rounding as directed, the solutions set is { }0.76 .

7. 5

, 6 6

π π

8. 5

2 , 2 , is any integer6 6

k k kπ πθ θ π θ π = + = +

9. False because of the circular nature of the functions.

10. False, 2 is outside the range of the sin function.

11. 2sin 3 2

2sin 1

1sin

2

θθ

θ

+ == −

= −

7 112 or 2 , is any integer

6 6k k kθ θπ π= + π = + π

On 0 2θ≤ < π , the solution set is 7 11

,6 6

π π

.

12. 1

1 cos21

1 cos2

1cos

2

θ

θ

θ

− =

− =

=

52 or 2 , is any integer

3 3k k kθ θπ π= + π = + π

On 0 2θ≤ < π , the solution set is 5

,3 3

π π

.

13. 2

2

4cos 1

1cos

41

cos2

θ

θ

θ

=

=

= ±

2or , is any integer

3 3k k kθ θπ π= + π = + π

On the interval 0 2θ≤ < π , the solution set is 2 4 5

, , ,3 3 3 3

π π π π

.

14. 2 1tan

3

1 3tan

3 3

θ

θ

=

= ± = ±


6 6k k kθ θπ π= + π = + π


, , ,6 6 6 6

π π π π

.

15. 2

2

2

2sin 1 0

2sin 1

1sin

2

1 2sin

2 2

θθ

θ

θ

− =

=

=

= ± = ±


4 4k k k

π πθ π θ π= + = +


, , ,4 4 4 4

π π π π

.


671


16. 2

2

2

4cos 3 0

4cos 3

3cos

4

3cos

2

θθ

θ

θ

− =

=

=

= ±


6 6k k kθ θπ π= + π = + π


, , ,6 6 6 6

π π π π

.

17. ( )sin 3 1

33 2

22

, is any integer2 3

k

kk

θ

θ

θ

= −π= + π

π π= +

On the interval 0 2θ≤ < π , the solution set is 7 11

, ,2 6 6

π π π

.

18. tan 32

, is any integer2 3

22 , is any integer

3

k k

k k

θ

θ π π

πθ π

=

= +

= +


3

π

.

19. ( ) 1cos 2

2θ = −

2 42 2 or 2 2

3 32

or ,3 3

k k

k k

π πθ π θ π

π πθ π θ π

= + = +

= + = + k is any integer


, , ,3 3 3 3

π π π π

.

20. ( )tan 2 1θ = −

32 , is any integer

43

, is any integer8 2

k k

kk

θ

θ

π= + π

π π= +

On the interval 0 2θ≤ < π , the solution set is 3 7 11 15

, , ,8 8 8 8

π π π π

.

21. 3

sec 22

θ = −

3 2 3 42 or 2

2 3 2 34 4 8 4

or ,9 3 9 3

k k

k k

θ π θ ππ π

π π π πθ θ

= + = +

= + = +

k is any integer On the interval 0 2θ≤ < π , the solution set is

4 8 16, ,

9 9 9

π π π

.

22. 2

cot 33

θ = −

2 5, is any integer

3 65 3

, is any integer4 2

k k

kk

θ π π

π πθ

= +

= +


4

π

.

23. 2sin 1 0

2sin 1

1sin

2

θθ

θ

+ == −

= −


6 6k k kθ θπ π= + π = + π


,6 6

π π

.

24. cos 1 0

cos 1

θθ

+ == −

2 , is any integerk kθ = π + π

On the interval 0 2θ≤ < π , the solution set is { }π .

25. tan 1 0

tan 1

θθ

+ == −

3, is any integer

4k kθ π= + π


,4 4

π π

.


672


26. 3 cot 1 0

3 cot 1

1 3cot

33

θθ

θ

+ =

= −

= − = −

2, is any integer

3k kθ π= + π


,3 3

π π

.

27. 4sec 6 2

4sec 8

sec 2

θθθ

+ = −= −= −


3 3k k kθ θπ π= + π = + π


,3 3

π π

.

28. 5csc 3 2

5csc 5

csc 1

θθθ

− ===

2 , is any integer2

k kθ π= + π


π

.

29. 3 2 cos 2 1

3 2 cos 3

1 2cos

22

θθ

θ

+ = −

= −

= − = −


4 4k k kθ θπ π= + π = + π


,4 4

π π

.

30. 4sin 3 3 3

4sin 2 3

2 3 3sin

4 2

θθ

θ

+ =

= −

= − = −


3 3k k kθ θπ π= + π = + π


,3 3

π π

.

31. cos 2 12

θ π − = −

2 22

32 2

23

, is any integer4

k

k

k k

θ

θ

θ

π− = π + π

π= + π

π= + π


,4 4

π π

.

32. sin 3 118

θ π + =

3 218 2

43 2

94 2

, is any integer27 3

k

k

kk

θ

θ

θ

π π+ = + π

π= + π

π π= +


, ,27 27 27

π π π

.

33. tan 12 3

θ π + =

2 3 4

2 12

2 , is any integer6

k

k

k k

θ

θ

θ

π π+ = + π

π= − + π

π= − + π


6

π

.

34. 1

cos3 4 2

θ π − =

52 or 2

3 4 3 3 4 37 23

2 or 23 12 3 12

7 236 or 6 ,

4 4

k k

k k

k k

θ θ

θ θ

θ θ

π π π π− = + π − = + π

π π= + π = + π

π π= + π = + π

k is any integer.


4

π

.


673


35. 1

sin2

θ =

52 or 2

6 6k k

π πθ θ π θ π = + = +

, k is any

integer. Six solutions are 5 13 17 25 29

, , , , ,6 6 6 6 6 6

θ π π π π π π= .

36. tan 1θ =

4kθ θ π = + π

, k is any integer

Six solutions are 5 9 13 17 21

, , , , ,4 4 4 4 4 4


37. 3

tan3

θ = −

5

6kθ θ π = + π

, k is any integer

Six solutions are 5 11 17 23 29 35

, , , , ,6 6 6 6 6 6


38. 3

cos2

θ = −

5 72 or 2

6 6k k

πθ θ θ π = + π = + π

, k is any

integer. Six solutions are 5 7 17 19 29 31

, , , , ,6 6 6 6 6 6


39. cos 0θ =

32 or = 2

2 2k kθ θ θ π π = + π + π

, k is any

integer


, , , , ,2 2 2 2 2 2


40. 2

sin2

θ =

32 or 2

4 4k k

πθ θ θ π π = + π = +

, k is any

integer


, , , , ,4 4 4 4 4 4


41. ( ) 1cos 2

2θ = −

2 42 2 or 2 2 , is any integer

3 3k k kθ θπ π= + π = + π

2 or , is any integer

3 3k k kθ θ θ π π = + π = + π


, , , , ,3 3 3 3 3 3


42. ( )sin 2 1θ = −

32 2 , is any integer

2k kθ π= + π

3, is any integer

4k kθ θ π = + π


, , , , ,4 4 4 4 4 4


43. 3

sin2 2

θ = −


2 3 2 3k k k

θ θπ π= + π = + π

8 10

4 or 43 3

k kθ θ θ π π = + π = + π

, k is any

integer. Six solutions are 8 10 20 22 32 34

, , , , ,3 3 3 3 3 3


44. tan 12

θ = −

3, is any integer

2 4k k

θ π π= +

32 , is any integer

2k k

πθ θ π = +


, , , , ,2 2 2 2 2 2


45.

( )1

sin 0.4

sin 0.4 0.41

θθ −

=

= ≈

0.41θ ≈ or 0.41 2.73θ π≈ − ≈ .

The solution set is { }0.41, 2.73 .


674


46.

( )1

cos 0.6

cos 0.6 0.93

θθ −

=

= ≈

0.93θ ≈ or 2 0.93 5.36θ π≈ − ≈ .


47.

( )1

tan 5

tan 5 1.37

θθ −

=

= ≈

1.37θ ≈ or 1.37 4.51θ π≈ + ≈ .


48.

1

cot 2

1tan

21

tan 0.462

θ

θ

θ −

=

=

= ≈

0.46θ ≈ or 0.46 3.61θ π≈ + ≈ .


49.

( )1

cos 0.9

cos 0.9 2.69

θθ −

= −

= − ≈

2.69θ ≈ or 2 2.69 3.59θ π≈ − ≈ .


50.

( )1

sin 0.2

sin 0.2 0.20

θθ −

= −

= − ≈ −

0.20 2

6.08

θ π≈ − +≈

or ( )0.20

3.34

θ π≈ − −≈

.


51.

1

sec 4

1cos

41

cos 1.824

θ

θ

θ −

= −

= −

= − ≈

1.82θ ≈ or 2 1.82 4.46θ π≈ − ≈ .


52.

1

csc 3

1sin

31

sin 0.343

θ

θ

θ −

= −

= −

= − ≈ −

0.34 2

5.94

θ π≈ − +≈

or ( )0.34

3.48

θ π≈ − −≈

.


53.

1

5 tan 9 0

5 tan 9

9tan

59

tan 1.0645

θθ

θ

θ −

+ == −

= −

= − ≈ −

1.064

2.08

θ π≈ − +≈

or 1.064 2

5.22

θ π≈ − +≈


54.

1

4cot 5

5cot

44

tan5

4tan 0.675

5

θ

θ

θ

θ −

= −

= −

= −

= − ≈ −

0.675

2.47

θ π≈ − +≈

or 0.675 2

5.61

θ π≈ − +≈

.


55.

1

3sin 2 0

3sin 2

2sin

32

sin 0.733

θθ

θ

θ −

− ==

=

= ≈

0.73θ ≈ or 0.73 2.41θ π≈ − ≈ .


56.

1

4cos 3 0

4cos 3

3cos

43

cos 2.424

θθ

θ

θ −

+ == −

= −

= − ≈

2.42θ ≈ or 2 2.42 3.86θ π≈ − ≈ .



675


57. 22cos cos 0

cos (2cos 1) 0

θ θθ θ

+ =+ =

cos 0 or 2cos 1 0

3 2cos 1,

2 2 1cos

22 4

,3 3

θ θθθθ

θ

= + =π π = −=

= −

π π=

The solution set is 2 4 3

, , , 2 3 3 2

π π π π

.

58. 2sin 1 0

(sin 1)(sin 1) 0

θθ θ

− =+ − =

sin 1 0 or sin 1 0

sin 1 sin 1

3

2 2

θ θθ θ

θ θ

+ = − == − =

π π= =


, 2 2

π π

.

59. 22sin sin 1 0

(2sin 1)(sin 1) 0

θ θθ θ

− − =+ − =

2sin 1 0 or sin 1 0

2sin 1 sin 1

1sin

2 27 11

,6 6

θ θθ θ

θ θ

θ

+ = − == − =

π= − =

π π=

The solution set is 7 11

, , 2 6 6

π π π

.

60. 22cos cos 1 0

(cos 1)(2cos 1) 0

θ θθ θ

+ − =+ − =

cos 1 0 or 2cos 1 0

cos 1 2cos 1

1cos

25

,3 3

θ θθ θθ θ

θ

+ = − == − == π =

π π=


, , 3 3

π ππ

.

61. (tan 1)(sec 1) 0θ θ− − =

tan 1 0 or sec 1 0

tan 1 sec 1

50,

4 4

θ θθ θ

θθ

− = − == =

π π ==


0, , 4 4

π π

.

62. 1

(cot 1) csc 02

θ θ + − =

1cot 1 0 or csc 0

2cot 1 1

csc3 7 2,4 4 (not possible)

θ θθ

θθ

+ = − == −

=π π=


, 4 4

π π

.

63.

( )

( )( )

2 2

2 2

2

2

sin cos 1 cos

1 cos cos 1 cos

1 2cos 1 cos

2cos cos 0

cos 2cos 1 0

θ θ θ

θ θ θ

θ θθ θ

θ θ

− = +

− − = +

− = +

+ =+ =

cos 0 or 2cos 1 0

3 1, cos

2 2 22 4

,3 3

θ θπ πθ θ

π πθ

= + =

= = −

=


, , , 2 3 3 2

π π π π

.

64.

( )

( )( )

2 2

2 2

2

2

cos sin sin 0

1 sin sin sin 0

1 2sin sin 0

2sin sin 1 0

2sin 1 sin 1 0

θ θ θ

θ θ θ

θ θθ θ

θ θ

− + =

− − + =

− + =

− − =+ − =

2sin 1 0 or sin 1 0

1 sin 1sin

27 11 2,6 6

θ θθθ

πθπ πθ

+ = − === −=

=


, , 2 6 6

π π π

.


676


65. ( )( )( )( )

( )( )

2

2

2

2

sin 6 cos 1

sin 6 cos 1

1 cos 6cos 6

cos 6cos 5 0

cos 5 cos 1 0

θ θ

θ θ

θ θθ θ

θ θ

= − +

= +

− = +

+ + =+ + =

cos 5 0 or cos 1 0

cos 5 cos 1

(not possible)

θ θθ θ

θ π

+ = + == − = −

=

The solution set is { }π .

66. ( )( )( )

( ) ( )

( )( )

2

2

2

2

2

2sin 3 1 cos

2sin 3 1 cos

2 1 cos 3 1 cos

2 2cos 3 3cos

2cos 3cos 1 0

2cos 1 cos 1 0

θ θ

θ θ

θ θ

θ θθ θ

θ θ

= − −

= −

− = −

− = −

− + =− − =

2cos 1 0 or cos 1 0

1 cos 1cos

2 05

,3 3

θ θθθθ

π πθ

− = − ====

=


0, , 3 3

π π

.

67. ( )( )

cos sin

cos sin

cos sin

sin1

costan 1

5,

4 4

θ θθ θθ θθθθ

θ

= − −

= − −=

=

=π π=


, 4 4

π π

.

68. ( )( )( )

cos sin 0

cos sin 0

cos sin 0

sin cos

sin1

costan 1

3 7,

4 4

θ θ

θ θθ θ

θ θθθθ

θ

− − =

− − =

+ == −

= −

= −π π=


, 4 4

π π

.

69. tan 2sin

sin2sin

cossin 2sin cos

0 2sin cos sin

0 sin (2cos 1)

θ θθ θθθ θ θ

θ θ θθ θ

=

=

== −= −

2cos 1 0 or sin 0

1 0,cos

25

, 3 3

θ θθθ

θ

− = == π=

π π=


0, , , 3 3

π ππ

.

70.

2

tan cot

1tan

tan

tan 1

tan 1

3 5 7, , ,

4 4 4 4

θ θ

θθ

θθ

θ

=

=

== ±

π π π π=


, , , 4 4 4 4

π π π π

.


677


71. 2

2

2

2

1 sin 2cos

1 sin 2(1 sin )

1 sin 2 2sin

2sin sin 1 0

(2sin 1)(sin 1) 0

θ θθ θθ θ

θ θθ θ

+ =

+ = −

+ = −

+ − =− + =

2sin 1 0 or sin 1 0

1 sin 1sin

2 35 2,

6 6

θ θθθθ

θ

− = + == −=

π=π π=


, , 6 6 2

π π π

.

72.

( )

2

2

2

2

sin 2cos 2

1 cos 2cos 2

cos 2cos 1 0

cos 1 0

cos 1 0

cos 1

θ θθ θ

θ θ

θθ

θθ

= +

− = +

+ + =

+ =+ =

= −= π

The solution set is { }π .

73.

( )( )22sin 5sin 3 0

2sin 3 sin 1 0

θ θθ θ

− + =− + =

2sin 3 0 or sin 1 0

3sin (not possible)

2 2

θ θ

θ θ

− = − =π= =


π

.

74.

( )( )22cos 7cos 4 0

2cos 1 cos 4 0

θ θθ θ

− − =+ − =

2cos 1 0 or cos 4 0

1 cos 4sin

2 (not possible)2 4

,3 3

θ θθθ

θ

+ = − === −

π π=


, 3 3

π π

.

75.

( )( )

2

2

2

3(1 cos ) sin

3 3cos 1 cos

cos 3cos 2 0

cos 1 cos 2 0

θ θθ θ

θ θθ θ

− =

− = −

− + =− − =

cos 1 0 or cos 2 0

cos 1 cos 2

0 (not possible)

θ θθ θθ

− = − == ==

The solution set is { }0 .

76.

( )( )

2

2

2

4(1 sin ) cos

4 4sin 1 sin

sin 4sin 3 0

sin 1 sin 3 0

θ θθ θ

θ θθ θ

+ =

+ = −

+ + =+ + =

sin 1 0 or sin 3 0

sin 1 sin 3

3 (not possible)

2

θ θθ θ

πθ

+ = + == − = −

=


2

π

.

77. 2

2

2

2

3tan sec

23

sec 1 sec2

2sec 2 3sec

2sec 3sec 2 0

(2sec 1)(sec 2) 0

θ θ

θ θ

θ θθ θ

θ θ

=

− =

− =

− − =+ − =

2sec 1 0 or sec 2 0

1 sec 2sec

2 5,

(not possible) 3 3

θ θθθθ

+ = − === −

π π=


, 3 3

π π

.

78. 2

2

2

csc cot 1

1 cot cot 1

cot cot 0

cot (cot 1) 0

θ θθ θ

θ θθ θ

= +

+ = +

− =− =

cot 0 or cot 1

3 5, ,

2 2 4 4

θ θ

θ θ

= =π π π π= =


, , , 4 2 4 2

π π π π

.


678


79. 2

2

sec tan 0

tan 1 tan 0

θ θθ θ

+ =

+ + =

This equation is quadratic in tanθ .

The discriminant is 2` 4 1 4 3 0b ac− = − = − < . The equation has no real solutions.

80. sec tan cotθ θ θ= +

2 2

1 sin cos

cos cos sin

1 sin cos

cos sin cos1 1

cos sin cossin cos

1cos

θ θθ θ θ

θ θθ θ θ

θ θ θθ θ

θ

= +

+=

=

=

sin 1

2

θ

θ

=π=

Since sec and tan2 2

π π

do not exist, the

equation has no real solutions.

81. 5cos 0x x+ = Find the zeros (x-intercepts) of 1 5cosY x x= + :

−10

10

−3π 3π

−10

10

−3π 3π

−10

10

−3π 3π

1.31, 1.98, 3.84x ≈ −

82. 4sin 0x x− = Find the zeros (x-intercepts) of 1 4sinY x x= − :

−10

10

−3π 3π

−10

10

−3π 3π

−10

10

−3π 3π

2.47, 0, 2.47x ≈ −

83. 22 17sin 3x x− = Find the intersection of 1 22 17sinY x x= − and

2 3Y = :

−5

5

−π π

0.52x ≈

84. 19 8cos 2x x+ = Find the intersection of 1 19 8cosY x x= + and

2 2Y = :

−5

5

−π π

0.30x ≈ −

85. sin cosx x x+ = Find the intersection of 1 sin cosY x x= + and

2Y x= :

−3

3

−π π

1.26x ≈


679


86. sin cosx x x− = Find the intersection of 1 sin cosY x x= − and

2Y x= :

−3

3

−π π

1.26x ≈ −

87. 2 2cos 0x x− =

Find the zeros (x-intercepts) of 21 2cosY x x= − :

−3

3

−π π

−3

3

−π π

1.02,1.02x ≈ −

88. 2 3sin 0x x+ =

Find the zeros (x-intercepts) of 21 3sinY x x= + :

−3

3

−π π

−3

3

−π π

1.72, 0x ≈ −

89. ( )2 2sin 2 3x x x− =

Find the intersection of ( )21 2sin 2Y x x= − and

2 3Y x= :

−3

12

−π 3π2

__

−3

12

−π 3π2

__

0, 2.15x ≈

90. 2 3cos(2 )x x x= +

Find the intersection of 21Y x= and

2 3cos(2 )Y x x= + :

−5

10

−2π 2π

−5

10

−2π 2π

0.62, 0.81x ≈ −

91. 6sin 2, 0xx e x− = >

Find the intersection of 1 6sin xY x e= − and

2 2Y = :

−6

6

0 2π

−6

6

0 2π

0.76,1.35x ≈

92. 4cos(3 ) 1, 0xx e x− = >

Find the intersection of 1 4cos(3 ) xY x e= − and

2 1Y = :

−6

6

0 π

0.31x ≈

93. ( )2

2

2

0

4sin 3 0

4sin 3

3sin

4

3 3sin

4 2

f x

x

x

x

x

=

− =

=

=

= ± = ±

2 or

3 3x k x k

π ππ π= + = + , k is any integer


680


On the interval [ ]0, 2π , the zeros of f are

2 4 5, , ,

3 3 3 3

π π π π.

94. ( )( )

( )

( )

0

2cos 3 1 0

2cos 3 1

1cos 3

2

f x

x

x

x

=

+ =

= −

= −

2 43 2 or 3 2

3 32 2 4 2

or ,9 3 9 3

x k x k

k kx x

π ππ π

π π π π

= + = +

= + = +

k is any integer On the interval [ ]0, π , the zeros of f are

2 4 8, ,

9 9 9

π π π.

95. a. ( ) 0

3sin 0

sin 0

f x

x

x

===

0 2 or 2 ,x k x kπ π π= + = + k is any integer

On the interval [ ]2 ,4π− π , the zeros of f are

2 , , 0, , 2 , 3 , 4− π −π π π π π .

b. ( ) 3sinf x x=

c. ( ) 3

23

3sin21

sin2

f x

x

x

=

=

=

52 or 2 ,

6 6x k x k

π ππ π= + = + k is any integer

On the interval [ ]2 ,4π− π , the solution set is

11 7 5 13 17, , , , ,

6 6 6 6 6 6

π π π π π π − −

.

d. From the graph in part (b) and the results of

part (c), the solutions of ( ) 3

2f x > on the

interval [ ]2 ,4π− π is 11 7

6 6x x

π π− < < −

5 13 17or or

6 6 6 6x x

π π π π< < < <

.

96. a. ( ) 0

2cos 0

cos 0

f x

x

x

===

32 or 2 ,

2 2x k x k

π ππ π= + = + k is any

integer On the interval [ ]2 ,4π− π , the zeros of f are

3 3 5 7, , , , ,

2 2 2 2 2 2

π π π π π π− − .

b. ( ) 2cosf x x=

c. ( ) 3

2cos 3

3cos

2

f x

x

x

= −

= −

= −

5 72 or 2 ,

6 6x k x k

π ππ π= + = + k is any

integer On the interval [ ]2 ,4π− π , the solution set is

7 5 5 7 17 19, , , , ,

6 6 6 6 6 6

π π π π π π − −

.

d. From the graph in part (b) and the results of

part (c), the solutions of ( ) 3f x < − on the


681


interval [ ]2 ,4π− π is 7 5

6 6x x

π π− < < −

5 7 17 19or or

6 6 6 6x x

π π π π< < < <

.

97. ( ) 4 tanf x x=

a. ( ) 4

4 tan 4

tan 1

f x

x

x

= −= −= −

, is any integer4

x x k kπ π = − +

b. ( ) 4

4 tan 4

tan 1

f x

x

x

< −< −< −

Graphing 1 tany x= and 2 1y = − on the

interval ,2 2

π π −

, we see that 1 2y y< for

2 4x

π π− < < − or ,2 4

π π − −

.

2

π2

π−

−6

6

_ _

98. ( ) cotf x x=

a. ( ) 3

cot 3

f x

x

= −

= −

5, is any integer

6x x k k

π π = +

b. ( ) 3

cot 3

f x

x

> −

> −

Graphing 1

1

tany

x= and 2 3y = − on the

interval ( )0,π , we see that 1 2y y> for

50

6x

π< < or 5

0,6

π

.

−6

6

0 π

99. a, d. ( ) ( )3sin 2 2f x x= + ; ( ) 7

2g x =

b. ( ) ( )

( )

( )

( )

73sin 2 2

23

3sin 221

sin 22

f x g x

x

x

x

=

+ =

=

=

52 2 or 2 2

6 65

or ,12 12

x k x k

x k x k

π ππ π

π ππ π

= + = +

= + = +

k is any integer

On [ ]0, π , the solution set is 5

,12 12

π π

.

c. From the graph in part (a) and the results of part (b), the solution of ( ) ( )f x g x> on

[ ]0,π is 5

12 12x x

π π< <

or 5

,12 12

π π

.

100. a, d. ( ) 2cos 32

xf x = + ; ( ) 4g x =


682


b. ( ) ( )

2cos 3 42

2cos 12

1cos

2 2

f x g x

x

x

x

=

+ =

=

=

52 or 2

2 3 2 32 10

4 or 4 ,3 3

x xk k

x k x k

π ππ π

π ππ π

= + = +

= + = +

k is any integer

On [ ]0, 4π , the solution set is 2 10

,3 3

π π

.

c. From the graph in part (a) and the results of part (b), the solution of ( ) ( )f x g x< on

[ ]0,4π is 2 10

3 3x x

π π< <

or 2 10

,3 3

π π

.

101. a, d. ( ) 4cosf x x= − ; ( ) 2cos 3g x x= +

b. ( ) ( )

4cos 2cos 3

6cos 3

3 1cos

6 2

f x g x

x x

x

x

=− = +− =

= = −−

2 4

2 or 2 ,3 3

x k x kπ ππ π= + = +

k is any integer

On [ ]0, 2π , the solution set is 2 4

,3 3

π π

.


[ ]0,2π is 2 4

3 3x x

π π < <

or 2 4

,3 3

π π

.

102. a, d. ( ) 2sinf x x= ; ( ) 2sin 2g x x= − +

b. ( ) ( )

2sin 2sin 2

4sin 2

2 1sin

4 2

f x g x

x x

x

x

== − +=

= =

5

2 or 2 ,6 6

x k x kπ ππ π= + = +

k is any integer

On [ ]0, 2π , the solution set is 5

,6 6

π π

.


[ ]0, 2π is 5

6 6x x

π π< <

or 5

,6 6

π π

.

103. ( ) 7100 20sin

3P t t

π = +

a. Solve ( ) 100P t = on the interval [ ]0,1 .

7100 20sin 100

3

720sin 0

3

7sin 0

3

t

t

t

π

π

π

+ = = =

3

7

7, is any integer

3

, is any integer

t k k

t k k

π π=

=

We need 370 1k≤ ≤ , or 7

30 k≤ ≤ .


683


For 0k = , 0t = sec.

For 1k = , 3

0.437

t = ≈ sec.

For 2k = , 6

0.867

t = ≈ sec.

The blood pressure will be 100 mmHg after 0 seconds, 0.43 seconds, and 0.86 seconds.

b. Solve ( ) 120P t = on the interval [ ]0,1 .

7100 20sin 120

3

720sin 20

3

7sin 1

3

t

t

t

π

π

π

+ = = =

( )12

72 , is any integer

3 2

3 2, is any integer

7

t k k

kt k

π ππ= +

+=

We need

( )12

712 3

1 112 61 114 12

3 20 1

7

0 2

2

k

k

k

k

+≤ ≤

≤ + ≤

− ≤ ≤

− ≤ ≤

For 0k = , 3

0.21 sec14

t = ≈

The blood pressure will be 120mmHg after 0.21 sec .

c. Solve ( ) 105P t = on the interval [ ]0,1 .

1

1

7100 20sin 105

3

720sin 5

3

7 3sin

3 4

7 3sin

3 4

3 3sin

7 4

t

t

t

t

t

π

π

π

π

π

−

−

+ = = =

=

=

On the interval [ ]0,1 , we get 0.03t ≈

seconds, 0.39t ≈ seconds, and 0.89t ≈ seconds. Using this information, along with

the results from part (a), the blood pressure will be between 100 mmHg and 105 mmHg for values of t (in seconds) in the interval

[ ] [ ] [ ]0,0.03 0.39,0.43 0.86,0.89∪ ∪ .

104. ( ) 125sin 0.157 1252

h t tπ = − +

a. Solve ( ) 125sin 0.157 125 1252

h t tπ = − + =

on the interval [ ]0,40 .

125sin 0.157 125 1252

125sin 0.157 02

sin 0.157 02

t

t

t

π

π

π

− + =

− = − =

0.157 , is any integer2

0.157 , is any integer2

2 , is any integer0.157

t k k

t k k

kt k

π π

ππ

ππ

− =

= +

+=

For 0

20, 10 seconds0.157

k t

π+= = ≈ .

For 21, 30 seconds0.157

k t

ππ += = ≈ .

For 2

22, 50 seconds0.157

k t

ππ += = ≈ .

So during the first 40 seconds, an individual on the Ferris Wheel is exactly 125 feet above the ground when 10 secondst ≈ and again when 30 secondst ≈ .

b. Solve ( ) 125sin 0.157 125 2502

h t tπ = − + =


125sin 0.157 125 2502

125sin 0.157 1252

sin 0.157 12

t

t

t

π

π

π

− + =

− = − =


684


0.157 2 , is any integer2 2

0.157 2 , is any integer

2, is any integer

0.157

t k k

t k k

kt k

π π π

π ππ π

− = +

= ++=

For 0, 20 seconds0.157

k tπ= = ≈ .

For 2

1, 60 seconds0.157

k tπ π+= = ≈ .

For 4

2, 100 seconds0.157

k tπ π+= = ≈ .

So during the first 80 seconds, an individual on the Ferris Wheel is exactly 250 feet above the ground when 20 secondst ≈ and again when 60 secondst ≈ .

c. Solve ( ) 125sin 0.157 125 1252

h t tπ = − + >


125sin 0.157 125 1252

125sin 0.157 02

sin 0.157 02

t

t

t

π

π

π

− + >

− > − >

Graphing 1 sin 0.1572

y xπ = −

and 2 0y =

on the interval [ ]0,40 , we see that 1 2y y> for

10 30x< < .

−1.5

0

1.5

40

So during the first 40 seconds, an individual on the Ferris Wheel is more than 125 feet above the ground for times between about 10 and 30 seconds. That is, on the interval 10 30x< < , or ( )10,30 .

105. ( ) ( )70sin 0.65 150d x x= +

a. ( ) ( )( )( )

0 70sin 0.65 0 150

70sin 0 150

150 miles

d = +

= +=

b. Solve ( ) ( )70sin 0.65 150 100d x x= + = on

the interval [ ]0,20 .

( )( )

( )

70sin 0.65 150 100

70sin 0.65 50

5sin 0.65

7

x

x

x

+ =

= −

= −

1

1

50.65 sin 2

7

5sin 2

70.65

x k

kx

π

π

−

−

= − + − + =

3.94 2 5.94 2 or ,

0.65 0.65 is any integer

k kx x

k

π π+ +≈ ≈

For 0k = , 3.94 0 5.94 0

or 0.65 0.65

6.06 min 8.44 min

x x+ +≈ ≈

≈ ≈

For 1k = , 3.94 2 5.94 2

or 0.65 0.65

15.72 min 18.11 min

x xπ π+ +≈ ≈

≈ ≈

For 2k = , 3.94 4 5.94 4

or 0.65 0.65

25.39 min 27.78 min

x xπ π+ +≈ ≈

≈ ≈

So during the first 20 minutes in the holding pattern, the plane is exactly 100 miles from the airport when 6.06 minutesx ≈ ,

8.44 minutesx ≈ , 15.72 minutesx ≈ , and 18.11 minutesx ≈ .

c. Solve ( ) ( )70sin 0.65 150 100d x x= + > on

the interval [ ]0,20 .

( )( )

( )

70sin 0.65 150 100

70sin 0.65 50

5sin 0.65

7

x

x

x

+ >

> −

> −

Graphing ( )1 sin 0.65y x= and 2

5

7y = − on

the interval [ ]0,20 , we see that 1 2y y> for

0 6.06x< < , 8.44 15.72x< < , and


685


18.11 20x< < .

−1.5

0

1.5

20

So during the first 20 minutes in the holding pattern, the plane is more than 100 miles from the airport before 6.06 minutes, between 8.44 and 15.72 minutes, and after 18.11 minutes.

d. No, the plane is never within 70 miles of the airport while in the holding pattern. The minimum value of ( )sin 0.65x is 1− . Thus,

the least distance that the plane is from the airport is ( )70 1 150 80− + = miles.

106. ( ) ( )672sin 2R θ θ=

a. Solve ( ) ( )672sin 2 450R θ θ= = on the

interval 0,2

π

.

( )

( )

1

1

672sin 2 450

450 225sin 2

672 336225

2 sin 2336

225sin 2

3362

k

k

θ

θ

θ π

πθ

−

−

=

= =

= + + =

0.7337 2 2.408 2 or ,

2 2 is any integer

k k

k

π πθ θ+ +≈ ≈

For 0k = , 0.7337 0

20.36685

21.02

θ +=

≈≈ °

or 2.408 0

21.204

68.98

θ +=

≈≈ °

For 1k = , 0.7337 2

2

3.508

200.99

πθ +=

≈≈ °

or 2.408 2

2

4.3456

248.98

πθ +=

≈≈ °

So the golfer should hit the ball at an angle of either 21.02° or 68.98° .

b. Solve ( ) ( )672sin 2 540R θ θ= = on the

interval 0,2

π

.

( )

( )

1

1

672sin 2 540

540 135sin 2

672 168135

2 sin 2168

135sin 2

1682

k

k

θ

θ

θ π

πθ

−

−

=

= =

= + + =

0.9333 2 2.2083 2 or ,

2 2 is any integer

k k

k

π πθ θ+ +≈ ≈

For 0k = , 0.9330 0

20.46665

26.74

θ +=

≈≈ °

or 2.2083 0

21.10415

63.26

θ +=

≈≈ °

For 1k = , 0.9330 2

2

3.608

206.72

πθ +=

≈≈ °

or 2.2083 2

2

4.246

243.28

πθ +=

≈≈ °

So the golfer should hit the ball at an angle of either 26.74° or 63.26° .

c. Solve ( ) ( )672sin 2 480R θ θ= ≥ on the

interval 0,2

π

.

( )

( )

( )

672sin 2 480

480sin 2

6725

sin 27

θ

θ

θ

≥

≥

≥

Graphing ( )1 sin 2y x= and 2

5

7y = on the

interval 0,2

π

and using INTERSECT, we

see that 1 2y y≥ when 0.3978 1.1730x≤ ≤

radians, or 22.79 67.21x° ≤ ≤ ° .

2

π

−1.5

1.5

_0


686


2

π

−1.5

1.5

_0

So, the golf ball will travel at least 480 feet

if the angle is between about 22.79° and 67.21° .

d. No; since the maximum value of the sine function is 1, the farthest the golfer can hit the ball is ( )672 1 672= feet.

107. Find the first two positive intersection points of

1Y x= − and 2 tanY x= .

−12

20 2π

−12

20 2π

The first two positive solutions are 2.03x ≈ and 4.91x ≈ .

108. a. Let L be the length of the ladder with x and y being the lengths of the two parts in each hallway. L x y= +

3cos

3

cos

x

x

θ

θ

=

=

4

sin

4

sin

y

y

θ

θ

=

=

3 4( ) 3sec 4csc

cos sinL θ θ θ

θ θ= + = +

3sec tan 4csc cot 0θ θ θ θ− =

3

3

3sec tan 4csc cot

sec tan 4

csc cot 34

tan3

4tan 1.10064

347.74º

θ θ θ θθ θθ θ

θ

θ

θ

=

=

=

= ≈

≈

b. ( ) ( ) ( )3 4

47.74ºcos 47.74º sin 47.74º

9.87 feet

L = +

≈

c. Graph 1

3 4

cos sinY

x x= + and use the

MINIMUM feature:

0

20

0 90

An angle of 47.74θ ≈ ° minimizes the length at 9.87 feetL ≈ .

d. For this problem, only one minimum length exists. This minimum length is 9.87 feet, and it occurs when 47.74θ ≈ ° . No matter if we find the minimum algebraically (using calculus) or graphically, the minimum will be the same.

109. a. ( )2(34.8) sin 2

1079.8

θ=

( )

( )

2

1

107(9.8)sin 2 0.8659

(34.8)

2 sin 0.8659

2 60º or 120º

30º or 60º

θ

θθθ

−

= ≈

≈≈≈

b. Notice that the answers to part (a) add up to 90° . The maximum distance will occur when the angle of elevation is 90 2 45° ÷ = ° :

( ) ( )2(34.8) sin 2 4545 123.6

9.8R

° ° = ≈

The maximum distance is 123.6 meters.

c. Let 2

1

(34.8) sin(2 )

9.8

xY =

0

125

0 90

d.


687


110. a. 2(40) sin(2 )

1109.8

θ=

( )2

1

110 9.8sin(2 ) 0.67375

40

2 sin 0.67375

2 42.4º or 137.6º

21.2º or 68.8º

θ

θθ

θ

−

⋅= ≈

≈≈

≈

b. The maximum distance will occur when the angle of elevation is 45° :

( ) [ ]2(40) sin 2(45 )45 163.3

9.8R

°° = ≈

The maximum distance is approximately 163.3 meter

c. Let 2

1

(40) sin(2 )

9.8

xY = :

0

170

0 90

d.

111.

( )

2

2

2

12

sin 401.33

sin

1.33sin sin 40

sin 40sin 0.4833

1.33

sin 0.4833 28.90

θθ

θ

θ −

° =

= °°= ≈

= ≈ °

112.

( )

2

2

2

12

sin 501.66

sin

1.66sin sin 50

sin 50sin 0.4615

1.66

sin 0.4615 27.48

θθ

θ

θ −

° =

= °°= ≈

= ≈ °

113. Calculate the index of refraction for each:

1 11 2

2 2

sin

sinsin10º

10º 8º 1.2477sin8º

sin 20º20º 15º 30 ' 15.5º 1.2798

sin15.5ºsin 30º

30º 22º30 ' 22.5º 1.3066sin 22.5ºsin 40º

40º 29º 0 ' 29º 1.3259sin 29ºsin 50º

50º 35º 0 ' 35º 1.3356sin 35ºsin 60º

60º 40º30 ' 40.5ºsin

v

v

θθ θθ

=

≈

= ≈

= ≈

= ≈

= ≈

= 1.333540.5º

sin 70º70º 45º30 ' 45.5º 1.3175

sin 45.5ºsin80º

80º 50º 0 ' 50º 1.2856sin 50º

≈

= ≈

= ≈

Yes, these data values agree with Snell’s Law. The results vary from about 1.25 to 1.34.

114. 8

18

2

2.998 101.56

1.92 10

v

v

×= ≈×

The index of refraction for this liquid is about 1.56.

115. Calculate the index of refraction:

11 2

2

sin sin 40º40º , 26º ; 1.47

sin sin 26º

θθ θθ

= = = ≈

116. The index of refraction of crown glass is 1.52.

( )

2

2

2

12

sin 30º1.52

sin

1.52sin sin 30

sin 30sin 0.3289

1.52

sin 0.3352 19.20

θθ

θ

θ −

≈

= °°= ≈

≈ ≈ °

The angle of refraction is about 19.20° .


688


117. If θ is the original angle of incidence and φ is

the angle of refraction, then 2

sin

sinn

θφ

= . The

angle of incidence of the emerging beam is also

φ , and the index of refraction is 2

1

n. Thus, θ is

the angle of refraction of the emerging beam. The two beams are parallel since the original angle of incidence and the angle of refraction of the emerging beam are equal.

118. Here we have 1 1.33n = and 2 1.52n = .

1 B 2

2

1

2

1

1 12

1

sin cos

sin

cos

tan

1.52tan tan 48.8

1.33

B

B

B

B

B

n n

n

n

n

n

n

n

θθθ

θ

θ − −

=

=

=

= = ≈ °

119. Answers will vary.

120. Since the range of siny x= is 1 1y− ≤ ≤ , then

5siny x x= + cannot be equal to 3 when

4x π> or x π< − since you are multiplying the result by 5 and adding x.

Section 7.4

1. True

2. True

3. identity; conditional

4. 1−

5. 0

6. True

7. False, you need to work with one side only.

8. True

9. sin 1 1

tan csccos sin cos

θθ θθ θ θ

⋅ = ⋅ =

10. cos 1 1

cot secsin cos sin

θθ θθ θ θ

⋅ = ⋅ =

11. ( )

( )2

2

cos 1 sincos 1 sin

1 sin 1 sin 1 sincos 1 sin

cos1 sin

cos

θ θθ θθ θ θ

θ θθ

θθ

++⋅ =− + −

+=

+=

12. ( )

( )2

2

sin 1 cossin 1 cos

1 cos 1 cos 1 cossin 1 cos

sin1 cos

sin

θ θθ θθ θ θ

θ θθ

θθ

−−⋅ =+ − −

−=

−=

13.

( )2

2 2

2 2

sin cos cos sin

cos sin

sin sin cos cos cos sin

sin cos

sin sin cos cos cos sin

sin cos

sin cos sin cos cos sin

sin cos

θ θ θ θθ θ

θ θ θ θ θ θθ θ



+ −+

+ + −=

+ + −=

+ + −=

1

sin cosθ θ=

14. ( ) ( )

2

2

1 1 1 cos 1 cos

1 cos 1 cos 1 cos 1 cos

2

1 cos2

sin

v v

v v v v

v

v

+ + −+ =− + − +

=−

=

Section 7.4: Trigonometric Identities

689


15. ( )( )

2 2

sin cos sin cos 1

sin cos

sin 2sin cos cos 1

sin cos

θ θ θ θθ θ

θ θ θ θθ θ

+ + −

+ + −=

2 2sin cos 2sin cos 1

sin cos1 2sin cos 1

sin cos2sin cos

sin cos2

θ θ θ θθ θ

θ θθ θ

θ θθ θ

+ + −=

+ −=

=

=

16. ( ) ( ) 2

2 2

2 2

tan 1 tan 1 sec

tan

tan 2 tan 1 sec

tan

tan 1 2 tan sec

tan

θ θ θθ

θ θ θθ

θ θ θθ

+ + −

+ + −=

+ + −=

2 2sec 2 tan sec

tan2 tan

tan2

θ θ θθ

θθ

+ −=

=

=

17. ( )( )( )( )

2

2

3sin 1 sin 13sin 4sin 1

sin 1 sin 1sin 2sin 1

3sin 1

sin 1

θ θθ θθ θθ θθ

θ

+ ++ + =+ ++ ++=

+

18. ( )( )

( )2

2

cos 1 cos 1cos 1

cos cos 1cos cos

cos 1

cos

θ θθθ θθ θ

θθ

+ −− =−−

+=

19. 1 cos

csc cos cos cotsin sin

θθ θ θ θθ θ

⋅ = ⋅ = =

20. 1 sin

sec sin sin tancos cos

θθ θ θ θθ θ

⋅ = ⋅ = =

21. 2 2 2 21 tan ( ) 1 ( tan ) 1 tan secθ θ θ θ+ − = + − = + =

22. 2 2 2 21 cot ( ) 1 ( cot ) 1 cot cscθ θ θ θ+ − = + − = + =

23.

2 2

sin coscos (tan cot ) cos

cos sin

sin coscos

cos sin

1cos

cos sin

1

sincsc

θ θθ θ θ θθ θθ θθ

θ θ

θθ θ

θθ

+ = + += =

=

=

24.

2 2

cos sinsin (cot tan ) sin

sin cos

cos sinsin

sin cos

1sin

sin cos

1

cossec

θ θθ θ θ θθ θ

θ θθθ θ

θθ θ

θθ

+ = + += =

=

=

25. 2 2

2

2

1tan cot cos tan cos

tan

1 cos

sin

u u u u uu

u

u

− = ⋅ −

= −

=

26. 2 2

2

2

1sin csc cos sin cos

sin

1 cos

sin

u u u u uu

u

u

− = ⋅ −

= −

=

27. 2 2(sec 1)(sec 1) sec 1 tanθ θ θ θ− + = − =

28. 2 2(csc 1)(csc 1) csc 1 cotθ θ θ θ− + = − =

29. 2 2(sec tan )(sec tan ) sec tan 1θ θ θ θ θ θ+ − = − =

30. 2 2(csc cot )(csc cot ) csc cot 1θ θ θ θ θ θ+ − = − =

31. 2 2 2 2

22

cos (1 tan ) cos sec

1cos

cos1

θ θ θ θ

θθ

+ = ⋅

= ⋅

=


690


32. 2 2 2 2

22

(1 cos )(1 cot ) sin csc

1sin

sin1

θ θ θ θ

θθ

− + = ⋅

= ⋅

=

33. 2 2

2 2

2 2

2 2

2 2

(sin cos ) (sin cos )

sin 2sin cos cos

sin 2sin cos cos

2sin 2cos

2(sin cos )

2 1

2

θ θ θ θθ θ θ θ


+ + −

= + +

+ − +

= +

= += ⋅=

34. 2 2 2 2

2 22 2

2 2

2 2

tan cos cot sin

sin coscos sin

cos sin

sin cos

1

θ θ θ θθ θθ θθ θθ θ

+

= ⋅ + ⋅

= +=

35. 4 2 2 2

2 2

4 2

sec sec sec (sec 1)

(tan 1) tan

tan tan

θ θ θ θθ θ

θ θ

− = −

= +

= +

36. 4 2 2 2

2 2

4 2

csc csc csc (csc 1)

(cot 1)cot

cot cot

θ θ θ θθ θ

θ θ

− = −

= +

= +

37.

2

1 sinsec tan

cos cos1 sin 1 sin

cos 1 sin

1 sin

cos (1 sin )

uu u

u uu u

u u

u

u u

− = −

− + = ⋅ + −=

+

2cos

cos (1 sin )

cos

1 sin

u

u u

u

u

=+

=+

38.

2

1 coscsc cot

sin sin1 cos 1 cos

sin 1 cos

1 cos

sin (1 cos )

uu u

u uu u

u u

u

u u

− = −

− + = ⋅ + −=

+

2sin

sin (1 cos )

sin

1 cos

u

u u

u

u

=+

=+

39. 2 2 2 2 2

2 2 2

2

2

3sin 4cos 3sin 3cos cos

3(sin cos ) cos

3 1 cos

3 cos

θ θ θ θ θθ θ θ

θθ

+ = + +

= + +

= ⋅ +

= +

40. 2 2 2 2 2

2 2 2

2

2

9sec 5 tan 4sec 5sec 5 tan

4sec 5(sec tan )

4sec 5 1

5 4sec

θ θ θ θ θθ θ θθ

θ

− = + −

= + −

= + ⋅

= +

41.

( )

2 2cos 1 sin1 1

1 sin 1 sin(1 sin )(1 sin )

11 sin

1 1 sin

1 1 sin

sin

θ θθ θ

θ θθ

θθ

θ

−− = −+ +

− += −+

= − −= − +=

42. 2 2sin 1 cos

1 11 cos 1 cos

(1 cos )(1 cos )1

1 cos1 (1 cos )

1 1 cos

cos

θ θθ θ

θ θθ

θθ

θ

−− = −− −

− += −−

= − += − −= −

43.

111 tan cot

11 tan 1cot

v vv

v

++ =− −

11 cot

cot1

1 cotcot

cot 1

cot 1

vv

vv

v

v

+ = −

+=−


691


44.

11csc 1 sin

1csc 1 1sin

v vv

v

−− =+ +

11 sin

sin1

1 sinsin

1 sin

1 sin

vv

vv

v

v

− = +

−=+

45.

1sec sin sincos

1csc cos cossinsin sin

cos costan tan

2 tan

θ θ θθθ θ θ

θθ θθ θθ θ

θ

+ = +

= +

= +=

46.

2

2

csc 1 csc 1 csc 1

cot cot csc 1

csc 1

cot (csc 1)

cot

cot (csc 1)

cot

csc 1

θ θ θθ θ θ

θθ θ

θθ θ

θθ

− − += ⋅+

−=+

=+

=+

47.

111 sin csc

11 sin 1csc

θ θθ

θ

++ =− −

csc 1csc

csc 1csc

csc 1 csc

csc csc 1csc 1

csc 1

θθ

θθ

θ θθ θ

θθ

+

=−

+= ⋅−

+=−

48.

11cos 1 sec

1cos 1 1sec1 sec

sec1 sec

sec1 sec

1 sec

θ θθ

θθ

θθ

θθθ

++ =− −

+

=−

+=−

49. 2 2

2 2

1 sin cos (1 sin ) cos

cos 1 sin cos (1 sin )

1 2sin sin cos

cos (1 sin )

v v v v

v v v v

v v v

v v

− − ++ =− −

− + +=−

1 2sin 1

cos (1 sin )

2 2sin

cos (1 sin )

2(1 sin )

cos (1 sin )

2

cos2sec

v

v v

v

v v

v

v v

vv

− +=−

−=−

−=−

=

=

50. 2 2

2 2

cos 1 sin cos (1 sin )

1 sin cos cos (1 sin )

cos 1 2sin sin

cos (1 sin )

v v v v

v v v v

v v v

v v

+ + ++ =+ +

+ + +=+

2 2sin

cos (1 sin )

2(1 sin )

cos (1 sin )

2

cos2sec

v

v v

v

v v

vv

+=+

+=+

=

=

51.

1sin sin sin

1sin cos sin cossin

1cos

1sin1

1 cot

θ θ θθ θ θ θ

θ

θθ

θ

= ⋅− −

=−

=−


692


52.

( )( )

( )

2 2sin 1 cos1 1

1 cos 1 cos1 cos 1 cos

11 cos

1 1 cos

cos

θ θθ θ

θ θθ

θθ

−− = −+ +

− += −

+= − −=

53. 2

2 2

2

2 2

(sec tan )

sec 2sec tan tan

1 1 sin sin2

cos coscos cos

θ θθ θ θ θ

θ θθ θθ θ

−

= − +

= − ⋅ ⋅ +

2

2

2

1 2sin sin

cos(1 sin )(1 sin )

1 sin(1 sin )(1 sin )

(1 sin )(1 sin )

1 sin

1 sin

θ θθ

θ θθ

θ θθ θ

θθ

− +=

− −=−

− −=− +

−=+

54. 2

2 2

2

2 2

(csc cot )

csc 2csc cot cot

1 1 cos cos2

sin sinsin sin

θ θθ θ θ θ

θ θθ θθ θ

−

= − +

= − ⋅ ⋅ +

2

2

2

1 2cos cos

sin(1 cos )(1 cos )

1 cos(1 cos )(1 cos )

(1 cos )(1 cos )

1 cos

1 cos

θ θθ

θ θθ

θ θθ θ

θθ

− +=

− −=−

− −=− +

−=+

55. cos sin

1 tan 1 cotcos sin

sin cos1 1

cos sincos sin

cos sin sin coscos sin

θ θθ θθ θ

θ θθ θθ θ

θ θ θ θθ θ

+− −

= +− −

= +− −

2 2

2 2

cos sin

cos sin sin cos

cos sin

cos sin(cos sin )(cos sin )

cos sinsin cos

θ θθ θ θ θθ θθ θθ θ θ θ

θ θθ θ

= +− −−=−− +=

−= +

56. cot tan

1 tan 1 cotcos sinsin cos

sin cos1 1

cos sincos sinsin cos

cos sin sin coscos sin

θ θθ θθ θθ θ


θ θ θ θθ θ

+− −

= +− −

= +− −

2 2

2 2

3 3

2 2

2 2

cos sin

sin (cos sin ) cos (sin cos )

cos cos sin sin

sin cos (sin cos )

sin cos

sin cos (sin cos )

(sin cos )(sin sin cos cos )

sin cos (sin cos )

sin sin cos cos

sin cos

si

θ θθ θ θ θ θ θ


θ θθ θ θ θθ θ θ θ θ θ


θ θ

= +− −

− ⋅ + ⋅=−

−=−

− + +=−

+ +=

=2 2n sin cos cos

sin cos sin cos sin cossin cos

1cos sin1 tan cot

θ θ θ θθ θ θ θ θ θθ θθ θ

θ θ

+ +

= + +

= + +


693


57.

2

cos sin costan

1 sin cos 1 sin

sin (1 sin ) cos

cos (1 sin )

θ θ θθθ θ θ

θ θ θθ θ

+ = ++ +

+ +=+

2 2sin sin cos

cos (1 sin )

sin 1

cos (1 sin )

1

cossec

θ θ θθ θθ

θ θ

θθ

+ +=+

+=+

=

=

58. 2

2 22 2

2

2

2

2

1(sin cos )

sin cos cos1cos sin (cos sin )

cossincossin

1cos

tan

1 tan

θ θθ θ θθ θ θ θ

θθθ

θθ

θθ

⋅=

− − ⋅

=−

=−

59. tan sec 1

tan sec 1tan (sec 1) tan (sec 1)

tan (sec 1) tan (sec 1)

θ θθ θ


+ −− +

+ − + −= ⋅− − + −

2 2

2 2

2 2

2 2

tan 2 tan (sec 1) sec 2sec 1

tan (sec 2sec 1)

sec 1 2 tan (sec 1) sec 2sec 1

sec 1 sec 2sec 1



+ − + − +=− − +

− + − + − +=− − + −

22sec 2sec 2 tan (sec 1)

2sec 22sec (sec 1) 2 tan (sec 1)

2sec 22(sec 1)(sec tan )

2(sec 1)

tan sec

θ θ θ θθ

θ θ θ θθ

θ θ θθ

θ θ

− + −=−

− + −=−

− +=−

= +

60. sin cos 1

sin cos 1(sin cos ) 1 (sin cos ) 1

(sin cos ) 1 (sin cos ) 1

θ θθ θ


− ++ −

− + + += ⋅+ − + +

2 2

2

sin cos sin cos sin cos 1

(sin cos ) 1


− + + + − +=+ −

2 2

2 2

2 2

2

sin cos 2sin 1

sin 2sin cos cos 1

sin (1 sin ) 2sin 1

2sin cos 1 1

2sin 2sin

2sin cos2sin (sin 1)

2sin cossin 1

cos

θ θ θθ θ θ θθ θ θ

θ θθ θθ θ

θ θθ θ

θθ

− + +=+ + −− − + +=

+ −+=

+=

+=

61.

2 2

2 2

2 2

2 2

sin costan cot cos sin

sin costan cotcos sin

sin coscos sin

sin coscos sin

sin cos

1

sin cos

θ θθ θ θ θ


θ θθ θ

θ θθ θ

θ θ

−− =+ +

−

=+

−=

= −

62.

2

2

1 cossec cos cos cossec cos 1 cos

cos cos

θθ θ θ θθ θ θ

θ θ

−− =+ +

2

2

2

2

2

2

1 coscos

1 coscos

1 cos

1 cos

sin

1 cos

θθ

θθ

θθ

θθ

−

=+

−=+

=+


694


63.

2 2

2 2

2 2

sin costan cot cos sin1 1


sin coscos sin 1

sin coscos sin

sin cos1

1

u uu u u u

u uu uu u

u u

u uu u

u u

u u

−− + = ++ +

−

= ++

−= +

2 2

2 2

2 2

2

sin cos 1

sin (1 cos )

sin sin

2sin

u u

u u

u u

u

= − +

= + −

= +

=

64. 2 2

2 2

22 2

sin costan cot cos sin2cos 2cos


sin coscos sin 2cos

sin coscos sin

u uu u u uu u

u uu uu u

u u

u u uu u

u u

−− + = ++ +

−

= ++

2 22

2 2

sin cos2cos

1

sin cos

1

u uu

u u

−= +

= +=

65.

1 sinsec tan cos cos

coscot cos cossin

1 sincos

cos cos sinsin

1 sin sin

cos cos (1 sin )

θθ θ θ θ

θθ θ θθ

θθ

θ θ θθ

θ θθ θ θ

++ =+ +

+

=+

+= ⋅+

sin 1

cos costan sec

θθ θθ θ

= ⋅

=

66.

1sec cos

11 sec 1cos

θ θθ

θ

=+ +

2

2

1cos

cos 1cos

1 1 cos

1 cos 1 cos

1 cos

1 cos1 cos

sin

θθ

θθ

θ θθθ

θθ

=+

− = ⋅ + − −=−−=

67. 2 2 2

2 2 2

2 2

2

2 2

2

2

1 tan 1 tan 1 tan1

1 tan 1 tan 1 tan

1 tan 1 tan

1 tan2 2

1 tan sec1

2sec

2cos

θ θ θθ θ θ

θ θθ

θ θ

θθ

− − ++ = ++ + +

− + +=+

= =+

= ⋅

=

68. 2 2

2 22 2

22

2 2

1 cot 1 cot2cos 2cos

1 cot csc

1 cot2cos

csc csc

θ θθ θθ θ

θ θθ θ

− −+ = ++

= − +

2

22 2

2

2 2 2

2 2

cos

sinsin 2cos1

sin

sin cos 2cos

sin cos

1

θθθ θ

θθ θ θθ θ

= − +

= − +

= +=

69. sec csc sec csc

sec csc sec csc sec csc1 1

csc secsin cos

θ θ θ θθ θ θ θ θ θ

θ θθ θ

− = −

= −

= −


695


70. 2

2

2

2

sin tan

cos cotsin

sincoscos

cossin

θ θθ θ

θθθθθθ

−−

−=

−

2

2

2

2

2

2

2

sin cos sincos

cos sin cossin

sin cos sin sin

cos cos sin cossin (sin cos 1) sin

cos cos (cos sin 1)

sin

cos

tan

θ θ θθ

θ θ θθ



θθθ

−

=−

−= ⋅−

−= ⋅−

=

=

71.

2

2

1sec cos cos

cos

1 cos

cos

sin

cossin

sincos

sin tan

θ θ θθ

θθ

θθ

θθθ

θ θ

− = −

−=

=

= ⋅

=

72.

2 2

sin costan cot

cos sin

sin cos

sin cos1

sin cos1 1

cos sinsec csc

θ θθ θθ θθ θθ θ

θ θ

θ θθ θ

+ = +

+=

=

= ⋅

=

73.

2

2

2

1 1 1 sin 1 sin

1 sin 1 sin (1 sin )(1 sin )

2

1 sin2

cos

2sec

θ θθ θ θ θ

θ

θθ

+ + −+ =− + − +

=−

=

=

74.

2 2

1 sin 1 sin

1 sin 1 sin

(1 sin ) (1 sin )

(1 sin )(1 sin )

θ θθ θ

θ θθ θ

+ −−− +

+ − −=− +

2 2

2

2

1 2sin sin (1 2sin sin )

1 sin4sin

cossin 1

4cos cos

4 tan sec

θ θ θ θθ

θθθθ θ

θ θ

+ + − − +=−

=

= ⋅ ⋅

=

75.

2

2

2

3

sec sec 1 sin

1 sin 1 sin 1 sin

sec (1 sin )

1 sinsec (1 sin )

cos1 1 sin

cos cos1 sin

cos

θ θ θθ θ θ

θ θθ

θ θθ

θθ θ

θθ

+ = ⋅ − − + +=

−+=

+= ⋅

+=

76. 1 sin (1 sin )(1 sin )

1 sin (1 sin )(1 sin )

θ θ θθ θ θ

+ + +=− − +

2

2

2

2

2

2

2

(1 sin )

1 sin

(1 sin )

cos

1 sin

cos

1 sin

cos cos

(sec tan )

θθ

θθθ

θ

θθ θ

θ θ

+=−+=

+ =

= +

= +


696


77. 2

2 2

2 2

2

(sec tan ) 1

csc (sec tan )

sec 2sec tan tan 1

csc (sec tan )

sec 2sec tan sec

csc (sec tan )

2sec 2sec tan

csc (sec tan )

v v

v v v

v v v v

v v v

v v v v

v v v

v v v

v v v

− +−

− + +=−

− +=−

−=−

2sec (sec tan )

csc (sec tan )

2sec

csc1

2cos1

sin1 sin

2cos 1sin

2cos

2 tan

v v v

v v v

v

v

v

vv

vv

vv

−=−

=

⋅=

= ⋅ ⋅

= ⋅

=

78. 2 2sec tan tan 1 tan

sec secsin

1cos1

coscos sin

cos1

coscos sin

v v v v

v vv

v

vv v

v

vv v

− + +=

+=

+

=

= +

79. sin cos sin cos

cos sinsin cos sin cos

cos cos sin sinsin cos

1 1cos sin

θ θ θ θθ θ

θ θ θ θθ θ θ θθ θθ θ

+ −−

= + − +

= + − +

2 2sin cos

cos sin1

cos sinsec csc

θ θθ θ

θ θθ θ

+=

=

=

80. sin cos cos sin

sin cossin cos cos sin

sin sin cos coscos sin

1 1sin cos

θ θ θ θθ θ


θ θθ θ

+ −−

= + − +

= + − +

2 2cos sin

cos sin1

cos sinsec csc

θ θθ θ

θ θθ θ

+=

=

=

81. 3 3

2 2

2 2

sin cos

sin cos


sin cos

sin cos sin cos

1 sin cos

θ θθ θ


θ θ θ θθ θ

++

+ − +=+

= + −= −

82. 3 3

2

2 2

2 2

sin cos

1 2cos


1 cos cos

θ θθ


+−

+ − +=− −

2 2

2 2

(sin cos )(sin cos sin cos )

sin cos(sin cos )(1 sin cos )

(sin cos )(sin cos )

11 sin cos cos

1sin coscos

1sin

cossin

1cos

sec sin

tan 1



θ θ θθ θ

θ

θθ

θθ

θ θθ

+ + −=−

+ −=+ −

−= ⋅−

−=

−

−=−

83.

( )

2 2 2 2

2 2

2

2 2

2 2

2

22 2

2 2

2

cos sin cos sin

1 tan sin1

cos

cos sin

cos sin

cos

coscos sin

cos sin

cos

θ θ θ θθ θ

θθ θθ θ

θθθ θ

θ θθ

− −=− −

−=−

= − ⋅−

=


697


84. 3 3

2

2

cos sin sin cos sin sin

sin sin sin sin

cot 1 sin

cot cos

θ θ θ θ θ θθ θ θ θ

θ θθ θ

+ − = + −

= + −

= +

85.

22 2 22 2

4 4 2 2 2 2

2cos (sin cos )(2cos 1)

cos sin (cos sin )(cos sin )

θ θ θθθ θ θ θ θ θ

− +− =− − +

2 2 2

2 2 2 2

2 2

2 2

2 2

2 2

2

(cos sin )

(cos sin )(cos sin )

cos sin

cos sin

cos sin

1 sin sin

1 2sin

θ θθ θ θ θ

θ θθ θθ θ

θ θθ

−=− +

−=+

= −

= − −

= −

86. 2 2 2

2 2

2 2

1 2cos 1 cos cos

sin cos sin cos

sin cos

sin cos

sin cos

sin cos sin cos

θ θ θθ θ θ θ

θ θθ θθ θ

θ θ θ θ

− − −=

−=

= −

sin cos

cos sintan cot

θ θθ θθ θ

= −

= −

87. 1 sin cos

1 sin cos(1 sin ) cos (1 sin ) cos

(1 sin ) cos (1 sin ) cos

θ θθ θ


+ ++ −

+ + + += ⋅+ − + +

2 2

2 2

2 2

2 2

2

1 2sin sin 2cos (1 sin ) cos

1 2sin sin cos

1 2sin sin 2cos (1 sin ) (1 sin )

1 2sin sin (1 sin )

2 2sin 2cos (1 sin )

2sin 2sin2(1 sin ) 2cos (1 sin )

2sin (1 sin )

2(1 sin )(1 co



θ θ θθ θ

θ θ θθ θ

θ

+ + + + +=+ + −

+ + + + + −=+ + − −

+ + +=+

+ + +=+

+ += s )

2sin (1 sin )

1 cos

sin

θθ θθ

θ

++=

88. 1 cos sin

1 cos sin(1 cos ) sin (1 cos ) sin

(1 cos ) sin (1 cos ) sin

θ θθ θ


+ ++ −

+ + + += ⋅+ − + +

2 2

2 2

2 2

2 2

1 2cos cos 2sin (1 cos ) sin

1 2cos cos sin

1 2cos cos 2sin (1 cos ) 1 cos

1 2cos cos (1 cos )



+ + + + +=+ + −

+ + + + + −=+ + − −

2

2 2cos 2sin (1 cos )

2cos 2cos2(1 cos ) 2sin (1 cos )

2cos (1 cos )

2(1 cos )(1 sin )

2cos (1 cos )

1 sin

cos1 sin

cos cossec tan

θ θ θθ θ

θ θ θθ θ

θ θθ θ

θθ

θθ θθ θ

+ + +=+

+ + +=+

+ +=+

+=

= +

= +

89. 2 2

2 2 2 2

2 2 2 2

2 2 2 2 2 2

2 2

( sin cos ) ( cos sin )

sin 2 sin cos cos

cos 2 sin cos sin

(sin cos ) (sin cos )

a b a b

a ab b

a ab b

a b

a b



+ + −

= + +

+ − +

= + + +

= +

90. 2 2 2 2 2(2 sin cos ) (cos sin )a aθ θ θ θ+ −

( )( )( )( )( )

2 2 2

2 4 2 2 4

2 2 2 4 2 2 4

2 4 2 2 4

22 2 2

22

2

4 sin cos

cos 2cos sin sin

4sin cos cos 2cos sin sin

cos 2cos sin sin

cos sin

1

a

a

a

a

a

a

a

θ θθ θ θ θ

θ θ θ θ θ θ

θ θ θ θ

θ θ

=

+ − +

= + − +

= + +

= +

=

=


698


91. tan tan tan tan

1 1cot cottan tan

tan tantan tantan tan

α β α βα β

α βα ββ αα β

+ +=+ +

+=+

tan tan

(tan tan )tan tan

tan tan

α βα βα β

α β

= + ⋅ + =

92. (tan tan )(1 cot cot )

(cot cot )(1 tan tan )

tan tan tan cot cot

tan cot cot cot cot

cot tan tan cot tan tan

tan tan cot cot cot

α β α βα β α β

α β α α ββ α β α β

α α β β α βα β β α α

+ −+ + −

= + −− + +

− −= + − − +

cot tan tan

0

β β α+ − −=

93. 2(sin cos ) (cos sin )(cos sin )α β β α β α+ + + − 2 2 2 2

2

sin 2sin cos cos cos sin

2sin cos 2cos

2cos (sin cos )

α α β β β αα β ββ α β

= + + + −

= += +

94. 2(sin cos ) (cos sin )(cos sin )α β β α β α− + + − 2 2 2 2

2

sin 2sin cos cos cos sin

2sin cos 2cos 2cos (sin cos )

α α β β β αα β β β α β

= − + + −

= − + = − −

95. 11

ln sec ln ln cos ln coscos

θ θ θθ

−= = = −

96. sin

ln tan ln ln sin ln coscos

θθ θ θθ

= = −

97.

( )2

2

ln 1 cos ln 1 cos

ln 1 cos 1 cos

ln 1 cos

ln sin

2 ln sin

θ θ

θ θ

θ

θ

θ

+ + −

= + ⋅ −

= −

=

=

98.

( )2 2

2 2

ln sec tan ln sec tan

ln sec tan sec tan

ln sec tan

ln tan 1 tan

ln 1

0

θ θ θ θ

θ θ θ θ

θ θ

θ θ

+ + −

= + ⋅ −

= −

= + −

==

99. ( ) sin tanf x x x= ⋅

( )

2

2

2

sinsin

cos

sin

cos

1 cos

cos

1 cos

cos cossec cos

xx

x

x

x

x

x

x

x xx x

g x

= ⋅

=

−=

= −

= −=

100. ( ) cos cotf x x x= ⋅

( )

2

2

2

coscos

sin

cos

sin

1 sin

sin

1 sin

sin sincsc sin

xx

x

x

x

x

x

x

x xx x

g x

= ⋅

=

−=

= −

= −=

Section 7.5: Sum and Difference Formulas

699


101. ( ) 1 sin cos

cos 1 sinf

θ θθθ θ

−= −+

( )( )( ) ( )

( )( )

( )

( )

( )

( )

2 2

2 2

1 sin 1 sin cos cos

cos 1 sin 1 sin cos

1 sin cos

cos 1 sin

1 sin cos

cos 1 sin

1 1

cos 1 sin

0

cos 1 sin

0

g


θ θθ θ

θ θθ θ

θ θ

θ θ

θ

− + ⋅= −+ + ⋅

− −=+

− +=

+−=+

=+

==

102. ( ) tan secf θ θ θ= +

( )

( )

( )

2

2

sin 1

cos cos1 sin

cos1 sin 1 sin

cos 1 sin

1 sin

cos 1 sin

cos

cos 1 sin

cos

1 sing

θθ θ

θθ

θ θθ θ

θθ θ

θθ θ

θθ

θ

= +

+=

+ −= ⋅−

−=−

=−

=−

=

103. ( )22

2

2 2

1 21200sec 2sec 1 1200 1

cos cos

1 2 cos1200

cos cos cos

θ θθ θ

θθ θ θ

− = −

= −

( )

( )

2

2

2

3

2

3

1 2 cos1200

cos cos

1200 1 1 cos

cos

1200 1 sin

cos

θθ θ

θ

θθ

θ

−=

+ −=

+=

104. ( )( )2

2

csc 1 sec tan4

csc seccsc 1 sec tan

4csc sec

tI A

A

θ θ θθ θ

θ θ θθ θ

− +=

− += ⋅

( ) ( )( )

( )

2

2

2 2

22 2

1 tan4 1 1

csc sec

4 1 sin 1 sin

4 1 sin

4 cos 2 cos

A

A

A

A A

θθ θ

θ θ

θ

θ θ

= − +

= − +

= −

= =

105. Answers will vary.

106. 2 2sin cos 1θ θ+ = 2 2tan 1 secθ θ+ =

2 21 cot cscθ θ+ =

107 – 108. Answers will vary.

Section 7.5

1. ( ) ( )( )22

2 2

5 2 1 3

3 4 9 16 25 5

− + − −

= + = + = =

2. 3

5−

3. a. 2 1 2

2 2 4⋅ =

b. 1 1

12 2

− =

4. 4, 5, 3 (Quadrant 2)

3cos

5

y r x

x

rα

= = = −

= = −

5. −

6. −

7. False

8. False

9. False

10. True


700


11.

( )

5 3 2sin sin

12 12 12

sin cos cos sin4 6 4 6

2 3 2 1

2 2 2 21

6 24

π π π = + π π π π= ⋅ + ⋅

= ⋅ + ⋅

= +

12.

( )

3 2sin sin

12 12 12


2 3 2 1

2 2 2 21

6 24

π π π = − π π π π= ⋅ − ⋅

= ⋅ − ⋅

= −

13.

( )

7 4 3cos cos

12 12 12

cos cos sin sin3 4 3 4

1 2 3 2

2 2 2 21

2 64

π π π = + π π π π= ⋅ − ⋅

= ⋅ − ⋅

= −

14. 7 3 4

tan tan12 12 12

π π π = +

tan tan4 3

1 tan tan4 3

1 3

1 1 3

1 3 1 3

1 3 1 3

1 2 3 3

1 3

4 2 3

2

2 3

π π+=

π π− ⋅

+=− ⋅

+ += ⋅ − +

+ +=−

+=−

= − −

15. ( )

( )

cos165º cos 120º 45º

cos120º cos 45º sin120º sin 45º

1 2 3 2

2 2 2 21

2 64

= += ⋅ − ⋅

= − ⋅ − ⋅

= − +

16. ( )

( )

sin105º sin 60º 45º

sin 60º cos 45º cos 60º sin 45º

3 2 1 2

2 2 2 21

6 24

= += ⋅ + ⋅

= ⋅ + ⋅

= +

17. tan15º tan(45º 30º )

tan 45º tan 30º

1 tan 45º tan 30º

31 33

331 1

3

3 3 3 3

3 3 3 3

= −−=

+ ⋅

−= ⋅

+ ⋅

− −= ⋅+ −

9 6 3 3

9 3

12 6 3

6

2 3

− +=−

−=

= −

18. tan195º tan(135º 60º )

tan135º tan 60º

1 tan135º tan 60º

1 3

1 ( 1) 3

1 3 1 3

1 3 1 3

= ++=

− ⋅− +=

− − ⋅

− + −= ⋅+ −

1 2 3 3

1 3

4 2 3

2

2 3

− + −=−

− +=−

= −


701


19.

( )

17 15 2sin sin

12 12 12

5 5sin cos cos sin

4 6 4 6

2 3 2 1

2 2 2 2

16 2

4

π π π = +

π π π π= ⋅ + ⋅

= − ⋅ + − ⋅

= − +

20. 19 15 4

tan tan12 12 12

π π π = +

5tan tan

4 35

1 tan tan4 3

1 3

1 1 3

1 3 1 3

1 3 1 3

1 2 3 3

1 3

4 2 3

2

2 3

π π+=

π π− ⋅

+=− ⋅+ += ⋅− ++ +=

−+=−

= − −

21. 1 1

sec3 412 cos cos

12 12 12

1


π − = = π π π − −

= π π π π⋅ + ⋅

1

2 1 2 32 2 2 2

1

2 64

4 2 6

2 6 2 6

4 2 4 6

2 6

4 2 4 6

4

6 2

=⋅ + ⋅

=+

−= ⋅+ −

−=−−=

−= −

22. 5 5 1

cot cot512 12 tan12

π π − − = − = π

1

3 2tan

12 12

1

tan tan4 6

1 tan tan4 6

−=π π +

−= π π+

π π− ⋅

1 tan tan4 6

tan tan4 61

1 133

1 313

3 1 3 1

3 1 3 1

π π − ⋅ = − π π +

− ⋅= − ⋅

+

− −= − ⋅+ −

3 3 3 1

3 1

4 2 3

2

2 3

− − += −−

−= −

= − +

23. sin 20º cos10º cos 20º sin10º sin(20º 10º )

sin 30º

1

2

⋅ + ⋅ = +=

=

24. sin 20º cos80º cos 20º sin80º sin(20º 80º )

sin( 60º )

sin 60º

3

2

⋅ − ⋅ = −= −= −

= −

25. cos 70º cos 20º sin 70º sin 20º cos(70º 20º )

cos90º

0

⋅ − ⋅ = +==

26. cos 40º cos10º sin 40º sin10º cos(40º 10º )

cos30º

3

2

⋅ + ⋅ = −=

=


702


27. ( )tan 20º tan 25ºtan 20º 25º

1 tan 20º tan 25ºtan 45º

1

+ = +−

==

28. ( )tan 40º tan10ºtan 40º 10º

1 tan 40º tan10ºtan 30º

3

3

− = −+

=

=

29. 7 7 7

sin cos cos sin sin12 12 12 12 12 12

6sin

12

sin2

1

π π π π π π ⋅ − ⋅ = −

π = −

π = −

= −

30. 5 7 5 7 5 7

cos cos sin sin cos12 12 12 12 12 12

12cos

12cos

1

π π π π π π ⋅ − ⋅ = +

π=

= π= −

31. 5 5 5

cos cos sin sin cos12 12 12 12 12 12

4cos

12

cos3

cos3

1

2

π π π π π π ⋅ + ⋅ = −

π = −

π = − π=

=

32. 5 5 5

sin cos cos sin sin18 18 18 18 18 18

6sin

18

sin3

3

2

π π π π π π ⋅ + ⋅ = +

π=

π=

=

33. 3

sin , 05 2

α α π= < <

2 5cos , 0

5 2β βπ= − < <

y

x

35

( , 3)x

xα

y

x

5y

β

( )2 5, y

2 5

2 2 2

2

3 5 , 0

25 9 16, 0

4

4 3cos , tan

5 4

x x

x x

x

α α

+ = >

= − = >=

= =

( )22 2

2

2 5 5 , 0

25 20 5, 0

5

5 5 1sin , tan

5 22 5

y y

y y

y

β β

+ = <

= − = <

= −

−= − = = −

a. sin( ) sin cos cos sin

3 2 5 4 5

5 5 5 5

6 5 4 5

25

2 5

25

α β α β α β+ = +

= ⋅ + ⋅ −

−=

=

b. cos( ) cos cos sin sin

4 2 5 3 5

5 5 5 5

8 5 3 5

25

11 5

25

α β α β α β+ = −

= ⋅ − ⋅ −

+=

=


703


c. sin( ) sin cos cos sin

3 2 5 4 5

5 5 5 5

6 5 4 5

25

10 5

25

2 5

5

α β α β α β− = −

= ⋅ − ⋅ −

+=

=

=

d. tan tan

tan( )1 tan tan

α βα βα β−− =

+ ⋅

3 14 2

3 11

4 2

54582

− − =

+ −

=

=

34. 5

cos , 05 24

sin , 05 2

α α

β β

π= < <

π= − − < <

y

x

5

α

y

( )5, y

5

−4

xy

x

5

( , )x −4

β

( )22 2

2

5 5 , 0

25 5 20, 0

20 2 5

2 5 2 5sin , tan 2

5 5

y y

y y

y

α α

+ = >

= − = >

= =

= = =

2 2 2

2

( 4) 5 , 0

25 16 9, 0

3

3 4 4cos , tan

5 3 3

x x

x x

x

β β

+ − = >

= − = >=

−= = = −


2 5 3 5 4

5 5 5 5

6 5 4 5

25

2 5

25

α β α β α β+ = +

= ⋅ + ⋅ −

−=

=


5 3 2 5 4

5 5 5 5

3 5 8 5

25

11 5

25

α β α β α β+ = −

= ⋅ − ⋅ −

+=

=


2 5 3 5 4

5 5 5 5

6 5 4 5

25

10 5

25

2 5

5

α β α β α β− = −

= ⋅ − ⋅ −

+=

=

=

d. tan tan

tan( )1 tan tan


+

42

34

1 23

103532

− − = + ⋅ −

=−

= −

35. 4

tan ,3 2

1cos , 0

2 2

α α

β β

π= − < < π

π= < <


704


y

x

4r

α−3

(−3, 4)

(1, )yy

x

2

β

y

1

2 2 2( 3) 4 25

5

4 3 3sin , cos

5 5 5

r

r

α α

= − + ==

−= = = −

2 2 2

2

1 2 , 0

4 1 3, 0

3

3 3sin , tan 3

2 1

y y

y y

y

β β

+ = >

= − = >

=

= = =


4 1 3 3

5 2 5 2

4 3 3

10

α β α β α β+ = +

= ⋅ + − ⋅

−=


3 1 4 3

5 2 5 2

3 4 3

10

α β α β α β+ = −

= − ⋅ − ⋅

− −=


4 1 3 3

5 2 5 2

4 3 3

10

α β α β α β− = −

= ⋅ − − ⋅

+=

d. tan tan

tan( )1 tan tan

43

34

1 33

4 3 33

3 4 33


+

− −=

+ − ⋅

− −

=−

4 3 3 3 4 3

3 4 3 3 4 3

48 25 3

39

25 3 48

39

− − += ⋅ − +

− −=−

+=

36. 5 3

tan ,12 2

1 3sin ,

2 2

α α

β β

π= π < <

π= − π < <

y

x−12

−5

α

r

(−12,−5)

βy

xx

1

( , )x −12

2 2 2( 12) ( 5) 169

13

5 5 12 12sin , cos

13 13 13 13

r

r

α α

= − + − ==

− −= = − = = −

2 2 2

2

( 1) 2 , 0

4 1 3, 0

3

3 1 3cos , tan

2 33

x x

x x

x

β β

+ − = <

= − = <

= −

−= − = =−


5 3 12 1

13 2 13 2

5 3 12 12 5 3

26 26

α β α β α β+ = +

= − ⋅ − + − ⋅ −

+ += =


705



12 3 5 1

13 2 13 2

12 3 5

26

α β α β α β+ = −

= − ⋅ − − − ⋅ −

−=


5 3 12 1

13 2 13 2

5 3 12

26

α β α β α β− = −

= − ⋅ − − − ⋅ −

−=

d. tan tan

tan( )1 tan tan

5 3 5 4 312 3 12

5 3 36 5 31

12 3 36


+ ⋅

−−= =

++ ⋅

15 12 3 36 5 3

36 5 3 36 5 3

540 507 3 180

1296 75

720 507 3

1221

240 169 3

407

− −= ⋅ + −

− +=−

−=

−=

37. 5 3

sin ,13 2

tan 3,2

α α

β β

π= − < < −π

π= − < < π

y

x

5

x α

13

( , )x 5

y

x−1

r

β

( )1, 3−

3

2 2 2

2

5 13 , 0

169 25 144, 0

12

12 12 5cos , tan

13 13 12

x x

x x

x

α α

+ = <

= − = <= −

−= = − = −

22 2( 1) 3 4

2

3 1 1sin , cos

2 2 2

r

r

β β

= − + ==

−= = = −


5 1 12 3

13 2 13 2

5 12 3 5 12 3 or

26 26

α β α β α β+ = +

= ⋅ − + − ⋅

− − += −


12 1 5 3

13 2 13 2

12 5 3

26

α β α β α β+ = −

= − ⋅ − − ⋅

−=


5 1 12 3

13 2 13 2

5 12 3

26

α β α β α β− = −

= ⋅ − − − ⋅

− +=

d.

( )( )

tan tantan( )

1 tan tan

53

125

1 312

5 12 312

12 5 312


+

− − −=

+ − ⋅ −

− +

=+

5 12 3 12 5 3

12 5 3 12 5 3

240 169 3

69

− + −= ⋅ + −

− +=


706


38. 1

cos , 02 21

sin , 03 2

α α

β β

π= − < <

π= < <

1y

x

2(1, )y

αy

x

β

y

x

31

( , 1)x

2 2 2

2

1 2 , 0

4 1 3, 0

3

3 3 3sin , tan 3

2 2 1

y y

y y

y

α α

+ = <

= − = <

= −

− −= = − = = −

2 2 2

2

1 3 , 0

9 1 8. 0

8 2 2

2 2 1 2cos , tan

3 42 2

x x

x x

x

β β

+ = >

= − = >

= =

= = =


3 2 2 1 1

2 3 2 3

1 2 6

6

α β α β α β+ = +

= − ⋅ + ⋅

−=


1 2 2 3 1

2 3 2 3

3 2 2

6

α β α β α β+ = −

= ⋅ − − ⋅

+=


3 2 2 1 1

2 3 2 3

1 2 6

6

α β α β α β− = −

= − ⋅ − ⋅

− −=

d.

( )

tan tantan( )

1 tan tan

23

42

1 34

4 3 24

4 64


+

− −=

+ − ⋅

− −

=−

4 3 2 4 6

4 6 4 6

16 3 4 2 4 18 12

16 6

18 3 16 2

10

9 3 8 2

5

− − += ⋅ − +

− − − −=−

− −=

− −=

39. 1

sin , in quadrant II3

θ θ=

a. 2cos 1 sinθ θ= − −2

11

3

11

9

8

9

2 2

3

= − −

= − −

= −

= −

b. sin sin cos cos sin6 6 6

1 3 2 2 1

3 2 3 2

3 2 2 2 2 3

6 6

π π πθ θ θ + = ⋅ + ⋅

= + −

− − += =

c. cos cos cos sin sin3 3 3

2 2 1 1 3

3 2 3 2

2 2 3

6

π π πθ θ θ − = ⋅ + ⋅

= − +

− +=


707


d. tan tan

4tan4 1 tan tan

41

12 2

11 1

2 2

πθπθ πθ

+ + = − ⋅

− +=

− − ⋅

1 2 2

2 22 2 1

2 2

2 2 1 2 2 1

2 2 1 2 2 1

8 4 2 1

8 1

4 2

7

− +

=+

− −= ⋅ + −

− +=−

9 −=

40. 1

cos , in quadrant IV4

θ θ=

a. 2sin 1 cosθ θ= − −

2

11

4

11

16

15

16

15

4

= − −

= − −

= −

= −

b. sin sin cos cos sin6 6 6

15 3 1 1

4 2 4 2

1 3 5

8

π π πθ θ θ − = ⋅ − ⋅

= − ⋅ − ⋅

− −=

c. cos cos cos sin sin3 3 3

1 1 15 3

4 2 4 2

1 3 5

8

π π πθ θ θ + = ⋅ − ⋅

= ⋅ − − ⋅

+=

d.

( )

tan tan4tan

4 1 tan tan4

15 1

1 15 1

1 15 1 15

1 15 1 15

θθ

θ

π−π − = π + ⋅

− −=+ − ⋅

− − += ⋅ − +

1 2 15 15

1 15

16 2 15

14

8 15

7

− − −=−

− −=−

+=

41. lies in quadrant Iα . Since 2 2 4x y+ = ,

4 2r = = . Now, ( , 1)x is on the circle, so 2 2

2 2

2

1 4

4 1

4 1 3

x

x

x

+ == −

= − =

Thus, 1

sin2

y

rα = = and

3cos

2

x

rα = = .

lies in quadrant IVβ . Since 2 2 1x y+ = ,

1 1r = = . Now, 1

,3

y

is on the circle, so

22

22

2

11

3

11

3

1 8 2 21

3 9 3

y

y

y

+ =

= −

= − − = − = −

Thus, 2 23 2 2

sin1 3

y

rβ

− −= = = and

13 1

cos1 3

x

rβ = = = . Thus,

( ) ( )sin

sin cos cos sin

1 1 3 2 2

2 3 2 3

1 2 6 1 2 6

6 6 6

f α β α βα β α β

+ = += ⋅ + ⋅

= + −

−= − =


708


42. From the solution to Problem 41, we have

1sin

2α = ,

3cos

2α = ,

2 2sin

3β −= , and

1cos

3β = . Thus,

( ) ( )cos

cos cos sin sin

3 1 1 2 2

2 3 2 3

3 2 2 3 2 2

6 6 6

g α β α βα β α β

+ = += ⋅ − ⋅

= − −

+= + =


1sin

2α = ,

3cos

2α = ,

2 2sin

3β −= , and

1cos

3β = . Thus,

( ) ( )cos

cos cos sin sin

3 1 1 2 2

2 3 2 3

3 2 2 3 2 2

6 6 6

g α β α βα β α β

− = −= ⋅ + ⋅

= + −

−= − =


1sin

2α = ,

3cos

2α = ,

2 2sin

3β −= , and

1cos

3β = . Thus,

( ) ( )sin

sin cos cos sin

1 1 3 2 2

2 3 2 3

1 2 6 1 2 6

6 6 6

f α β α βα β α β

− = −= ⋅ − ⋅

= − −

+= + =


1sin

2α = ,

3cos

2α = ,

2 2sin

3β −= , and

1cos

3β = . Thus,

1sin 1 32tancos 33 3

2

ααα

= = = = and

2 2sin 3tan 2 2

1cos3

βββ

−= = = − . Finally,

( ) ( )

( )( )

tan tantan

1 tan tan

32 2

33

1 2 23

32 2 33

32 61

3

3 6 2 3 2 6

3 2 6 3 2 6

3 3 6 2 18 2 24 3

9 6 6 6 6 24

27 3 24 2 8 2 9 3

15 5

hα βα β α β

α β++ = + =

−

+ −=

− −

−= ⋅

+

− −= ⋅+ −

− − +=− + −− −= =

−


3tan

3α = and tan 2 2β = − . Thus,

( ) ( )

( )( )

tan tantan

1 tan tan

32 2

33

1 2 23

32 2 33

32 61

3

3 6 2 3 2 6

3 2 6 3 2 6

3 3 6 2 18 2 24 3

9 6 6 6 6 24

27 3 24 2 8 2 9 3

15 5

hα βα β α β

α β−− = − =

+

− −=

+ −

+= ⋅

−

+ += ⋅− +

+ + +=+ − −+ += = −

−


709


47. sin sin cos cos sin2 2 2

1 cos 0 sin

cos

θ θ θ

θ θθ

π π π + = ⋅ + ⋅

= ⋅ + ⋅=

48. cos cos cos sin sin2 2 2

0 cos 1 sin

sin

θ θ θ

θ θθ

π π π + = ⋅ − ⋅

= ⋅ − ⋅= −

49. ( )( )

sin sin cos cos sin

0 cos 1 sin

sin

θ θ θθ θ

θ

π − = π ⋅ − π ⋅

= ⋅ − −=

50. ( )cos cos cos sin sin

1 cos 0 sin

cos

θ θ θθ θ

θ

π − = π⋅ + π⋅= − ⋅ + ⋅= −

51. ( )( )

sin sin cos cos sin

0 cos 1 sin

sin

θ θ θθ θ

θ

π + = π⋅ + π ⋅

= ⋅ + −= −

52. ( )cos cos cos sin sin

1 cos 0 sin

cos

θ θ θθ θ

θ

π + = π ⋅ − π⋅= − ⋅ − ⋅= −

53. ( ) tan tantan

1 tan tan0 tan

1 0 tantan

1tan

θθθ

θθ

θ

θ

π −π − =+ π⋅

−=+ ⋅

−=

= −

54. ( ) tan 2 tantan 2

1 tan 2 tan0 tan

1 0 tantan

1tan

θθθ

θθ

θ

θ

π −π − =+ π⋅

−=+ ⋅

−=

= −

55. 3 3 3

sin sin cos cos sin2 2 2

1 cos 0 sin

cos

θ θ θ

θ θθ

π π π + = ⋅ + ⋅

= − ⋅ + ⋅= −

56. 3 3 3

cos cos cos sin sin2 2 2

0 cos ( 1) sin

sin

θ θ θ

θ θθ

π π π + = ⋅ − ⋅

= ⋅ − − ⋅=

57. sin( ) sin( )

sin cos cos sin

sin cos cos sin

2sin cos


α β α βα β

+ + −= +

+ −=

58. cos( ) cos( )

cos cos sin sin

cos cos sin sin

2cos cos


α β α βα β

+ + −= −

+ +=

59. sin( ) sin cos cos sin

sin cos sin cos

sin cos cos sin

sin cos sin cos

1 cot tan

α β α β α βα β α β


α β

+ +=

= +

= +


cos cos cos cos

sin cos cos sin

cos cos cos cos

tan tan


α β α βα β α βα β

+ +=

= +

= +

61. cos( ) cos cos sin sin

cos cos cos cos

cos cos sin sin

cos cos cos cos

1 tan tan



α β

+ −=

= −

= −


sin cos sin cos

cos cos sin sin

sin cos sin cos

cot tan


α β α βα β α βα β

− +=

= +

= +


710



sin( ) sin cos cos sin

α β α β α βα β α β α β

+ +=− −

sin cos cos sincos cos

sin cos cos sincos cos

sin cos cos sincos cos cos cossin cos cos sincos cos cos cos

tan tan

tan tan

α β α βα β

α β α βα β

α β α βα β α βα β α βα β α βα βα β

+

=−

+=

−

+=−


cos( ) cos cos sin sin


+ −=− +

cos cos sin sincos cos


cos cos sin sincos cos cos coscos cos sin sincos cos cos cos

1 tan tan

1 tan tan

α β α βα β

α β α βα β

α β α βα β α βα β α βα β α β

α βα β

−

=+

−=

+

−=+

65. cos( )

cot( )sin( )

cos cos sin sin

sin cos cos sin

α βα βα βα β α βα β α β

++ =+

−=+

cos cos sin sinsin sin

sin cos cos sinsin sin

cos cos sin sinsin sin sin sinsin cos cos sinsin sin sin sin

cot cot 1

cot cot

α β α βα β

α β α βα β

α β α βα β α βα β α βα β α βα ββ α

−

=+

−=

+

−=+

66. cos( )

cot( )sin( )

cos cos sin sin

sin cos cos sin

α βα βα βα β α βα β α β

−− =−

+=−


sin cos cos sinsin sin

cos cos sin sinsin sin sin sinsin cos cos sinsin sin sin sin

cot cot 1

cot cot

α β α βα β

α β α βα β

α β α βα β α βα β α βα β α βα ββ α

+

=−

+=

−

+=−

67. 1

sec( )cos( )

1

cos cos sin sin

α βα β

α β α β

+ =+

=−

1sin sin


1 1sin sin

cos cos sin sinsin sin sin sin

csc csc

cot cot 1

α βα β α β

α β

α βα β α βα β α βα β

α β

=−

⋅=

−

=−

68. 1

sec( )cos( )

1

cos cos sin sin

α βα β

α β α β

− =−

=+

1cos cos


1 1cos cos

cos cos sin sincos cos cos cos

sec sec

1 tan tan

α βα β α β

α β

α βα β α βα β α βα β

α β

=+

⋅=

+

=+


711


69.

( )( )2 2 2 2

2 2 2 2

2 2 2 2 2 2

2 2

sin( )sin( )

sin cos cos sin sin cos cos sin

sin cos cos sin

sin (1 sin ) (1 sin )sin

sin sin sin sin sin sin

sin sin

α β α βα β α β α β α β

α β α βα β α βα α β β α βα β

− += − +

= −

= − − −

= − − +

= −

70.

( )( )2 2 2 2

2 2 2 2

2 2 2 2 2 2

2 2

cos( )cos( )

cos cos sin sin cos cos sin sin

cos cos sin sin

cos (1 sin ) (1 cos )sin

cos cos sin sin cos sin

cos sin

α β α βα β α β α β α β

α β α βα β α βα α β β α βα β

− += + −

= −

= − − −

= − − +

= −


(sin )( 1) (cos )(0)

( 1) sin , any integer

k

k

k k k

k

θ θ θθ θ

θ

+ π = ⋅ π + ⋅ π

= − +

= −


(cos )( 1) (sin )(0)

( 1) cos , any integer

k

k

k k k

k

θ θ θθ θ

θ

+ π = ⋅ π − ⋅ π

= − −

= −

73. 1 11sin sin cos 0 sin

2 6 2

2sin

3

3

2

π π

π

− − + = +

=

=

74. 1 13sin sin cos 1 sin 0

2 3

sin3

3

2

− − π + = + π=

=

75. 1 13 4sin sin cos

5 5− − − −

Let 1 3sin

5α −= and 1 4

cos5

β − = −

. α is in

quadrant I; β is in quadrant II. Then 3

sin5

α = ,

02

πα≤ ≤ , and 4

cos5

β = − , 2

π β π≤ ≤ .

2

2

cos 1 sin

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

β β= −

= − − = − = =

( )1 13 4sin sin cos sin

5 5

sin cos cos sin

3 4 4 3

5 5 5 5

12 12

25 2524

25

α β

α β α β

− − − − = − = −

= ⋅ − − ⋅

= − −

= −

76. 1 14 3sin sin tan

5 4− − − −

Let 1 4sin

5α − = −

and 1 3

tan4

β −= . α is in

quadrant IV; β is in quadrant I. Then

4sin

5α = − , 0

2απ− ≤ ≤ , and

3tan

4β = ,

02

β π< < .

2

2

cos 1 sin

4 16 9 31 1

5 25 25 5

α α= −

= − − = − = =

2

2

sec 1 tan

3 9 25 51 1

4 16 16 4

β β= +

= + = − = =

4

cos5

β =


712


2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

β β= −

= − = − = =

( )

1 14 3sin sin tan

5 4

sin

sin cos cos sin

4 4 3 3

5 5 5 5

16 9 25

25 25 251

α βα β α β

− − − − = −= −

= − ⋅ − ⋅

= − − = −

= −

77. 1 14 5cos tan cos

3 13− − +

Let 1 4tan

3α −= and 1 5

cos13

β −= . α is in

quadrant I; β is in quadrant I. Then 4

tan3

α = ,

02

πα< < , and 5

cos13

β = , 02

πβ≤ ≤ .

2

2

sec 1 tan

4 16 25 51 1

3 9 9 3

α α= +

= + = + = =

3

cos5

α =

2

2

sin 1 cos

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

2

2

sin 1 cos

5 25 144 121 1

13 169 169 13

β β= −

= − = − = =

( )

1 14 5cos tan cos

3 13

cos

cos cos sin sin

3 5 4 12

5 13 5 13

15 48 33

65 65 65

α βα β α β

− − +

= += −

= ⋅ − ⋅

= − = −

78. 1 15 3cos tan sin

12 5− − − −

Let 1 5tan

12α −= and 1 3

sin5

β − = −

. α is in

quadrant I; β is in quadrant IV. Then

5tan

12α = , 0

2α π< < , and

3sin

5β = − ,

02

απ− < < .

2

2

sec 1 tan

5 25 169 131 1

12 144 144 12

α α= +

= + = + = =

12

cos13

α =

2

2

sin 1 cos

12 144 25 51 1

13 169 169 13

α α= −

= − = − = =

2

2

cos 1 sin

3 9 16 41 1

5 25 25 5

β β= −

= − − = − = =

( )

1 15 3cos tan sin

12 5

cos

cos cos sin sin

12 4 5 3 48 15 33

13 5 13 5 65 65 65

α βα β α β

− − − − = −= +

= ⋅ + ⋅ − = − =

79. 1 15 3cos sin tan

13 4− − −

Let 1 5sin

13α −= and 1 3

tan4

β −= . α is in


713



sin13

α = ,

02

α π≤ ≤ , and 3

tan4

β = , 02

β π< < .

2

2

cos 1 sin

5 25 144 121 1

13 169 169 13

α α= −

= − = − = =

2

2

sec 1 tan

3 9 25 51 1

4 16 16 4

β β= +

= + = + = =

4

cos5

β =

2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

β β= −

= − = − = =

( )

1 15 3cos sin tan

13 4

cos

cos cos sin sin

12 4 5 3

13 5 13 548 15

65 6563

65

α βα β α β

− − − = −= +

= ⋅ + ⋅

= +

=

80. 1 14 12cos tan cos

3 13− − +

Let 1 4tan

3α −= and 1 12

cos13

β −= . α is in


tan3

α = ,

02

α π< < , and 12

cos13

β = , 02

β π≤ ≤ .

2

2

sec 1 tan

4 16 25 51 1

3 9 9 3

α α= +

= + = + = =

3

cos5

α =

2

2

sin 1 cos

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

2

2

sin 1 cos

12 144 25 51 1

13 169 169 13

β β= −

= − = − = =

1 14 12cos tan cos

3 13− − +

( )cos

cos cos sin sin

3 12 4 5

5 13 5 13

36 20

65 6516

65

α βα β α β

= += −

= ⋅ − ⋅

= −

=

81. 1 3tan sin

5 6

π− +

Let 1 3sin

5α −= . α is in quadrant I. Then

3sin

5α = , 0

2

πα≤ ≤ .

2

2

cos 1 sin

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

3sin 3 5 35tan =

4cos 5 4 45

ααα

= = ⋅ =


714


1

1

1

3tan sin tan

3 5 6tan sin

35 6 1 tan sin tan5 6

3 34 3

3 31

4 3

9 312

12 3 312

9 3 12 3 3

12 3 3 12 3 3

ππ

π

−

−

−

+ + = − ⋅

+=

− ⋅

+

=−

+ += ⋅− +

108 75 3 36

144 27

144 75 3

117

48 25 3

39

+ +=−

+=

+=

82. 1 3tan cos

4 5

π − −

Let 1 3cos


3cos

5α = , 0

2

πα≤ ≤ .

2

2

sin 1 cos

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

4sin 4 5 45tan =

3cos 5 3 35

ααα

= = ⋅ =

1

1

1

3tan tan cos

3 4 5tan cos

34 51 tan tan cos

4 5

4 11 1 3 13 3

4 7 3 7 71 13 3

ππ

π

−

−

−

− − = + ⋅

− −= = = − ⋅ = −

+ ⋅

83. 1 14tan sin cos 1

5− − +

Let 1 4sin

5α −= and 1cos 1β −= ; α is in

quadrant I. Then 4

sin5

α = , 02

πα≤ ≤ , and

cos 1β = , 0 β π≤ ≤ . So, 1cos 1 0β −= = .

2

2

cos 1 sin

4 16 9 31 1

5 25 25 5

α α= −

= − = − = =

4sin 4 5 45tan =

3cos 5 3 35

ααα

= = ⋅ =

( )

( )

1 1

1 1

1 1

4tan sin cos 1

5

4tan sin tan cos 1

54

1 tan sin tan cos 15

4 40 43 3

4 1 31 03

− −

− −

− −

− + = − ⋅

+= = =

− ⋅

84. 1 14tan cos sin 1

5− − +

Let 1 4cos

5α −= and -1sin 1β = ; α is in

quadrant I. Then 4

cos5

α = , 02

πα≤ ≤ , and

sin 1β = , 2 2

π πβ− ≤ ≤ . So, 1sin 12

πβ −= = .

2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

α α= −

= − = − = =

3sin 3 5 35tan =

4cos 5 4 45

ααα

= = ⋅ = , but tan2

π is

undefined. Therefore, we cannot use the sum formula for tangent. Rewriting using sine and cosine, we obtain:


715


( )( )

1 1

1 1

1 1

4sin cos sin 1

4 5tan cos sin 1

45tan cos sin 1

5

sin

cos

α βα β

− −

− −

− −

+ + = +

+=

+

sin cos cos sin

cos cos sin sin

3 4(0) (1)

5 54 3

(0) (1)5 5

445

3 35


+=−

+ = −

= = −−

85. ( )1 1cos cos sinu v− −+

Let 1cos uα −= and 1sin vβ −= .

Then cos , 0uα α= ≤ ≤ π , and

sin ,2 2

vβ βπ π= − ≤ ≤

1 1u− ≤ ≤ , 1 1v− ≤ ≤ 2 2sin 1 cos 1 uα α= − = − 2 2cos 1 sin 1 vβ β= − = −

( )1 1

2 2

cos cos sin cos( )

cos cos sin sin

1 1

u v

u v v u

α β

α β α β

− −+ = +

= −

= − − −

86. ( )1 1sin sin cosu v− −−

Let 1sin uα −= and 1cos vβ −= . Then

sin , 2 2

uα απ π= − ≤ ≤ , and

cos , 0vβ β= ≤ ≤ π .

1 1u− ≤ ≤ , 1 1v− ≤ ≤ 2 2cos 1 sin 1 uα α= − = − 2 2sin 1 cos 1 vβ β= − = −

( )1 1

2 2

sin sin cos sin( )

sin cos cos sin

1 1

u v

uv u v

α β

α β α β

− −− = −

= −

= − − −

87. ( )1 1sin tan sinu v− −−

Let 1tan uα −= and 1sin vβ −= . Then

tan , 2 2

uα απ π= − < < , and

sin ,2 2

vβ βπ π= − ≤ ≤ .

u−∞ < < ∞ , 1 1v− ≤ ≤

2 2sec tan 1 1uα α= + = +

2

1cos

1uα =

+

2 2cos 1 sin 1 vβ β= − = −

2

2

2

2

2

2

2

sin 1 cos

11

1

1 1

1

1

1

u

u

u

u

uu

u

α α= −

= −+

+ −=+

=+

=+

( )1 1

2

2 2

2

2

sin tan sin

sin( )

sin cos cos sin

11

1 1

1

1

u v

uv v

u u

u v v

u

α βα β α β

− −−

= −= −

= ⋅ − − ⋅+ +

− −=+

88. ( )1 1cos tan tanu v− −+

Let 1tan uα −= and 1tan vβ −= . Then

tan , 2 2


tan ,2 2

vβ βπ π= − < < .

u−∞ < < ∞ , v−∞ < < ∞

2 2sec tan 1 1uα α= + = +

2

1cos

1uα =

+


716


2

2

2

2

2

2

2

sin 1 cos

11

1

1 1

1

1

1

u

u

u

u

uu

u

α α= −

= −+

+ −=+

=+

=+

2 2sec tan 1 1vβ β= + = +

2

1cos

1vβ =

+

2

2

2

2

2

2

2

sin 1 cos

11

1

1 1

1

1

1

v

v

v

v

vv

v

β β= −

= −+

+ −=+

=+

=+

( )1 1

2 2 2 2

2 2

cos tan tan

cos( )

cos cos sin sin

1 1

1 1 1 11

1 1

u v

u v

u v u vuv

u v

α βα β α β

− −+

= += −

= ⋅ − ⋅+ + + +

−=+ ⋅ +

89. ( )1 1tan sin cosu v− −−

Let 1sin uα −= and 1cos vβ −= . Then

sin , 2 2

uα απ π= − ≤ ≤ , and

cos , 0vβ β= ≤ ≤ π .

1 1u− ≤ ≤ , 1 1v− ≤ ≤

2 2cos 1 sin 1 uα α= − = −

2

sintan

cos 1

u

u

ααα

= =−

2 2sin 1 cos 1 vβ β= − = −

2sin 1

tancos

v

v

βββ

−= =

( )1 1

2

2

2

2

tan sin cos tan( )

tan tan

1 tan tan

1

1

11

1

u v

u v

vu

u v

vu

α β

α βα β

− −− = −

−=+

−−−=

−+ ⋅−

2 2

2

2 2

2

2 2

2 2

1 1

1

1 1

1

1 1

1 1

uv u v

v u

v u u v

v u

uv u v

v u u v

− − −

−=− + −

−

− − −=− + −

90. ( )1 1sec tan cosu v− −+

Let 1tan uα −= and 1cos vβ −= . Then

tan , 2 2


cos , 0vβ β= ≤ ≤ π .

u−∞ < < ∞ , 1 1v− ≤ ≤

2 2sec tan 1 1uα α= + = +

2

1cos

1uα =

+

2

2

2

2

2

2

2

sin 1 cos

11

1

1 1

1

1

1

u

u

u

u

uu

u

α α= −

= −+

+ −=+

=+

=+

2 2sin 1 cos 1 vβ β= − = −


717


( )1 1sec tan cos

sec( )

1

cos( )

1

cos cos sin sin

u v

α β

α β

α β α β

− −+

= +

=+

=−

2

2 2

2

2 2

2

2

2

2

11

11 1

1

1

1 11

1

1

1

1

uv v

u u

v u v

u u

v u v

u

u

v u v

=⋅ − ⋅ −

+ +

=−−

+ +

=− −

+

+=− −

91. sin 3 cos 1θ θ− = Divide each side by 2:

1 3 1sin cos

2 2 2θ θ− =

Rewrite in the difference of two angles form

using 1

cos2

φ = , 3

sin2

φ = , and 3

φ π= :

1sin cos cos sin

21

sin( )2

θ φ θ φ

θ φ

− =

− =

5 or

6 65

3 6 3 67

2 6

θ φ θ φ

θ θ

θ θ

π π− = − =

π π π π− = − =

π π= =


, 2 6

π π

.

92. 3 sin cos 1θ θ+ = Divide each side by 2:

3 1 1sin cos

2 2 2θ θ+ =

Rewrite in the sum of two angles form using

3cos

2φ = ,

1sin

2φ = , and

6φ π= :

1sin cos cos sin

21

sin( )2

θ φ θ φ

θ φ

+ =

+ =

5 or

6 65

or 6 6 6 6

20 or

3

θ φ θ φ

θ θ

θ θ

π π+ = + =

π π π π+ = + =

π= =


0, 3

π

.

93. sin cos 2θ θ+ =

Divide each side by 2 : 1 1

sin cos 12 2

θ θ+ =

Rewrite in the sum of two angles form using 1

cos2

φ = , 1

sin2

φ = , and 4

φ π= :

sin cos cos sin 1

sin( ) 1

2

4 2

4

θ φ θ φθ φ

θ φ

θ

θ

+ =+ =

π+ =

π π+ =

π=


π

.


718


94. sin cos 2θ θ− = −


sin cos 12 2

θ θ− = −

Rewrite in the sum of two angles form using 1

cos2

φ = , 1

sin2

φ = , and 4

φ π= :

sin cos sin cos 1

sin( ) 1

3

23

4 27

4

θ φ φ θθ φ

θ φ

θ

θ

− = −− = −

π− =

π π− =

π=

The solution set is { }74π .

95. tan 3 sec

sin 13

cos cos

sin 3 cos 1

θ θθθ θ

θ θ

+ =

+ =

+ =

sin 3 cos 1θ θ+ = Divide each side by 2:

1 3 1sin cos

2 2 2θ θ+ =


using 1

cos2

φ = , 3

sin2

φ = , and 3

φ π= :

1sin cos cos sin

21

sin( )2

θ φ θ φ

θ φ

+ =

+ =

5 or

6 65

3 63 611

26 6

θ φ θ φ

π π θθ

π π θθ

π π+ = + =

π π=+ = +

π== − =

But since 2

π is not in the domain of the tangent

function then the solution set is 11

6

π

.

96. cot csc 3

cos 13

sin sin

cos 1 3 sin

θ θθθ θ

θ θ

+ = −

+ = −

+ = −

3 sin cos 1θ θ+ = − Divide each side by 2:

3 1 1sin cos

2 2 2θ θ+ = −


3cos

2φ = ,

1sin

2φ = , and

6φ π= :

1sin cos cos sin

21

sin( )2

θ φ θ φ

θ φ

+ = −

+ = −

7 11 or

6 67 11

or 6 6 6 6

5 or

3

θ φ θ φ

θ θ

θ π θ

π π+ = + =

π π π π+ = + =

π= =

But since π is not in the domain of the cotangent function then the solution set is

5

3

π

.

97. Let 1sin vα −= and 1cos vβ −= . Then

sin cosvα β= = , and since

sin cos2

α απ = −

, cos cos2

α βπ − =

. If

0v ≥ , then 02

α π≤ ≤ , so that 2

απ −

and β

both lie in the interval 0,2

π

. If 0v < , then

02

απ− ≤ < , so that 2

απ −

and β both lie in

the interval ,2

π π . Either way,

cos cos2

α βπ − =

implies 2

α βπ − = , or

2α β π+ = . Thus, 1 1sin cos

2v v− − π+ = .


719


98. Let 1tan vα −= and 1cot vβ −= . Then

tan cotvα β= = , and since

tan cot2

α απ = −

, cot cot2

α βπ − =

. If

0v ≥ , then 02

α π≤ < , so that 2

απ −

and β

both lie in the interval 0,2

π

. If 0v < , then

02

απ− < < , so that 2

απ −

and β both lie in

the interval ,2

π π

. Either way,

cot cot2

α βπ − =

implies 2

α βπ − = , or

2α β π+ = . Thus, 1 1tan cot

2v v− − π+ = . Note

that 0v ≠ since 1cot 0− is undefined.

99. Let 1 1tan

vα − =

and 1tan vβ −= . Because

1

v

must be defined, 0v ≠ and so , 0α β ≠ . Then

1 1tan cot

tanvα β

β= = = , and since

tan cot2

α απ = −

, cot cot2

α βπ − =

.

Because 0v > , 02

α π< < and so 2

απ −

and

β both lie in the interval 0,2

π

. Then

cot cot2

α βπ − =

implies 2

α βπ − = or

2α βπ= − . Thus,

1 11tan tan , if 0

2v v

v− −π = − >

.

100. Let 1tan veθ − −= . Then tan veθ −= , so 1

cot vv

ee

θ −= = . Because 02

πθ< < , we know

that 0ve− > , which means

( )1 1 1cot cot cot tanv ve eθ θ− − − −= = = .

101. ( )( ) ( )

( ) ( )

1 1

1 1

1 1

2 2

2 2

sin sin cos

sin sin cos cos

cos sin sin cos

1 1

1

1

v v

v v

v v

v v v v

v v

− −

− −

− −

+

=

+

= ⋅ + − −

= + −=

102. ( )( ) ( )

( ) ( )

1 1

1 1

1 1

2 2

cos sin cos

cos sin cos cos

sin sin sin cos

1 1

0

v v

v v

v v

v v v v

− −

− −

− −

+

=

−

= − ⋅ − ⋅ −=

103. ( ) ( )f x h f x

h

+ −

( )

sin( ) sin

sin cos cos sin sin

cos sin sin sin cos

cos sin sin 1 cos

sin 1 coscos sin

x h x

hx h x h x

hx h x x h

hx h x h

hh h

x xh h

+ −=

+ −=

− +=

− −=

−= ⋅ − ⋅

104. ( ) ( )f x h f x

h

+ −

( )

cos( ) cos

cos cos sin sin cos

sin sin cos cos cos

sin sin cos 1 cos

sin 1 cossin cos

x h x

hx h x h x

hx h x h x

hx h x h

hh h

x xh h

+ −

− −=

− + −=

− − −=

−= − ⋅ − ⋅


720


105. a. ( ) ( )( ) ( ) ( )( ) ( )

( ) ( )( ) ( )( ) ( )

( ) ( )

1 1 1

1 1 1 1 1 1

1 1 1

1 1

1 1

1 1

1 1

tan tan 1 tan 2 tan tan 3tan tan 1 tan 2 tan 3 tan tan 1 tan 2 tan 3

1 tan tan 1 tan 2 tan tan 3

tan tan 1 tan tan 23 1 2

31 tan tan 1 tan tan 21 1 2

1tan tan 1 tan tan 2 11 3

1 tan tan 1 tan tan 2

− − −− − − − − −

− − −

− −

− −

− −

− −

+ ++ + = + + =

− +

++ + +− − ⋅= =

+ −− ⋅

−

33 3 3 01 0

2 3 1 9 103 1 31 1 2 1

+ − +−= = = =+ +⋅ − ⋅

− ⋅ −

b. From the definition of the inverse tangent function we know 10 tan 12

π−< < , 10 tan 22

π−< < , and

10 tan 32

π−< < . Thus, 1 1 1 30 tan 1 tan 2 tan 3

2

π− − −< + + < . On the interval 3

0,2

π

, tan 0θ = if and only if

θ π= . Therefore, from part (a), 1 1 1tan 1 tan 2 tan 3 π− − −+ + = .

106. ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

2cos sin sin sin cos sin cos sin sin cos

sin sin cos cos sin

sin sin

t t t t t t

t t t

t t

φ ω φ ω ω ω φ ω φ ω

ω ω φ ω φ

ω ω φ

− = −

= −

= −

107. Note that 2 1θ θ θ= − .

Then ( ) 2 1 2 12 1

2 1 2 1

tan tantan tan

1 tan tan 1

m m

m m

θ θθ θ θθ θ− −

= − = =+ +

108. sin( )sin( )sin( )α θ β θ γ θ− − −

( )( )( )

3

sin cos cos sin sin cos cos sin sin cos cos sin

cos cos cossin sin cos sin sin cos sin sin cos

sin sin sin

cos cos cossin sin sin

sin sin s

α θ α θ β θ β θ γ θ γ θ

θ θ θθ α α θ β β θ γ γθ θ θ

θ α θθ α βθ α

= − − −

= − − −

= −

cos cos cossin

in sin sin sin

β θ γγθ β θ γ

− −

( )( ) ( )( ) ( )( )( )( )( )

3

3

3

3

sin sin cot cot sin cot cot sin cot cot

sin sin sin sin cot cot cot cot cot cot

cos cos cos cos cos cossin sin sin sin


sin(sin sin sin sin

θ α θ α β θ β γ θ γ

θ α β γ β γ α γ α β

β γ α γ α βθ α β γβ γ α γ α βγθ α β γ

= − − −

= + + +

= + + +

= ) sin( ) sin( )


β γ α β αβ γ α γ α β

+ + +

3

3

3

sin(180º ) sin(180º ) sin(180º )sin sin sin sin


sin sin sinsin sin sin sin


sin

α β γθ α β γβ γ α γ α βα β γθ α β γ

β γ α γ α βθ

− − −= =

=

Section 7.6: Double-angle and Half-angle Formulas

721


109. If tan 1xα = + and tan 1xβ = − , then

( ) ( )

( )

( )( )( )( )

( )( )2

2

2

12cot 2

tan

2tan tan

1 tan tan

2 1 tan tan

tan tan

2 1 1 1

1 1

2 1 1

1 1

2

2

x x

x x

x

x x

x

x

α βα β

α βα β

α βα β

− = ⋅−

= −+

+=

−+ + −

=+ − −

+ −=

+ − +

=

=

110. The first step in the derivation,

tan tan2tan

2 1 tan tan2

θθ

θ

π+π + = π − ⋅, is impossible

because tan2

π is undefined.

111. If formula (7) is used, we obtain

tan tan2tan

2 1 tan tan2

θθ

θ

π −π − = π + ⋅. However, this is

impossible because tan2

π is undefined. Using

formulas (3a) and (3b), we obtain

sin2

tan2

cos2

cos

sincot

θθ

θ

θθθ

π − π − = π −

=

=

.

Section 7.6

1. 2sin θ , 22cos θ , 22sin θ

2. 1 cosθ−

3. sinθ

4. True

5. False, only the first one is equivalent.

6. False, you cannot add the arguments or tan.

7. 3

sin , 05 2

θ θ π= < < . Thus, 02 4

θ π< < , which

means 2

θ lies in quadrant I.

3, 5y r= = 2 2 2

2

3 5 , 0

25 9 16, 0

4

x x

x x

x

+ = >

= − = >=

So, 4

cos5

θ = .

a. 3 4 24

sin(2 ) 2sin cos 25 5 25

θ θ θ= = ⋅ ⋅ =

b. 2 2

2 2

cos(2 ) cos sin

4 3 16 9 7

5 5 25 25 25

θ θ θ= −

= − = − =

c. 1 cos

sin2 2

4 11 1 1 10 105 5

2 2 10 1010 10

θ θ−=

−= = = = =

d. 1 cos

cos2 2

4 91 9 3 10 3 105 5

2 2 10 1010 10

θ θ+=

+= = = = =


722


8. 3

cos , 05 2

θ θ π= < < . Thus, 02 4

θ π< < , which

means 2


3, 5x r= = 2 2 2

2

3 5 , 0

25 9 16, 0

4

y y

y y

y

+ = >

= − = >=

So, 4

sin5

θ = .

a. 4 3 24

sin(2 ) 2sin cos 25 5 25

θ θ θ= = ⋅ ⋅ =

b. 2 2

2 2

cos(2 ) cos sin

3 4 9 16 7

5 5 25 25 25

θ θ θ= −

= − = − = −

c. 1 cos

sin2 2

3 21 1 1 5 55 5

2 2 5 55 5

θ θ−=

−= = = = =

d. 1 cos

cos2 2

3 81 4 2 5 2 55 5

2 2 5 55 5

θ θ+=

+= = = = =

9. 4 3

tan ,3 2

θ θ π= π < < . Thus, 3

2 2 4

θπ π< < ,

which means 2

θ lies in quadrant II.

3, 4x y= − = −

2 2 2( 3) ( 4) 9 16 25

5

r

r

= − + − = + ==

4 3sin , cos

5 5θ θ= − = −

a. sin(2 ) 2sin cos

4 3 242

5 5 25

θ θ θ=

= ⋅ − ⋅ − =

b. 2 2

2 2

cos(2 ) cos sin

3 4 9 16 7

5 5 25 25 25

θ θ θ= −

= − − − = − = −

c. 1 cos

sin2 2

31

52

84 2 5 2 55

2 5 55 5

θ θ−=

− − =

= = = =

d. 1 cos

cos2 2

31

52

21 1 5 55

2 5 55 5

θ θ+= −

+ − = −

= − = − = − = −

10. 1 3

tan ,2 2

θ θ π= π < < . Thus, 3

2 2 4

θπ π< < ,

which means 2

θ lie in quadrant II.

2, 1x y= − = − 2 2 2( 2) ( 1) 4 1 5

5

r

r

= − + − = + =

=

1 5 2 2 5sin , cos

5 55 5θ θ= − = − = − = −

a. sin(2 ) 2sin cos

5 2 5 42

5 5 5

θ θ θ=

= ⋅ − ⋅ − =

b. 2 2

2 2

cos(2 ) cos sin

2 5 5

5 5

20 5 15 3

25 25 25 5

θ θ θ= −

= − − −

= − = =

c. 1 cos

sin2 2

θ θ−=

2 51

5

2

5 2 552

5 2 5

10

−− =

+

=

+=


723


d. 1 cos

cos2 2

θ θ+= −

2 51

5

2

5 2 552

5 2 5

10

−+ = −

−

= −

−= −

11. 6

cos ,3 2

θ θπ= − < < π . Thus, 4 2 2

θπ π< < ,

which means 2


6, 3x r= − =

( )22 2

2

6 3

9 6 3

3

y

y

y

− + =

= − =

=

3sin

3θ =

a. sin(2 ) 2sin cos

3 62

3 3

2 18 6 2 2 2

9 9 3

θ θ θ=

= ⋅ ⋅ −

= − = − = −

b. 2 2

2 2

cos(2 ) cos sin

6 3

3 3

6 3 3 1

9 9 9 3

θ θ θ= −

= − −

= − = =

c. 1 cos

sin2 2

θ θ−=

61

3

2

3 632

3 6

6

− − =

+

=

+=

d. 1 cos

cos2 2

θ θ+=

61

3

2

3 632

3 6

6

+ − =

−

=

−=

12. 3 3

sin , 23 2

θ θπ= − < < π . Thus, 3

4 2

θπ < < π ,

which means 2


3, 3y r= − =

( )22

2

3 3

9 3 6

6

x

x

x

+ − =

= − =

=

6cos

3θ =

a. sin(2 ) 2sin cos

3 62

3 3

2 18 6 2 2 2

9 9 3

θ θ θ=

= ⋅ − ⋅

= − = − = −

b. 2 2

2 2

cos(2 ) cos sin

6 3

3 3

6 3 3 1

9 9 9 3

θ θ θ= −

= − −

= − = =

c. 1 cos

sin2 2

θ θ−=

61

32

3 632

3 6

6

−=

−

=

−=


724


d. 1 cos

cos2 2

θ θ+= −

61

32

3 632

3 6

6

+= −

+

= −

+= −

13. sec 3, sin 0θ θ= > , so 02

θ π< < . Thus,

02 4

θ π< < , which means 2


1cos

3θ = , 1x = , 3r = .

2 2 2

2

1 3

9 1 8

8 2 2

y

y

y

+ =

= − =

= =

2 2sin

3θ =

a. 2 2 1 4 2

sin(2 ) 2sin cos 23 3 9

θ θ θ= = ⋅ ⋅ =

b. 2 2

22

cos(2 ) cos sin

1 2 2 1 8 7

3 3 9 9 9

θ θ θ= −

= − = − = −

c. 1 cos

sin2 2

1 21 1 1 3 33 3

2 2 3 33 3

θ θ−=

−= = = = =

d. 1 cos

cos2 2

1 41 2 2 3 63 3

2 2 3 33 3

θ θ+=

+= = = = =

14. csc 5, cos 0θ θ= − < , so 3

2θ ππ < < . Thus,

3

2 2 4

θπ π< < , which means 2


1 5sin

55θ −= = − , 5, 1r y= = −

( )22 2

2

( 1) 5

5 1 4

2

x

x

x

+ − =

= − == −

2 2 5cos

55θ −= = −

a. sin(2 ) 2sin cos

5 2 5 42

5 5 5

θ θ θ=

= ⋅ − ⋅ − =

b. 2 2

2 2

cos(2 ) cos sin

2 5 5

5 5

20 5 15 3

25 25 25 5

θ θ θ= −

= − − −

= − = =

c. 1 cos

sin2 2

θ θ−=

2 51

5

2

5 2 552

5 2 5

10

− − =

+

=

+=

d. 1 cos

cos2 2

θ θ+= −

2 51

5

2

5 2 552

5 2 5

10

+ − = −

−

= −

−= −


725


15. cot 2, sec 0θ θ= − < , so 2

θπ < < π . Thus,

4 2 2

θπ π< < , which means 2


2, 1x y= − = 2 2 2( 2) 1 4 1 5

5

r

r

= − + = + =

=

1 5sin

55θ = = ,

2 2 5cos

55θ −= = −

a. sin(2 ) 2sin cos

5 2 5 20 42

5 5 25 5

θ θ θ=

= ⋅ ⋅ − = − = −

b. 2 2

2 2

cos(2 ) cos sin

2 5 5

5 5

20 5 15 3

25 25 25 5

θ θ θ= −

= − −

= − = =

c. 1 cos

sin2 2

θ θ−=

2 51

5

2

5 2 552

5 2 5

10

− − =

+

=

+=

d. 1 cos

cos2 2

θ θ+=

2 51

5

2

5 2 552

5 2 5

10

+ − =

−

=

−=

16. sec 2, csc 0θ θ= < , so 3

22

π θ π< < . Thus,

3

4 2

π θ π< < , which means 2


1cos

2θ = , 1, 2x r= =

2 2 2

2

1 2

4 1 3

3

y

y

y

+ =

= − =

=

3sin

2θ = −

a. sin(2 ) 2sin cos

3 1 32

2 2 2

θ θ θ=

= ⋅ − ⋅ = −

b. 2 2

22

cos(2 ) cos sin

1 3 1 3 1

2 2 4 4 2

θ θ θ= −

= − − = − = −

c. 1 cos

sin2 2

1 11 1 12 2

2 2 4 2

θ θ−=

−= = = =

d. 1 cos

cos2 2

1 31 3 32 2

2 2 4 2

θ θ+= −

+= − = − = − = −

17. tan 3, sin 0θ θ= − < , so 3

22

θπ < < π . Thus,

3

4 2

θπ < < π , which means 2


1, 3x y= = − 2 2 21 ( 3) 1 9 10

10

r

r

= + − = + =

=

3 3 10sin

1010θ −= = − ,

1 10cos

1010θ = =

a. ( )sin 2 2sin cos

3 10 102

10 10

6 3

10 5

θ θ θ=

= ⋅ − ⋅

= − = −


726


b. 2 2

2 2

cos(2 ) cos sin

10 3 10

10 10

10 90 80 4

100 100 100 5

θ θ θ= −

= − −

= − = − = −

c. 1 cos

sin2 2

θ θ−=

101

102

10 10102

10 10

20

1 10 10

2 5

−=

−

=

−=

−=

d. 1 cos

cos2 2

θ θ+= −

101

102

10 10102

10 10

20

1 10 10

2 5

+= −

+

= −

+= −

+= −

18. cot 3, cos 0θ θ= < , so 3

2π θ π< < . Thus,

3

2 2 4

θπ π< < which means 2

θ is in quadrant II.

3, 1x y= − = − 2 2 2( 3) ( 1) 9 1 10

10

r

r

= − + − = + =

=

1 10sin

1010θ = − = − ,

3 3 10cos

1010θ = − = −

a. ( )sin 2 2sin cos

10 3 10 6 32

10 10 10 5

θ θ θ=

= ⋅ − ⋅ − = =

b. 2 2

2 2

cos(2 ) cos sin

3 10 10

10 10

90 10 80 4

100 100 100 5

θ θ θ= −

= − − −

= − = =

c. 1 cos

sin2 2

θ θ−=

3 101

10

2

10 3 10102

10 3 10

20

1 10 3 10

2 5

− − =

+

=

+=

+=

d. 1 cos

cos2 2

θ θ+= −

3 101

10

2

10 3 10102

10 3 10

20

1 10 3 10

2 5

+ − = −

−

= −

−= −

−= −

19. 45

sin 22.5 sin2

° ° =

1 cos 45

2

21 2 2 2 22

2 4 2

− °=

− − −= = =

20. 45

cos 22.5 cos2

° ° =

1 cos 45

2

21 2 2 2 22

2 4 2

+ °=

+ + += = =


727


21.

77 4tan tan8 2

π π =

71 cos

47

1 cos4

21 22

221

2

2 2 2 2

2 2 2 2

π−= −

π+

−= − ⋅

+

− −= − ⋅ + −

( )

( )

22 2

2

2 2

2

2 1

1 2

−= −

−= −

= − −

= −

22.

99 4tan tan8 2

π π =

91 cos

49

1 cos4

21 22

221

2

2 2 2 2

2 2 2 2

π−=

π+

−= ⋅

+

− −= ⋅ + −

( )22 2

2

2 2

2

2 1

1 2

−=

−=

= −

= − +

23. 330

cos165 cos2

° ° =

1 cos330

2

31 2 3 2 32

2 4 2

+ °= −

+ + += − = − = −

24. 390

sin195 sin2

° ° =

1 cos390

2

31

22

2 3

4

2 3

2

− °= −

−= −

−= −

−= −

25. 15 1

sec158 cos

8

π =π

115

4cos2

1

151 cos

42

=π

=π+

1

21

22

1

2 24

=

+

=+

2

2 2

2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2

=+

+ = ⋅ + + + − = ⋅ + −

( )

( )

2 2 2 2 2

2

2 2 2 2

− +=

= − +


728


26. 7 1

csc78 sin8

π = π174sin21

71 cos

42

=π

=π−

1

21

22

1

2 24

=

−

=−

( )

( )

2

2 2

2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2

2

2 2 2 2

=−

− = ⋅ − − − + = ⋅ − +

+ −=

= + −

27. 4sin sin8 2

π − π − =

1 cos4

2

21 2 2 2 22

2 4 2

π − − = −

− − −= − = − = −

28.

33 4cos cos8 2

π − π − =

31 cos

42

21

2 2 2 2 2

2 4 2

π + − =

+ − − − = = =

29. θ lies in quadrant II. Since 2 2 5x y+ = , 5r = .

Now, the point ( , 2)a is on the circle, so 2 2

2 2

2

2 5

5 2

5 2 1 1

a

a

a

+ == −

= − − = − = −

(a is negative because θ lies in quadrant II.)

Thus, 2 2 5

sin55

b

rθ = = = and

1 5cos

55

a

rθ −= = = − . Thus,

( ) ( )2 sin 2 2sin cos

2 5 5 20 42

5 5 25 5

f θ θ θ θ= =

= ⋅ ⋅ − = − = −


2 5sin

5θ = and

5cos

5θ = − .

Thus, ( ) ( ) 2 2

2 2

2 cos 2 cos sin

5 2 5

5 5

5 20 15 3

25 25 25 5

g θ θ θ θ= = −

= − −

= − = − = −

31. Note: Since θ lies in quadrant II, 2

θ must lie in

quadrant I. Therefore, cos2

θ is positive. From the

solution to Problem 29, we have 5

cos5

θ = − .

Thus, 1 cos

cos2 2 2

gθ θ θ+ = =

( )

51

5

2

5 552

5 5

10

10 5 5

10

−+ =

−

=

−=

−=


729


32. Note: Since θ lies in quadrant II, 2

θ must lie in

quadrant I. Therefore, sin2

θ is positive. From the

solution to Problem 29, we have 5

cos5

θ = − .

Thus, 1 cos

sin2 2 2

fθ θ θ− = =

( )

51

5

2

5 552

5 5

10

10 5 5

10

−− =

+

=

+=

+=

33. θ lies in quadrant II. Since 2 2 5x y+ = , 5r = .

Now, the point ( , 2)a is on the circle, so 2 2

2 2

2

2 5

5 2

5 2 1 1

a

a

a

+ == −

= − − = − = −

(a is negative because θ lies in quadrant II.)

Thus, 2

tan 21

b

aθ = = = −

−.

( ) ( )

( )( )

2

2

2 tan 2

2 tan

1 tan2 2 4 4 4

1 4 3 31 2

h θ θθ

θ

=

=−

− − −= = = =− −− −


2 5sin

5θ = and

5cos

5θ = − . Thus,

51

51 costan

2 2 sin 2 55

hθ θ θ

θ

−− − = = =

5 55

2 55

5 5

2 5

5 5 5

2 5 5

5 5 5

10

5 1 1 5

2 2

+

=

+=

+= ⋅

+=

+ += =

35. α lies in quadrant III. Since 2 2 1x y+ = ,

1 1r = = . Now, the point 1

,4

b −

is on the

circle, so 2

2

22

2

11

4

11

4

1 15 151

4 16 4

b

b

b

− + =

= − −

= − − − = − = −

(b is negative because α lies in quadrant III.)

Thus,

114cos

1 4

a

rα

−= = = − and

15154sin

1 4

b

rα

−= = = − . Thus,

( ) ( ) 2 2

22

2 cos 2 cos sin

1 15

4 4

1 15 14 7

16 16 16 8

g α α α α= = −

= − −

= − = − = −


730



15sin

4α = − and

1cos

4α = − . Thus,

( ) ( )2 sin 2

2sin cos

15 1 152

4 4 8

f α αα α

==

= ⋅ − ⋅ − =

37. Note: Since α lies in quadrant III, 2

α must lie in

quadrant II. Therefore, sin2

α is positive. From

the solution to Problem 35, we have 1

cos4

α = − .

Thus, sin2 2

1 cos

2

fα α

α

=

−=

11

42

55 5 2 10 104

2 8 8 2 16 4

− − =

= = = ⋅ = =

38. Note: Since α lies in quadrant III, 2

α must lie in

quadrant II. Therefore, cos2

α is negative. From

the solution to Problem 35, we have 1

cos4

α = − .

Thus,

cos2 2

1 cos

2

gα α

α

=

+= −

11

42

33 3 2 6 64

2 8 8 2 16 4

+ − = −

= − = − = − ⋅ = − = −


15sin

4α = − and

1cos

4α = − . Thus,

1 costan

2 2 sinh

α α αα

− = =

11

4

154

541545

15

5 15

15 15

5 15

15

15

3

− − =

−

=−

= −

= − ⋅

= −

= −

40. α lies in quadrant III. Since 2 2 1x y+ = ,

1 1r = = . Now, the point 1

,4

b −

is on the

circle, so 2

2

22

2

11

4

11

4

1 15 151

4 16 4

b

b

b

− + =

= − −

= − − − = − = −

(b is negative because α lies in quadrant III.)

Thus,

154tan 1514

b

aθ

−= = =

−.

( ) ( )

( )( )

2

2

2 tan 2

2 tan

1 tan

2 15 2 15 2 15 15

1 15 14 71 15

h α αα

α

=

=−

= = = = −− −−


731


41. ( )( )

( ) ( )

24 2

2

2

sin sin

1 cos 2

2

11 2cos 2 cos 2

4

θ θ

θ

θ θ

=

− =

= − +

( ) ( )

( ) ( )

( ) ( )

( ) ( )

21 1 1cos 2 cos 2

4 2 4

1 cos 41 1 1cos 2

4 2 4 2

1 1 1 1cos 2 cos 4

4 2 8 83 1 1

cos 2 cos 48 2 8

θ θ

θθ

θ θ

θ θ

= − +

+ = − +

= − + +

= − +

42. ( ) ( )( ) ( )

( )( )

( ) ( )( )( )

2

2

2

3

sin 4 sin 2 2

2sin 2 cos 2

2(2sin cos ) 1 2sin

4sin cos 1 2sin

cos 4sin 1 2sin

cos 4sin 8sin

θ θθ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

= ⋅

=

= −

= −

= −

= −

43.

( ) ( )cos(3 ) cos(2 )

cos 2 cos sin 2 sin

θ θ θθ θ θ θ

= += −

( )

( )

2

3 2

3 2

3 3

3

2cos 1 cos 2sin cos sin

2cos cos 2sin cos

2cos cos 2 1 cos cos

2cos cos 2cos 2cos

4cos 3cos

θ θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θθ θ

= − −

= − −

= − − −

= − − +

= −

44. ( ) ( )

( )( )

2

22

4 2

4 2

4 2

cos 4 cos 2 2

2cos (2 ) 1

2 2cos 1 1

2 4cos 4cos 1 1

8cos 8cos 2 1

8cos 8cos 1

θ θ

θ

θ

θ θ

θ θθ θ

= ⋅

= −

= − −

= − + −

= − + −

= − +

45. We use the result of problem 42 to help solve this problem:

( )sin 5 sin(4 )θ θ θ= +

( ) ( )( ) ( )( ) ( )( )

( )( )( )( )

3

2 3 2

2 3

2

3 5

sin 4 cos cos 4 sin

cos 4sin 8sin cos cos 2(2 ) sin

cos 4sin 8sin 1 2sin 2 sin

1 sin 4sin 8sin

sin 1 2 2sin cos

4sin 12sin 8sin

θ θ θ θ

θ θ θ θ θ θ

θ θ θ θ θ

θ θ θ

θ θ θ

θ θ θ

= +

= − +

= − + −

= − −

+ −

= − +

( )

( )

2 2

3 5

3 2

3 5 3 5

5 3

sin 1 8sin cos

4sin 12sin 8sin

sin 8sin 1 sin

5sin 12sin 8sin 8sin 8sin

16sin 20sin 5sin

θ θ θ

θ θ θθ θ θ


+ −

= − +

+ − −

= − + − +

= − +

46. We use the results from problems 42 and 44 to help solve this problem:

( ) ( )( )

( )( )4 2

3

5 3

2 4

5 3

cos(5 ) cos(4 )

cos 4 cos sin 4 sin

8cos 8cos 1 cos

cos 4sin 8sin sin

8cos 8cos cos

4cos sin 8cos sin

8cos 8cos cos

4cos (1

θ θ θθ θ θ θ

θ θ θ

θ θ θ θ

θ θ θθ θ θ θ

θ θ θθ

= += −

= − +

− −

= − +

− +

= − +

− − 2 2 2cos ) 8cos (1 cos )θ θ θ+ − 5 3

3 2 4

5 3

3 5

5 3

8cos 8cos cos 4cos

4cos 8cos (1 2cos cos )

8cos 4cos 3cos

8cos 16cos 8cos

16cos 20cos 5cos


θ θ θθ θ θ

θ θ θ

= − + −

+ + − +

= − −

+ − +

= − +

47. ( )( )( )

( )

4 4 2 2 2 2cos sin cos sin cos sin

1 cos 2

cos 2

θ θ θ θ θ θ

θθ

− = + −

= ⋅

=


732


48.

cos sincot tan sin cos

cos sincot tansin cos

θ θθ θ θ θ

θ θθ θθ θ

−− =+ +

( )

2 2

2 2

2 2

2 2

2 2

cos sinsin cos

cos sinsin cos

cos sin sin cos

sin cos cos sin

cos sin

1cos 2

θ θθ θ

θ θθ θ

θ θ θ θθ θ θ θθ θ

θ

−

=+

−= ⋅+

−=

=

49.

2

1 1cot(2 )

2 tantan(2 )1 tan

θ θθθ

= =

−

2

2

2

2

2

2

2

1 tan

2 tan1

1cot2

cot

cot 1

cot2

cot

cot 1 cot

2cot

cot 1

2cot

θθ

θ

θθ

θ

θθ θ

θθ

θ

−=

−=

−

=

−= ⋅

−=

50. 1

cot(2 )tan(2 )

θθ

=

( )

2

2

2

12 tan

1 tan

1 tan

2 tan

1 1 tan

2 tan tan

1cot tan

2

θθθ

θθ

θ θ

θ θ

=

−−=

= −

= −

51. 1

sec(2 )cos(2 )

θθ

=2

2

2

2

2

2

1

2cos 11

21

sec1

2 sec

sec

sec

2 sec

θ

θ

θθθ

θ

=−

=−

=−

=−

52. ( ) ( )1

csc 2sin 2

θθ

= 1

2sin cos1 1 1

2 cos sin1

sec csc2

θ θ

θ θ

θ θ

=

= ⋅ ⋅

=

53. [ ]2 2cos (2 ) sin (2 ) cos 2(2 ) cos(4 )u u u u− = =

54.

( )( )

2

2

(4sin cos )(1 2sin )

2(2sin cos )(1 2sin )

2sin 2 cos 2

sin 2 2

sin 4

u u u

u u u

u u

u

u

−

= −== ⋅

=

55. 2 2

2 2

cos(2 ) cos sin

1 sin(2 ) 1 2sin cos


cos sin 2sin cos

θ θ θθ θ θ


−=+ +

− +=+ +



cos sin

cos sincos sin

sincos sin

sincos sinsin sincos sinsin sincot 1

cot 1


θ θθ θθ θ

θθ θ

θθ θθ θθ θθ θθθ

− +=+ +

−=+−

=+

−=

+

−=+


733


56. ( )( )

( )

( )

( )

2 2 2 2

2

2

1sin cos 4sin cos

41

2sin cos41

sin 24

1 cos 41

4 2

11 cos 4

8

θ θ θ θ

θ θ

θ

θ

θ

=

=

=

− = ⋅

= −

57. 2

2

1 1 2sec

1 cos2 1 coscos

22

θθθ θ

= = = + +

58. 2

2

1 1 2csc

1 cos2 1 cossin

22

θθθ θ

= = = − −

59. 2

2

1cot

2tan

2

v

v =

11 cos1 cos1 cos

1 cos1

1sec

11

sec

v

vv

v

v

v

=−++=−

+=

−

sec 1sec

sec 1sec

sec 1 sec

sec sec 1sec 1

sec 1

v

vv

vv v

v vv

v

+

=−

+= ⋅−

+=−

60. 1 cos 1 cos

tan csc cot2 sin sin sin

v v vv v

v v v

−= = − = −

61.

2

2

1 cos11 tan

1 cos21 cos

1 tan 12 1 cos

1 cos (1 cos )1 cos

1 cos 1 cos1 cos

θθθ

θ θθ

θ θθ

θ θθ

−−−+=−+ ++

+ − −+=

+ + −+

2cos1 cos

21 cos2cos 1 cos

1 cos 2cos

θθ

θθ θθ

θ

+=

++= ⋅

+=

62.

( )( )

( ) ( )

( )

3 3

2 2

2 2

2 2

sin cos

sin cos

sin cos sin sin cos cos

sin cos

sin sin cos cos

1sin cos 2sin cos

21

1 sin 22

θ θθ θ


θ θ θ θ

θ θ θ θ

θ

++

+ − +=

+= − +

= + −

= −

63. ( ) ( )sin 3 cos cos 3 sinsin(3 ) cos(3 )

sin cos sin cossin(3 )

sin cossin 2

sin cos2sin cos

sin cos2

θ θ θ θθ θθ θ θ θ

θ θθ θ

θθ θ

θ θθ θ

−− =

−=

=

=

=


734


64. ( ) ( )

( )( )

2 2cos sin cos sincos sin cos sin

cos sin cos sin cos sin cos sin

θ θ θ θθ θ θ θθ θ θ θ θ θ θ θ

+ − −+ −− =− + − +

( )

( )

( )( )( )( )

2 2 2 2

2 2

2 2 2 2

2 2

cos 2cos sin sin cos 2cos sin sin

cos sin

cos 2cos sin sin cos 2cos sin sin

cos sin4cos sin

cos 2

2(2sin cos )

cos 2

2sin 2

cos 2

2 tan 2

θ θ θ θ θ θ θ θθ θ

θ θ θ θ θ θ θ θθ θ

θ θθθ θ

θθθθ

+ + − − +=

−+ + − + −=

−

=

=

=

=

65. ( )

( )( )

3

3 2 32 2

2 2 2 2 2

2 2

tan 3 tan(2 )

2 tan 2 tan tan tantantan 2 tan 3tan tan 1 tan 3tan tan1 tan 1 tan

2 tan1 tan 2 tan 1 tan 2 tan 1 tan 1 3tan 1 3tan1 tan1 tan 1 tan

θ θ θ

θ θ θ θθθ θ θ θ θ θ θθ θθθ θ θ θ θ θ θθ

θ θ

= +

+ −++ − − −− −= = = = ⋅ =− − − − − −− ⋅

− −

66.

( ) ( )

tan tan( 120º ) tan( 240º )

tan tan120º tan tan 240ºtan

1 tan tan120º 1 tan tan 240º

tan 3 tan 3tan

1 tan 3 1 tan 3

tan 3 tan 3tan

1 3 tan 1 3 tan

θ θ θθ θθ

θ θθ θθθ θ

θ θθθ θ

+ + + ++ += + +

− −− += + +

− − −

− += + ++ −

( ) ( ) ( ) ( )( )

( )

( )

2

2

3 2 2

2

3

2

3

2

tan 1 3tan tan 3 1 3 tan tan 3 1 3 tan

1 3tan

tan 3tan tan 3 tan 3 3tan tan 3 tan 3 3tan

1 3tan

3tan 9 tan

1 3tan

3 3tan tan

1 3tan3tan 3 (from Problem 65)

θ θ θ θ θ θ

θθ θ θ θ θ θ θ θ

θθ θ

θθ θ

θθ

− + − − + + +=

−− + − − + + + + +=

−− +=

−−

=−

=


735


67. ( )( )

( )

( )

1/ 2

1/ 22

1ln 1 cos 2 ln 2

21 1 cos 2

ln2 2

1 cos 2ln

2

ln sin

ln sin

θ

θ

θ

θ

θ

⋅ − −

−= ⋅

− =

=

=

68. ( )( )

( )

( )

1/ 2

1/ 22

1ln 1 cos 2 ln 2

21 1 cos 2

ln2 2

1 cos 2ln

2

ln cos

ln cos

θ

θ

θ

θ

θ

⋅ + −

+= ⋅

+ =

=

=

69. ( ) 2

2 2

2

cos 2 6sin 4

1 2sin 6sin 4

4sin 3

θ θ

θ θθ

+ =

− + =

=

2 3sin

4

3sin

22 4 5

, , ,3 3 3 3

θ

θ

π π π πθ

=

= ±

=


, , , 3 3 3 3

π π π π

.

70. ( ) 2

2 2

cos 2 2 2sin

1 2sin 2 2sin

1 2 (not possible)

θ θ

θ θ

= −

− = −=

The equation has no real solution.

71. 2

2

cos(2 ) cos

2cos 1 cos

2cos cos 1 0

(2cos 1)(cos 1) 0

θ θθ θ

θ θθ θ

=

− =

− − =+ − =

2cos 1 0 or cos 1 0

1 cos 1cos

2 02 4

, 3 3

θ θθθθ

θ

+ = − === −=

π π=


0, , 3 3

π π

.

72. sin(2 ) cos

2sin cos cos

2sin cos cos 0

(cos )(2sin 1) 0

θ θθ θ θ

θ θ θθ θ

==

− =− =

cos 0 or 2sin 1

cos 0 1sin

23,

52 2 ,6 6

θ θθ θ

θθ

= == =

π π= π π=


, , , 6 2 6 2

π π π π

.

73.

( )

sin(2 ) sin(4 ) 0sin(2 ) 2sin(2 )cos(2 ) 0

sin(2 ) 1 2cos(2 ) 0

θ θθ θ θ

θ θ

+ =+ =

+ =

sin(2 ) 0 or 1 2cos(2 ) 01

cos(2 )2

θ θ

θ

= + =

= −

2 0 2 or 2 2 or

2

k kk

k

θ θθ θ

= + π = π + π= π π= + π

2 42 2 or 2 2

3 32

3 3

k k

k k

θ θ

θ θ

π π= + π = + π

π π= + π = + π

On the interval 0 2θ π≤ < , the solution set is 2 4 3 5

0, , , , , , , 3 2 3 3 2 3

π π π π π ππ

.


736


74.

( ) ( )[ ]

( )( )( )

2 2

2

2 2 2

2 4 2

2 4 2

4 2

4 2

2 2

cos(2 ) cos(4 ) 0

2cos 1 2cos (2 ) 1 0

2cos 1 2 cos(2 )cos(2 ) 1 0

2cos 2 2cos ( ) 1 2cos ( ) 1 2 0

2cos 1 2 4cos 4cos 1 1 0

2cos 1 8cos 8cos 2 1 0

8cos 6cos 0

4cos 3cos 0

cos 4cos

θ θ

θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θθ θθ θ

θ

+ =

− + − =

− + − =

+ − − − =

− + − + − = − + − + − =

− =

− =

( )3 0θ − =

2 2

2

cos ( ) 0 or 4cos 3 0

3cos 0 or cos

4

3 cos

2

θ θ

θ θ

θ

= − =

= =

= ±

3 5 7 11, or , , ,

2 2 6 6 6 6θ θπ π π π π π= =

On the interval 0 2θ π≤ < , the solution set is

5 7 3 11, , , , ,

6 2 6 6 2 6

π π π π π π

.

75. 2

2

3 sin cos(2 )

3 sin 1 2sin

2sin sin 2 0

θ θθ θ

θ θ

− =

− = −

− + =

This equation is quadratic in sinθ .

The discriminant is 2` 4 1 16 15 0b ac− = − = − < . The equation has no real solutions.

76. 2

2

cos(2 ) 5cos 3 0

2cos 1 5cos 3 0

2cos 5cos 2 0

(2cos 1)(cos 2) 0

θ θθ θ

θ θθ θ

+ + =

− + + =

+ + =+ + =

2cos 1 or cos 2

1 (not possible)cos

22 4

,3 3

θ θ

θ

θ

= − = −

= −

π π=


, 3 3

π π

.

77. tan(2 ) 2sin 0θ θ+ =

( )( )

2

2

2

sin(2 )2sin 0

cos(2 )

sin 2 2sin cos 20

cos 2

2sin cos 2sin (2cos 1) 0

2sin cos 2cos 1 0

2sin 2cos cos 1 0

2sin (2cos 1)(cos 1) 0

θ θθ

θ θ θθ

θ θ θ θ

θ θ θ

θ θ θ

θ θ θ

+ =

+ =

+ − =

+ − =

+ − =

− + =

2cos 1 0 or 2sin 0 or

1 sin 0cos

2 0,5

,3 3

θ θθθθ

θ

− = ==== π

π π=

cos 1 0

cos 1

θθθ

+ == −= π


0, , , 3 3

π ππ

.

78. tan(2 ) 2cos 0θ θ+ =

( )( )

( )( )

2

2

2

sin(2 )2cos 0

cos(2 )

sin 2 2cos cos 20

cos 2

2sin cos 2cos (1 2sin ) 0

2cos sin 1 2sin 0

2cos 2sin sin 1 0

2cos (2sin 1)(sin 1) 0

θ θθ

θ θ θθ

θ θ θ θ

θ θ θ

θ θ θ

θ θ θ

+ =

+=

+ − =

+ − =

− − − =

− + − =

2cos 0 or 2sin 1 0 or

cos 0 1sin

23,

7 112 2 ,6 6

θ θθ θθ

θ

− = + == = −

π π= π π=

sin 1 0

sin 1

2

θθ

θ

− ==

π=


, , , 2 6 2 6

π π π π

.


737


79. 1 1 3sin 2sin sin 2 sin

2 6 3 2

π π− = ⋅ = =

80. 1 3 2 3sin 2sin sin 2 sin

2 3 3 2

π π− = ⋅ = =

81. 1 2 13 3cos 2sin 1 2sin sin

5 5− − = −

23

1 25

181

257

25

= −

= −

=

82. 1 2 14 4cos 2cos 2cos cos 1

5 5− − = −

24

2 15

321

257

25

= −

= −

=

83. 1 3tan 2cos

5− −

Let 1 3cos

5α − = −

. α lies in quadrant II.

Then 3

cos , 5 2

πα α= − ≤ ≤ π .

5sec

3α = −

2

2

tan sec 1

5 25 16 41 1

3 9 9 3

α α= − −

= − − − = − − = − = −

1 3tan 2cos tan 2

5α− − =

2

2

2 tan

1 tan4

23

41

3

893

16 919

αα

=− − = − −

−= ⋅

−

24

9 1624

724

7

−=−

−=−

=

84.

1

1

2 1

2

32 tan tan

3 4tan 2 tan

341 tan tan

4

32

4

31

4

−

−

−

= −

⋅ = −

3162

9 16116

24

16 924

7

= ⋅−

=−

=

85. 1 4sin 2cos

5−

Let 1 4cos

5α −= . α is in quadrant I.

Then 4

cos , 05 2

α α π= ≤ ≤ .


738


2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

α α= −

= − = − = =

1 4sin 2cos sin 2

5

3 4 242sin cos 2

5 5 25

α

α α

− =

= = ⋅ ⋅ =

86. 1 4cos 2 tan

3− −

Let 1 4tan

3α − = −

. α is in quadrant IV.

Then 4

tan , 03 2

α απ= − − < < .

2

2

sec tan 1

4 16 25 51 1

3 9 9 3

α α= +

= − + = + = =

3cos

5α =

1 4cos 2 tan cos 2

3α− − =

2

2

2cos 1

32 1

5

181

257

25

α= −

= −

= −

= −

87.

1

2 1

31 cos cos

1 3 5sin cos

2 5 2

−

−

− =

31

52

2521

5

−=

=

=

88. 2 11 3cos sin

2 5−

Let 1 3sin


3sin , 0

5 2α α π= < < .

2

2

cos 1 sin

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

2 1 21 3 1cos sin cos

2 5 2

4 911 cos 95 5

2 2 2 10

α

α

− = ⋅

++= = = =

89. 1 3sec 2 tan

4−

Let 1 3tan

4α − =

. α is in quadrant I.

Then 3

tan , 04 2

α α π= < < .

2

2

sec tan 1

3 9 25 51 1

4 16 16 4

α α= +

= + = + = =

4cos

5α =

( )1 3sec 2 tan sec 2

4α− =

( )

2

2

1

cos 2

1

2cos 11

42 1

5

α

α

=

=−

= −

1

321

25172525

7

=−

=

=

90. 1 3csc 2sin

5− −

Let 1 3sin

5α − = −

. α is in quadrant IV.

Then 3

sin , 05 2

α απ= − − ≤ ≤ .


739


2cos 1 sinα α= −2

31

5

91

25

16

254

5

= − −

= −

=

=

( )1 3csc 2sin csc 2

5α− − = ( )

1

sin 2

1

2sin cos13 4

25 5

α

α α

=

=

= −

1242525

24

=−

= −

91. ( )( )

( )

0

sin 2 sin 02sin cos sin 0

sin 2cos 1 0

f x

x xx x xx x

=− =− =

− =

sin 0 or 2cos 1 00, 1

cos2

5,

3 3

x xx

x

x

π

π π

= − == =

=

The zeros on 0 2x π≤ < are 5

0, , ,3 3

π ππ .

92. ( )( )

( )( )

2

2

0

cos 2 cos 0

2cos 1 cos 02cos cos 1 0

2cos 1 cos 1 0

f x

x x

x xx x

x x

=+ =

− + =+ − =

− + =

2cos 1 0 or cos 1 01 cos 1

cos2

5,

3 3

x xx

xx

x

ππ π

− = + == −= =

=


, ,3 3

π ππ .

93. ( )( ) 2

2 2 2

2

0

cos 2 sin 0

cos sin sin 0cos 0cos 0

3,

2 2

f x

x x

x x x

xx

xπ π

=+ =

− + ===

=


,2 2

π π.

94. a. cos(2 ) cos 0θ θ+ = , 0º 90ºθ< < 2

2

2cos 1 cos 0

2cos cos 1 0

(2cos 1)(cos 1) 0

θ θθ θ

θ θ

− + =

+ − =− + =

2cos 1 0 or cos 1 0

1 cos 1cos

2 180º 60º , 300º

θ θθθθ

θ

− = + == −==

=

On the interval 0º 90ºθ< < , the solution is 60˚.

b. ( ) ( )

2 2

(60º ) 16sin 60º cos 60º 1

3 116 1

2 2

12 3 in 20.78 in

A = +

= ⋅ +

= ≈

c. Graph ( )1 16sin cos 1Y x x= + and use the

MAXIMUM feature:

0

25

0 90

The maximum area is approximately 220.78 in. when the angle is 60˚.


740


95. a.

( )

12

csc cot2 csc cot

WD

W Dθ θ

θ θ

=−

= −

1 cos 1 coscsc cot

sin sin sin

tan2

θ θθ θθ θ θθ

−− = − =

=

Therefore, 2 tan2

W Dθ= .

b. Here we have 15D = and 6.5W = .

( )

1

1

6.5 2 15 tan2

13tan

2 6013

tan2 60

132 tan 24.45

60

θ

θ

θ

θ

−

−

=

=

=

= ≈ °

96. ( )( )( ) ( )( )

2 2

2 2

sin cos sin cos cos sin

sin cos cos sin

1sin 2 cos 2

2

sin 2 cos 22

x y xy

x y xy

x y xy

x yxy

I I I

I I I

I I I

I II

θ θ θ θ θ θ

θ θ θ θ

θ θ

θ θ

− + −

= − + −

= − +

−= +

97. a. 20

220

2( ) cos (sin cos )

16

2(cos sin cos )

16

vR

v

θ θ θ θ

θ θ θ

= −

= −

( ) ( )

( ) ( )

220

20

20

20

2 1(2cos sin 2cos )

16 2

2 1 cos 2sin 2 2

32 2

2sin 2 1 cos 2

32

2sin 2 cos 2 1

32

v

v

v

v

θ θ θ

θθ

θ θ

θ θ

= ⋅ −

+ = −

= − −

= − −

b. sin(2 ) cos(2 ) 0θ θ+ =


sin(2 ) cos(2 ) 02 2

θ θ+ =


1 1cos and sin and

42 2φ φ φ π= = = :

sin(2 )cos cos(2 )sin 0

sin(2 ) 0

2 0

2 04

24

8 23

67.5º8

k

k

k

k

θ φ θ φθ φθ φ

θ

θ

θ

θ

+ =+ =+ = + π

π+ = + π

π= − + π

π π= − +

π= =

c. ( )

( ) ( )( )

( )( )

232 2sin(2 67.5º ) cos(2 67.5º ) 1

32

32 2 sin 135º cos 135º 1

2 232 2 1

2 2

32 2 2 1

32 2 2 feet 18.75 feet

R = ⋅ − ⋅ −

= − −

= − − −

= −

= − ≈

d. Graph ( )2

1

32 2sin(2 ) cos(2 ) 1

32Y x x= − − and

use the MAXIMUM feature:

0

20

45 90

The angle that maximizes the distance is 67.5˚, and the maximum distance is 18.75 feet.


741


98.

[ ]

[ ]

( )2

1 1sin(2 ) sin(4 )

2 41 1

sin(2 ) sin(2 2 )2 41 1

sin(2 ) 2sin(2 )cos(2 )2 41 1

sin(2 ) sin(2 )cos(2 )2 21 1

sin(2 ) sin(2 ) 2cos ( ) 12 2

y x x

x x

x x x

x x x

x x x

π π

π π

π π π

π π π

π π π

= +

= + ⋅

= +

= +

= + ⋅ −

2

2

1 1sin(2 ) sin(2 )cos ( ) sin(2 )

2 2

sin(2 )cos ( )

x x x x

x x

π π π π

π π

= + −

=

99. Let b represent the base of the triangle.

cos2

cos2

h

s

h s

θ

θ

=

=

/ 2

sin2

2 sin2

b

s

b s

θ

θ

=

=

2

2

1

21

2 sin cos2 2 2

sin cos2 2

1sin

2

A b h

s s

s

s

θ θ

θ θ

θ

= ⋅

= ⋅

=

=

100. sin ; cos1 1

y xy xθ θ= = = =

a. 2 2cos sin 2sin cosA xy θ θ θ θ= = =

b. 2sin cos sin(2 )θ θ θ=

c. The largest value of the sine function is 1. Solve: sin 2 1

22

454

θπθ

πθ

=

=

= = °

d. 2 2

cos sin4 2 4 2

x yπ π= = = =

The dimensions of the largest rectangle are

22 by

2.

101. ( )sin 2 2sin cosθ θ θ=2

2

2

2

2sin cos

cos 1sin

2cos1

cos2 tan

sec2 tan 4

41 tan

θ θθ

θθ

θθθ

θθ

= ⋅

⋅=

=

= ⋅+

2

2

4(2 tan )

4 (2 tan )

4

4

x

x

θθ

=+

=+

102. ( )2 2

2 22 2

cos sincos 2 cos sin

cos sin

θ θθ θ θθ θ

−= − =+

( )( )

2 2

2

2 2

2

2

2

2

2

2

2

2

2

cos sin

coscos sin

cos

1 tan 4

41 tan

4 4 tan

4 4 tan

4 2 tan

4 2 tan

4

4

x

x

θ θθ

θ θθ

θθ

θθθθ

−

=+

−= ⋅+−=+−

=+

−=+

103. ( )21 1sin cos 2

2 4x C x⋅ + = − ⋅

( )

( )( )

( )

2

2

2 2

1 1cos 2 sin

4 21

cos 2 2sin41

1 2sin 2sin41

(1)41

4

C x x

x x

x x

= − ⋅ − ⋅

= − ⋅ +

= − ⋅ − +

= − ⋅

= −


742


104. ( )21 1cos cos 2

2 4x C x⋅ + = ⋅

( )

( )

2

2 2

2 2

1 1cos 2 cos

4 21 1

2cos 1 cos4 21 1 1

cos cos2 4 2

1

4

C x x

x x

x x

= ⋅ − ⋅

= ⋅ − −

= − −

= −

105. If tan2

zα =

, then

22

2

2

2 tan2 2

1 1 tan2

2 tan2

sec2

2 tan cos2 2

z

z

α

α

α

α

α α

=

+ +

= =

2

2sin2

cos2

cos2

2sin cos2 2

sin 22

sin

αα

α

α α

α

α

= ⋅ =

= =

106. If tan2

zα =

, then

22

22

1 tan1 2

1 1 tan2

1 cos1

1 cos1 cos

11 cos

z

z

α

α

αααα

− − =+ +

−−+=−++

1 cos (1 cos )1 cos

1 cos 1 cos1 cos

1 cos (1 cos )

1 cos 1 cos2cos

2cos

α αα

α αα

α αα α

α

α

+ − −+=

+ + −+

+ − −=+ + −

=

=

107. ( )2 1 cos 2

( ) sin2

xf x x

−= =

Starting with the graph of cosy x= , compress

horizontally by a factor of 2, reflect across the x-axis, shift 1 unit up, and shrink vertically by a factor of 2.

108. 2( ) cosg x x=( )1 cos 2

2

x+=

Starting with the graph of cosy x= , compress

horizontally by a factor of 2, reflect across the x-axis, shift 1 unit up, and shrink vertically by a factor of 2.


743


109.

( )( )

1 cos12 12sin sin

24 2 2

11 6 2

1 146 2

2 2 8

π π − π = =

− + = = − +

( ) ( )

( )( )

8 2 6 28 2 6 2

16 4

2 4 6 2 24 6 2

4 4

− +− += =

− += = − −

( )( )

1 cos12 12cos cos

24 2 2

11 6 2

1 146 2

2 2 8

π π + π = =

+ + = = + +

( ) ( )

( )

8 2 6 28 2 6 2

16 4

2 4 6 2 24 6 2

4 4

+ ++ += =

+ += = + +

110. 1 cos

4 4cos cos8 2 2

21 2 22

2 4

2 2

2

π π + π = =

+ += =

+=

1 cos

8 8sin sin16 2 2

2 21 2 2 22

2 4

2 2 2

2

π π − π = =

+− − += =

− +=

1 cos

8 8cos cos16 2 2

2 21 2 2 22

2 4

2 2 2

2

π π + π = =

++ + += =

+ +=

111. 3 3 3sin sin ( 120º ) sin ( 240º )θ θ θ+ + + +

( ) ( )( ) ( ) ( )( )3 33

3 3

3

sin sin cos 120º cos sin 120º sin cos 240º cos sin 240º

1 3 1 3sin sin cos sin cos

2 2 2 2

θ θ θ θ θ

θ θ θ θ θ

= + + + +

= + − ⋅ + ⋅ + − ⋅ − ⋅

( )( )

3 3 2 2 3

3 2 2 3

1sin sin 3 3 sin cos 9sin cos 3 3 cos

81

sin 3 3 sin cos 9sin cos 3 3 cos8

θ θ θ θ θ θ θ

θ θ θ θ θ θ

= + ⋅ − + − +

− + + +

3 3 2 2 3

3 2 2 3

1 3 3 9 3 3sin sin sin cos sin cos cos

8 8 8 8

1 3 3 9 3 3 sin sin cos sin cos cos

8 8 8 8

θ θ θ θ θ θ θ

θ θ θ θ θ θ

= − ⋅ + ⋅ − ⋅ + ⋅

− ⋅ − ⋅ − ⋅ − ⋅

( ) ( )

( ) ( )

3 2 3 2 3 3

3

3 9 3 3sin sin cos sin 3sin 1 sin sin 3sin 3sin

4 4 4 43 3

4sin 3sin sin 3 (from Example 2)4 4

θ θ θ θ θ θ θ θ θ

θ θ θ

= ⋅ − ⋅ = ⋅ − − = ⋅ − +

= ⋅ − = − ⋅


744


112.

3

2

tan tan 33

3tan tan3 3 (from problem 65)

1 3tan3

θθ

θ θ

θ

= ⋅

−=

−

2

2

3 2

2 2

2 2

tan 3 tan3 3

tan3 1 3tan

3

3tan tan tan 1 3tan3 3 3 3

3 tan 1 3tan3 3

3 tan 3 tan3 3

a

a

a

a a

θ θθ

θ

θ θ θ θ

θ θ

θ θ

− =

−

− = −

− = −

− = −

( )

2 2

2

2

3 tan tan 33 3

3 1 tan 33

3tan

3 3 1

3tan

3 3 1

a a

a a

a

a

a

a

θ θ

θ

θ

θ

− = −

− = −

−=−

−= ±−

113. Answers will vary. Section 7.7

1. sin(195 )cos(75 ) sin(150 45 )cos(30 45 )° ° = ° + ° ° + °

( )( )sin(150 45 )cos(30 45 )

sin150 cos 45 cos150 sin 45 cos30 cos 45 sin 30 sin 45

1 2 3 2 3 2 1 2

2 2 2 2 2 2 2 2

2 6 6 2

4 4 4 4

° + ° ° + ° == ° ° + ° ° ° ° − ° °

= + − −

= − −

12 4 36 12

16 16 16 16

2 3 2 6 2 3 3 1 3 3 2 3 4 3 1 1 31

16 16 16 16 8 8 8 8 8 8 4 2 2 2

= − − +

= − − + = − − + = − = − = −

Section 7.7: Product-to-Sum and Sum-to-Product Formulas

745


2. cos(285 )cos(195 ) cos(240 45 )cos(240 45 )° ° = ° + ° ° − °

( )( )( ) ( ) ( ) ( )2 2 2 2

2 2 22

cos(240 45 )cos(240 45 )

cos 240 cos 45 sin 240 sin 45 cos 240 cos 45 sin 240 sin 45

cos 240 cos 45 sin 240 sin 45

1 2 3 2 1 2 3 2

2 2 2 2 4 4 4 4

1 3

8 8

° + ° ° − ° == ° ° − ° ° ° ° + ° °

= ° ° − ° °

= − − − = −

= − 1

4= −

3. sin(285 )sin(75 ) sin(240 45 )sin(30 45 )° ° = ° + ° ° + °

( )( )sin(240 45 )sin(30 45 )

sin 240 cos 45 cos 240 sin 45 sin 30 cos 45 cos30 sin 45

3 2 1 2 1 2 3 2

2 2 2 2 2 2 2 2

6 2 2 6

4 4 4 4

° + ° ° + ° == ° ° + ° ° ° ° + ° °

= − + − +

= − − +

12 36 4 12

16 16 16 16

2 3 6 2 2 3 3 3 1 3 2 3 4 3 1 1 31

16 16 16 16 8 8 8 8 8 8 4 2 2 2

= − − − −

= − − − − = − − − − = − − = − − = − +

4.

[ ] [ ]sin(75 ) sin(15 ) sin(45 30 ) sin(45 30 )

sin(45 )cos(30 ) cos(45 )sin(30 ) sin(45 )cos(30 ) cos(45 )sin(30 )

2sin(45 )cos(30 )

2 3 62

2 2 2

° + ° = ° + ° + ° − °= ° ° + ° ° + ° ° − ° °= ° °

= =

5.

[ ] [ ]cos(255 ) cos(195 ) cos(225 30 ) cos(225 30 )

cos(225 )cos(30 ) sin(225 )sin(30 ) cos(225 )cos(30 ) sin(225 )sin(30 )

2sin(225 )sin(30 )

2 1 22

2 2 2

° − ° = ° + ° − ° − °= ° ° − ° ° − ° ° + ° °= − ° °

= − − =

6.

[ ] [ ]sin(255 ) sin(15 ) sin(135 120 ) sin(135 120 )



° − ° = ° + ° − ° − °= ° ° + ° ° − ° ° − ° °= ° ° + ° ° − ° ° + ° °

2cos(135 )sin(120 )

2 3 62

2 2 2

= ° °

= − = −


746


7. [ ]

( ) ( )

1sin(4 )sin(2 ) cos(4 2 ) cos(4 2 )

21

cos 2 cos 62

θ θ θ θ θ θ

θ θ

= − − +

= −

8. [ ]

( )

1cos(4 )cos(2 ) cos(4 2 ) cos(4 2 )

21

cos(2 ) cos 62

θ θ θ θ θ θ

θ θ

= − + +

= +

9. [ ]

( ) ( )

1sin(4 )cos(2 ) sin(4 2 ) sin(4 2 )

21

sin 6 sin 22

θ θ θ θ θ θ

θ θ

= + + −

= +

10. [ ]

( )

( ) ( )

1sin(3 )sin(5 ) cos(3 5 ) cos(3 5 )

21

cos( 2 ) cos 821

cos 2 cos 82

θ θ θ θ θ θ

θ θ

θ θ

= − − +

= − −

= −

11. [ ]

( )

( ) ( )

1cos(3 )cos(5 ) cos(3 5 ) cos(3 5 )

21

cos( 2 ) cos 821

cos 2 cos 82

θ θ θ θ θ θ

θ θ

θ θ

= − + +

= − +

= +

12. [ ]

( )

( ) ( )

1sin(4 )cos(6 ) sin(4 6 ) sin(4 6 )

21

sin 10 sin( 2 )21

sin 10 sin 22

θ θ θ θ θ θ

θ θ

θ θ

= + + −

= + −

= −

13. [ ]

( )

( )

1sin sin(2 ) cos( 2 ) cos( 2 )

21

cos( ) cos 321

cos cos 32

θ θ θ θ θ θ

θ θ

θ θ

= − − +

= − −

= −

14. [ ]

( )

( )

1cos(3 )cos(4 ) cos(3 4 ) cos(3 4 )

21

cos( ) cos 721

cos cos 72

θ θ θ θ θ θ

θ θ

θ θ

= − + +

= − +

= +

15.

( )

3 1 3 3sin cos sin sin

2 2 2 2 2 2 2

1sin 2 sin

2

θ θ θ θ θ θ

θ θ

= + + −

= +

16.

( )

( ) ( )

5 1 5 5sin cos sin sin

2 2 2 2 2 2 2

1sin 3 sin( 2 )

21

sin 3 sin 22

θ θ θ θ θ θ

θ θ

θ θ

= + + −

= + −

= −

17.

( )

4 2 4 2sin(4 ) sin(2 ) 2sin cos

2 2

2sin cos 3

θ θ θ θθ θ

θ θ

− + − =

=

18.

( )

4 2 4 2sin(4 ) sin(2 ) 2sin cos

2 2

2sin 3 cos

θ θ θ θθ θ

θ θ

+ − + =

=

19.

( )( )

2 4 2 4cos(2 ) cos(4 ) 2cos cos

2 2

2cos 3 cos( )

2cos 3 cos

θ θ θ θθ θ

θ θθ θ

+ − + =

= −

=

20.

( )

5 3 5 3cos(5 ) cos(3 ) 2sin sin

2 2

2sin 4 sin

θ θ θ θθ θ

θ θ

+ − − = −

= −

21.

( )( )

3 3sin sin(3 ) 2sin cos

2 2

2sin 2 cos( )

2sin 2 cos

θ θ θ θθ θ

θ θθ θ

+ − + =

= −

=


747


22.

( )( )

3 3cos cos(3 ) 2cos cos

2 2

2cos 2 cos( )

2cos 2 cos

θ θ θ θθ θ

θ θθ θ

+ − + =

= −

=

23.

3 33 2 2 2 2cos cos 2sin sin

2 2 2 2

2sin sin2

2sin sin2

2sin sin2

θ θ θ θθ θ

θθ

θθ

θθ

+ − − = −

= − −

= − −

=

24.

3 33 2 2 2 2sin sin 2sin cos

2 2 2 2

2sin cos2

2sin cos2

θ θ θ θθ θ

θ θ

θ θ

− + − =

= −

= −

25.

3 32sin cos

sin sin(3 ) 2 22sin(2 ) 2sin(2 )

2sin(2 )cos( )

2sin(2 )

cos( )

cos

θ θ θ θθ θ

θ θθ θ

θθ

θ

+ − + =

−=

= −=

26.

3 32cos cos

cos cos(3 ) 2 22cos(2 ) 2cos(2 )

2cos(2 )cos( )

2cos(2 )

cos( )

cos

θ θ θ θθ θ

θ θθ θ

θθ

θ

+ − + =

−=

= −=

27.

4 2 4 22sin cos

sin(4 ) sin(2 ) 2 2cos(4 ) cos(2 ) cos(4 ) cos(2 )

2sin(3 )cos

2cos(3 )cos

sin(3 )

cos(3 )

tan(3 )

θ θ θ θθ θθ θ θ θ

θ θθ θ

θθθ

+ − + =

+ +

=

=

=

28.

3 32sin sin

cos cos(3 ) 2 23 3sin(3 ) sin

2sin cos2 2

2sin(2 )sin( )

2sin cos(2 )

( sin )sin(2 )

sin cos(2 )

tan(2 )

θ θ θ θθ θ

θ θ θ θθ θ

θ θθ θθ θ

θ θθ

+ − − − =− +−

− −=

− −=

=

29.

3 32sin sin

cos cos(3 ) 2 23 3sin sin(3 )

2sin cos2 2

2sin(2 )sin( )

2sin(2 )cos( )

( sin )

costan

θ θ θ θθ θ

θ θ θ θθ θ

θ θθ θθ

θθ

+ − − − =+ −+

− −=−

− −=

=

30.

( )( )

5 52sin sin

cos cos(5 ) 2 25 5sin sin(5 )

2sin cos2 2

2sin(3 )sin( 2 )

2sin(3 )cos( 2 )

( sin 2 )

cos 2

tan 2

θ θ θ θθ θ

θ θ θ θθ θ

θ θθ θ

θθ

θ

+ − − − =+ −+

− −=−

− −=

=


748


31. [ ]

[ ][ ]

[ ]

[ ]

sin sin sin(3 )

3 3sin 2sin cos

2 2

sin 2sin(2 )cos( )

cos 2sin(2 )sin

1cos 2 cos cos(3 )

2

cos cos cos(3 )

θ θ θ

θ θ θ θθ

θ θ θθ θ θ

θ θ θ

θ θ θ

+

+ − = = −

=

= ⋅ − = −

32. ( )

[ ][ ]

sin sin 3 sin(5 )

3 5 3 5sin 2sin cos

2 2

sin 2sin(4 )cos( )

cos 2sin(4 )sin

θ θ θ

θ θ θ θθ

θ θ θθ θ θ

+ + − =

= −

=

( )

( )

1cos 2 cos 3 cos(5 )

2

cos cos 3 cos(5 )

θ θ θ

θ θ θ

= ⋅ − = −

33. sin(4 ) sin(8 )

cos(4 ) cos(8 )

θ θθ θ

++

4 8 4 82sin cos

2 24 8 4 8

2cos cos2 2

2sin(6 )cos( 2 )

2cos(6 )cos( 2 )

sin(6 )

cos(6 )

tan(6 )

θ θ θ θ

θ θ θ θ

θ θθ θ

θθθ

+ − =

+ −

−=−

=

=

34. sin(4 ) sin(8 )

cos(4 ) cos(8 )

θ θθ θ

−−

4 8 4 82sin cos

2 24 8 4 8

2sin sin2 2

2sin( 2 )cos(6 )

2sin(6 )sin( 2 )

cos(6 )

sin(6 )

cot(6 )

θ θ θ θ

θ θ θ θ

θ θθ θ

θθθ

− + =

+ − − −=

− −

=−

= −

35. sin(4 ) sin(8 )

sin(4 ) sin(8 )

θ θθ θ

+−

4 8 4 82sin cos

2 24 8 4 8

2sin cos2 2

2sin(6 )cos( 2 )

2sin( 2 )cos(6 )

sin(6 )cos(2 )

sin(2 )cos(6 )

tan(6 )cot(2 )

tan(6 )

tan(2 )

θ θ θ θ

θ θ θ θ

θ θθ θ

θ θθ θθ θθθ

+ − =

− + −

−=−

=−

= −

= −

36. cos(4 ) cos(8 )

cos(4 ) cos(8 )

θ θθ θ

−+

4 8 4 82sin sin

2 24 8 4 8

2cos cos2 2

2sin(6 )sin( 2 )

2cos(6 )cos( 2 )

sin(6 ) sin( 2 )

cos(6 ) cos( 2 )

tan(6 ) tan( 2 )

tan(2 ) tan(6 )

θ θ θ θ

θ θ θ θ

θ θθ θθ θθ θθ θ

θ θ

+ − − =

+ −

− −=−−= − ⋅−

= − −=

37. 2sin cos

sin sin 2 2sin sin

2sin cos2 2

sin cos2 2

cos sin2 2

tan cot2 2

α β α βα β

α β α βα β

α β α β

α β α β

α β α β

+ − + =

− +−

+ − = ⋅

+ −

+ − =


749


38. 2cos cos

cos cos 2 2cos cos

2sin sin2 2

cos cos2 2

sin sin2 2

cot cot2 2

α β α βα β

α β α βα β

α β α β

α β α β

α β α β

+ − + =

+ −− −

+ − = − ⋅

+ −

+ − = −

39. 2sin cos

sin sin 2 2cos cos

2cos cos2 2

sin2

cos2

tan2

α β α βα β

α β α βα β

α β

α β

α β

+ − + =

+ −+

+ =

+

+ =

40. 2sin cos

sin sin 2 2cos cos

2sin sin2 2

cos2

sin2

cot2

α β α βα β

α β α βα β

α β

α β

α β

− + − =

+ −− −

+ = −

+

+ = −

41.

0 6 0 6 2 4 2 4

2 2 2 2

1 cos(2 ) cos(4 ) cos(6 ) cos 0 cos(6 ) cos(2 ) cos(4 )

2cos cos 2cos cosθ θ θ θ θ θ

θ θ θ θ θ θ+ − + −

+ + + = + + +

= +

[ ]

[ ]

2

2cos(3 )cos( 3 ) 2cos(3 )cos( )

2cos (3 ) 2cos(3 )cos

2cos(3 ) cos(3 ) cos

3 32cos(3 ) 2cos cos

2 2

2cos(3 ) 2cos(2 )cos

4cos cos(2 )cos(3 )

θ θ θ θθ θ θ

θ θ θ

θ θ θ θθ

θ θ θθ θ θ

= − + −

= += +

+ − = ==

42. [ ] [ ]0 6 0 6 2 4 2 4

2 2 2 2

1 cos(2 ) cos(4 ) cos(6 ) cos 0 cos(6 ) cos(4 ) cos(2 )

2sin sin 2sin sinθ θ θ θ θ θ

θ θ θ θ θ θ+ − + −

− + − = − + −

= − −

[ ]

[ ]

2

2sin(3 )sin( 3 ) 2sin(3 )sin( )

2sin (3 ) 2sin(3 )sin

2sin(3 ) sin(3 ) sin

3 32sin(3 ) 2sin cos

2 2

2sin(3 ) 2sin cos(2 )

4sin cos(2 )sin(3 )

θ θ θ θθ θ θ

θ θ θ

θ θ θ θθ

θ θ θθ θ θ

= − − −

= −= −

− + = ==


750


43.

( )

sin(2 ) sin(4 ) 0sin(2 ) 2sin(2 )cos(2 ) 0

sin(2 ) 1 2cos(2 ) 0

θ θθ θ θ

θ θ

+ =+ =

+ =

sin(2 ) 0 or 1 2cos(2 ) 01

cos(2 )2

θ θ

θ

= + =

= −

2 0 2 or 2 2 or

2

k kk

k

θ θθ θ

= + π = π + π= π π= + π

2 42 2 or 2 2

3 32

3 3

k k

k k

θ θ

θ θ

π π= + π = + π

π π= + π = + π


0, , , , , , , 3 2 3 3 2 3

π π π π π ππ

.

44.

( )

cos(2 ) cos(4 ) 0

2 4 2 42cos cos 0

2 2

2cos(3 )cos( ) 0

2cos 3 cos 0

θ θθ θ θ θ

θ θθ θ

+ =+ − =

− =

=

cos(3 ) 0 or cos 0θ θ= =

33 2 or 3 2 or

2 22 2

6 3 2 3

k k

k k

θ θ

θ θ

π π= + π = + π

π π π π= + = +

3

2 or 22 2

k kθ θπ π= + π = + π


, , , , , 6 2 6 6 2 6

π π π π π π

.

45.

( )

4 6 4 6

2 2

cos(4 ) cos(6 ) 0

2sin sin 0

2sin(5 )sin( ) 0

2sin 5 sin 0

θ θ θ θ

θ θ

θ θθ θ

+ −

− =

− =

− − ==

sin(5 ) 0 or sin 0θ θ= =

5 0 2 or 5 2 or

2 2

5 5 5

k k

k k

θ θ

θ θ

= + π = π + ππ π π= = +

0 2 or 2k kθ θ= + π = π + π

On the interval 0 2θ π≤ < , the solution set is

2 3 4 6 7 8 90, , , , , , , , ,

5 5 5 5 5 5 5 5

π π π π π π π ππ

.

46.

4 6 4 6

2 2

sin(4 ) sin(6 ) 0

2sin cos 0

2sin( ) cos(5 ) 0

2sin cos(5 ) 0

θ θ θ θ

θ θ

θ θθ θ

− +

− =

=

− =− =

cos(5 ) 0 or sin 0θ θ= =

0 2 or 2 ork kθ θ π= + π = + π

3

5 2 or 5 22 2

2 3 2

10 5 10 5

k k

k k

θ θ

θ θ

π π= + π = + π

π π π π= + = +

On the interval 0 2θ π≤ < , the solution set is 3 7 9 11 13 3 17 19

0, , , , , , , , , , ,10 10 2 10 10 10 10 2 10 10

π π π π π π π π π ππ

.

47. a. [ ] [ ]sin 2 (852) sin 2 (1209)y t tπ π= +

2 (852) 2 (1209) 2 (852) 2 (1209)

2 22sin cos

2sin(2061 )cos( 357 )

2sin(2061 )cos(357 )

t t t t

t t

t t

π π π π

π ππ π

+ − =

= −=

b. Because sin 1θ ≤ and cos 1θ ≤ for all θ , it follows that sin(2061 ) 1tπ ≤ and cos(357 ) 1tπ ≤ for all

values of t. Thus, 2sin(2061 )cos(357 ) 2 1 1 2y t tπ π= ≤ ⋅ ⋅ = . That is, the maximum value of y is 2.


751


c. Let 1 2sin(2061 )cos(357 )Y x xπ π= .

48. a. [ ] [ ]sin 2 (941) sin 2 (1477)y t tπ π= +

2 (941) 2 (1477) 2 (941) 2 (1477)

2 22sin cos

2sin(2418 )cos( 536 )

2sin(2418 )cos(536 )

t t t t

t t

t t

π π π π

π ππ π

+ − =

= −=

b. Because sin 1θ ≤ and cos 1θ ≤ for all θ , it follows that sin(2418 ) 1tπ ≤ and cos(2418 ) 1tπ ≤ for all

values of t. Thus, 2sin(2418 )cos(536 ) 2 1 1 2y t tπ π= ≤ ⋅ ⋅ = . That is, the maximum value of y is 2.

c. Let 1 2sin(2418 )cos(536 )Y x xπ π= .

49. 2 2cos sin 2 sin cos

cos 2 1 1 cos 22sin cos

2 2

cos 2cos 2sin 2

2 2 2 2

cos 2 sin 22 2

u x y xy

x y xy

y yx xxy

x y x yxy

I I I I

I I I

I II II

I I I II

θ θ θ θ

θ θ θ θ

θθ θ

θ θ

= + −

+ − = + −

= + + − −

+ −= + −

2 2sin cos 2 sin cos

1 cos 2 cos 2 12sin cos

2 2

cos 2cos 2sin 2

2 2 2 2

cos 2 sin 22 2

v x y xy

x y xy

y yx xxy

x y x yxy

I I I I

I I I

I II II

I I I II

θ θ θ θ

θ θ θ θ

θθ θ

θ θ

= + +

− + = + +

= − + + +

+ −= − +


752


50. a. Since φ and 0v are fixed, we need to maximize ( )sin cosθ θ φ− .

( ) ( )( ) ( )( )( )

1sin cos sin sin

21

sin 2 sin2

θ θ φ θ θ φ θ θ φ

θ φ φ

− = + − + − −

= − +

This quantity will be maximized when ( )sin 2 1θ φ− = . So,

( ) ( )( )

( )( )( ) ( )

2 2 2 20 0 0 0max 2 2

12 1 sin 1 sin 1 sin2

1 sin 1 sin 1 sincos 1 sin

v v v vR

g gg g

φ φ φφ φ φφ φ

⋅ ⋅ + + += = = =

− + −−

b. ( )

( )

2

max

50598.24

9.8 1 sin 35R = ≈

− °

The maximum range is about 598 meters.

51. ( ) ( ) ( )sin 2 sin 2 sin 2α β γ+ +

( )

[ ]

2 2 2 22sin cos sin 2

2 2

2sin( ) cos( ) 2sin cos

2sin( ) cos( ) 2sin cos

2sin cos( ) 2sin cos

2sin cos( ) cos

2sin 2cos cos2 2

24sin cos

2

α β α β γ

α β α β γ γγ α β γ γ

γ α β γ γγ α β γ

α β γ α β γγ

π βγ

+ − = +

= + − += π − − += − += − +

− + − − = − =

2cos

2

4sin cos cos2 2

4sin sin sin

4sin sin sin

α π

π πγ β α

γ β αα β γ

−

= − −

==


753


52. sin sin sin

tan tan tancos cos cos

α β γα β γα β γ

+ + = + +

sin cos cos sin cos cos sin cos cos

cos cos cos

cos (sin cos cos sin ) sin cos cos

cos cos cos

cos sin( ) sin cos cos cos sin( ) sin cos cos

cos cos cos cos cos cos

α β γ β α γ γ α βα β γ

γ α β α β γ α βα β γ

γ α β γ α β γ γ γ α βα β γ α β γ

+ +=

+ +=

+ + π − += =

( ) [ ]

( )

cos sin sin cos cos sin (cos cos cos )


sin cos ( ) cos cos sin cos( ) cos cos


sin cos cos sin sin cos cos

cos cos cos

sin (sin sin )

γ γ γ α β γ γ α βα β γ α β γ

γ α β α β γ α β α βα β γ α β γ

γ α β α β α βα β γ

γ α β

+ += =

π − + + − + + = =

− + +=

= tan tan tancos cos cos

α β γα β γ

=

53. Add the sum formulas for sin( ) and sin( )α β α β+ − and solve for sin cosα β :

[ ]



sin( ) sin( ) 2sin cos

1sin cos sin( ) sin( )

2


α β α β α β

α β α β α β

+ = +− = −

+ + − =

= + + −

54.

( )

2 2 2 2

2 22sin cos

12 sin sin

2

2 2sin sin

2 2

sin sin

sin sin

α β α β α β α β

α β α β

α β

α βα β

− + − ++ −

− + = ⋅ +

− = +

= + −= −

Thus, 2 2

sin sin 2sin cosα β α βα β − + − =

.

55.

2 2 2 2

2 22cos cos

12 cos cos

2


α β α β

+ − + −− +

+ − = ⋅ +

2 2cos cos

2 2

cos cos

β α

β α

= +

= +

Thus, cos cos 2cos cos2 2

α β α βα β + − + =

.

56.

2 2 2 2

2 22sin sin

12 cos cos

2


α β α β

+ − + −− +

+ − − = − ⋅ −

2 2cos cos

2 2

cos cos

β α

α β

= − − = −

Thus, cos cos 2sin sin2 2

α β α βα β + − − = −

.


754


Chapter 7 Review Exercises

1. 1sin 1−



equals 1.

sin 1,2 2

2

θ θ

θ

π π= − ≤ ≤

π=

Thus, ( )1sin 12

− π= .

2. 1cos 0− Find the angle , 0 ,θ θ≤ ≤ π whose cosine

equals 0. cos 0, 0

2

θ θ

θ

= ≤ ≤ ππ=

Thus, ( )1cos 02

− π= .

3. 1tan 1−



equals 1.

tan 1,2 2

4

θ θ

θ

π π= − < <

π=

Thus, ( )1tan 14

− π= .

4. 1 1sin

2− −



equals 1

2− .

1sin ,

2 2 2

6

θ θ

θ

π π= − − ≤ ≤

π= −

Thus, 1 1sin

2 6− π − = −

.

5. 1 3cos

2−

−


equals 3

2− .

3cos , 0

25

6

θ θ

θ

= − ≤ ≤ π

π=

Thus, 1 3 5cos

2 6− π− =

.

6. ( )1tan 3− −



equals 3− .

tan 3,2 2

3

θ θ

θ

π π= − − < <

π= −

Thus, ( )1tan 33

− π− = − .

7. 1sec 2− Find the angle , 0 ,θ θ≤ ≤ π whose secant

equals 2 .

sec 2, 0

4

θ θ

θ

= ≤ ≤ ππ=

Thus, 1sec 24

− π= .

8. ( )1cot 1− −

Find the angle , 0 ,θ θ π< < whose cotangent

equals 1− . cot 1, 0

3

4

θ θ ππθ

= − < <

=

Thus, ( )1 3cot 1

4

π− − = .


755


9. 1 3sin sin

8

π−



3

8

π is in the interval ,

2 2

π π −

, we can apply


1 3 3sin sin

8 8

π π− = .

10. 1 3cos cos

4

π−


( )( ) ( )( )1 1cos cosf f x x x− −= = . Since 3

4

π is


directly and get 1 3 3cos cos

4 4

π π− =

.

11. 1 2tan tan

3

π−


equation ( )( ) ( )( )1 1tan tanf f x x x− −= = but we


3

π is not


π π −



π π −

for which

2tan tan

3

π θ =

. The angle 2

3

π is in quadrant

II so tangent is negative. The reference angle of 2

3

π is

3

π and we want θ to be in quadrant IV

so tangent will still be negative. Thus, we have

2tan tan

3 3

π π = −

. Since 3

π− is in the

interval ,2 2

π π −


above and get

1 12tan tan tan tan

3 3 3

π π π− − = − = − .

12. 1sin sin8



8


π π −

, we can apply


1sin sin8 8

π π− − = − .

13. 1 15cos cos

7

π−


equation ( )( ) ( )( )1 1cos cosf f x x x− −= = , but


7

π is

not in the interval 0,π . We need to find an

angle θ in the interval 0,π for which

15cos cos

7

π θ =

. The angle 15

7

π is in

quadrant I so the reference angle of 15

7

π is

7

π.

Thus, we have 15

cos cos7 7

π π =

. Since 7

π is


above and get

1 115cos cos cos cos

7 7 7

π π π− − = = .

14. 1 8sin sin

9


equation ( )( ) ( )( )1 1sin sinf f x x x− −= = , but we


9

π− is not


π π −



π π −

for which

8sin sin

9

π θ − =

. The angle 8

9

π− is in

quadrant III so sine is negative. The reference


756


angle of 8

9

π− is 9

π and we want θ to be in

quadrant IV so sine will still be negative. Thus,

we have 8

sin sin9 9

π π − = −

. Since 9

π− is


π π −

, we can apply the

equation above and get

1 18sin sin sin sin

9 9 9

π π π− − − = − = − .

15. ( )1sin sin 0.9− follows the form of the equation

( )( ) ( )( )1 1sin sinf f x x x− −= = . Since 0.9 is in

the interval 1,1− , we can apply the equation

directly and get ( )1sin sin 0.9 0.9− = .

16. ( )1cos cos 0.6− follows the form of the equation

( )( ) ( )( )1 1cos cosf f x x x− −= = . Since 0.6 is

in the interval 1,1− , we can apply the equation

directly and get ( )1cos cos 0.6 0.6− = .

17. ( )( )1cos cos 0.3− − follows the form of the

equation ( )( ) ( )( )1 1cos cosf f x x x− −= = .

Since 0.3− is in the interval 1,1− , we can

apply the equation directly and get

( )( )1cos cos 0.3 0.3− − = − .

18. ( )1tan tan 5− follows the form of the equation

( )( ) ( )( )1 1tan tanf f x x x− −= = . Since 5 is a


and get ( )1tan tan 5 5− = .

19. Since there is no angle θ such that cos 1.6θ = − ,

the quantity ( )1cos 1.6− − is not defined. Thus,

( )( )1cos cos 1.6− − is not defined.

20. Since there is no angle θ such that sin 1.6θ = ,

the quantity 1sin 1.6− is not defined. Thus,

( )1sin sin 1.6− is not defined.


3 2 6− −π π = − = −

22. ( )1 13cos tan cos 1

4− −π = − = π

23. ( )1 17tan tan tan 1

4 4− −π π = − = −

24. 1 17 3 5cos cos cos

6 2 6− − π π = − =

25. 1 3tan sin

2−

−



equals 3

2− .

3sin ,

2 2 2

3

θ θ

θ

π π= − − ≤ ≤

π= −

So, 1 3sin

2 3− π− = −

.

Thus, 1 3tan sin tan 3

2 3−

π − = − = − .

26. 1 1tan cos

2− −


equals 1

2− .

1cos , 0

22

3

θ θ

θ

= − ≤ ≤ π

π=

So, 1 1 2cos

2 3− π − =

Thus, 1 1 2tan cos tan 3

2 3− π − = = −

.


757


27. 1 3sec tan

3−


θ θπ π− < < whose tangent is

3

3

3tan ,

3 2 2

6

θ θ

θ

π π= − < <

π=

So, 1 3tan

3 6− π= .

Thus, 1 3 2 3sec tan sec

3 6 3− π = =

.

28. 1 3csc sin

2−


θ θπ π− ≤ ≤ whose sine equals

32

.

3sin ,

2 2 2

3

θ θ

θ

π π= − ≤ ≤

π=

So, 1 3sin

2 3− π= .

Thus, 1 3 2 3csc sin csc

2 3 3− π = =

.

29. 1 3sin cot

4−

Since 3

cot , 04

θ θ π= < < , θ is in quadrant I.

Let 3x = and 4y = . Solve for r: 2

2

9 16

25

5

r

r

r

+ =

==

Thus, 1 3 4sin tan sin

4 5yr

θ− = = =

.

30. 1 5cos csc

3−

Since 5

csc ,3 2 2

θ θπ π= − ≤ ≤ , let 5r = and 3y = .

Solve for x: 2

2

9 25

16

4

x

x

x

+ =

== ±


Thus, 1 5 4cos csc cos

3 5xr

θ− = = =

.

31. 1 4tan sin

5− −

Since 4

sin ,5 2 2

θ θπ π= − − ≤ ≤ , let 4y = − and

5r = . Solve for x: 2

2

16 25

9

3

x

x

x

+ === ±


Thus, 1 44 4tan sin tan

5 3 3yx

θ− − − = = = = −

32. 1 3tan cos

5− −

Since 3

cos , 05

θ θ= − ≤ ≤ π , let 3x = − and

5r = . Solve for y: 2

2

9 2516

4

y

yy

+ === ±


Thus, 1 3 4 4tan cos tan

5 3 3

y

xθ− − = = = = − −

33. ( ) ( )2sin 3f x x=

( )( )

( )

( )

1

1 1

2sin 3

2sin 3

sin 32

3 sin2

1sin

3 2

y x

x y

xy

xy

xy f x

−

− −

=

=

=

= = =



758


( )1f x− and is 6 6

xπ π− ≤ ≤ , or ,

6 6

π π −

in



function is 2


1,1− . That is,

1 12

2 2

x

x

− ≤ ≤

− ≤ ≤



domain of a function is the range of its inverse and the domain of the inverse is the range of the function. Therefore, the range of ( )f x is

[ ]2,2− .

34. ( ) ( )tan 2 3 1f x x= + −

( )( )( )

( )( )

( ) ( )

1

1

1 1

tan 2 3 1

tan 2 3 1

1 tan 2 3

2 3 tan 1

2 tan 1 3

1 3tan 1

2 2

y x

x y

x y

y x

y x

y x f x

−

−

− −

= + −

= + −

+ = +

+ = +

= + −

= + − =


( )1f x− and is 3 3

2 4 2 4x

π π− − < < − + , or

3 3,

2 4 2 4

π π − − − +

in interval notation. To find

the domain of ( )1f x− we note that the argument

of the inverse tangent function can be any real

number. Thus, the domain of ( )1f x− is all real

numbers, or ( ),−∞ ∞ in interval notation. Recall

that the domain of a function is the range of its inverse and the domain of the inverse is the range of the function. Therefore, the range of

( )f x is ( ),−∞ ∞ .

35. ( ) cos 3f x x= − +

( ) ( )1 1

cos 3

cos 3

3 cos

3 cos

cos 3

y x

x y

x y

x y

y x f x− −

= − += − +

− = −− =

= − =


( )1f x− and is 0 x π≤ ≤ , or 0,π in interval


that the argument of the inverse cosine function is 3 x− and that it must lie in the interval

1,1− . That is,

1 3 1

4 2

4 2

2 4

x

x

x

x

− ≤ − ≤− ≤ − ≤ −

≥ ≥≤ ≤

The domain of ( )1f x− is { }| 2 4x x≤ ≤ , or

2,4 in interval notation. Recall that the


2,4 .

36. ( ) ( )2sin 1f x x= − +

( )( )

( )

1

1

1

2sin 1

2sin 1

sin 12

1 sin2

sin 12

1 sin2

y x

x y

xy

xy

xy

xy

−

−

−

= − +

= − +

= − +

− + = − = −

= −

The domain of ( )f x equals the range of ( )1f x−

and is 1 12 2

xπ π− ≤ ≤ + , or 1 ,1

2 2

π π − +

in



function is 2



759


1,1− . That is,

1 12

2 2

x

x

− ≤ ≤

− ≤ ≤




2,2− .

37. Let 1sin uθ −= so that sin uθ = , 2 2

π πθ− ≤ ≤ ,

1 1u− ≤ ≤ . Then,

( )1 2

2 2

cos sin cos cos

1 sin 1

u

u

θ θ

θ

− = =

= − = −


π πθ− ≤ ≤ ,

1u ≥ . Then,

( )1

2 2

2

sincos csc cos cos cot sin

sin

cot cot csc 1

csc csc csc

1

u

u

u

θθ θ θ θθ

θ θ θθ θ θ

− = = ⋅ =

−= = =

−=


π πθ− ≤ ≤

and 0θ ≠ , 1u ≥ . Then,

( )1 1 1sin csc sin

cscu

uθ

θ− = = =


π πθ− ≤ ≤

and 0θ ≠ , 1u ≥ . Then,

( )1 2

22

2

tan csc tan tan

1sec 1

csc 1

1

1

u

u

θ θ

θθ

− = =

= − =−

=−

41. 2 2

2

2

1tan cot sin tan sin

tan1 sin

cos

θ θ θ θ θθ

θθ

− = ⋅ −

= −=

42. 2 2

2

2

1sin csc sin sin sin

sin1 sin

cos

θ θ θ θ θθ

θθ

− = ⋅ −

= −=

43. 2 2 2 2

22

sin (1 cot ) sin csc1

sinsin

1

θ θ θ θ

θθ

+ = ⋅

= ⋅

=

44. 2 2 2 2

22

(1 sin )(1 tan ) cos sec1

coscos

1

θ θ θ θ

θθ

− + = ⋅

= ⋅

=

45.

( )2 2 2 2 2

2 2 2

2

2

5cos 3sin 2cos 3cos 3sin

2cos 3 cos sin

2cos 3 13 2cos

θ θ θ θ θθ θ θθ

θ

+ = + += + +

= + ⋅= +

46. ( )2 2 2 2

2 2

2

4sin 2cos 4 1 cos 2cos

4 4cos 2cos4 2cos

θ θ θ θθ θθ

+ = − +

= − += −

47. 2 2

2 2

1 cos sin (1 cos ) sin

sin 1 cos sin (1 cos )1 2cos cos sin

sin (1 cos )


θ θ θθ θ

− − ++ =− −

− + +=−

1 2cos 1

sin (1 cos )2 2cos

sin (1 cos )2(1 cos )

sin (1 cos )2

sin2csc

θθ θ

θθ θ

θθ θ

θθ

− +=−

−=−

−=−

=

=


760


48. 2 2

2 2

sin 1 cos sin (1 cos )

1 cos sin sin (1 cos )sin 1 2cos cos

sin (1 cos )


θ θ θθ θ

+ + ++ =+ +

+ + +=+

1 2cos 1

sin (1 cos )2 2cos

sin (1 cos )2(1 cos )

sin (1 cos )2

sin2csc

θθ θ

θθ θ

θθ θ

θθ

+ +=+

+=+

+=+

=

=

49.

1cos cos cos

1cos sin cos sincos

1sin

1cos1

1 tan

θ θ θθ θ θ θ

θ

θθ

θ

= ⋅− −

=−

=−

50.

( )

2 2

2

2

sin 1 cos sin1

1 cos 1 cos1 cos 1 cos

1 coscos cos

1 coscos (1 cos )

1 coscos

θ θ θθ θ

θ θθ

θ θθ

θ θθ

θ

+ −− =+ +

+ − −=

++=

++=

+=

51.

2

2

1csc sinsin

11 csc sin1sin1

sin 11 1 sin

1 sin 1 sin1 sin

1 sin1 sin

cos

θ θθθ θ

θ

θθ

θ θθθ

θθ

= ⋅+ +

=+

−= ⋅+ −−=−−=

52.

2

2

111 sec coscos

1sec coscos

1 cos

11 cos 1 cos

1 1 cos1 cos

1 cossin

1 cos

θ θθθ θ

θθ

θ θθ

θθ

θθ

++ = ⋅

+=

+ −= ⋅−

−=−

=−

53.

2

2

1csc sin sin

sin1 sin

sincos

sincos

cossin

cos cot

θ θ θθ

θθθ

θθθθ

θ θ

− = −

−=

=

= ⋅

=

54.

2

2

3

1csc 1 cossin

1 cos 1 cos 1 cos1 1 cos

sin 1 cos1 cos

sin sin1 cos

sin

θ θθθ θ θ

θθ θ

θθ θ

θθ

+= ⋅− − +

+= ⋅−

+=

+=

55.

( )

( )

2

2

3

1 sincos (1 sin )

sec1 sin

cos (1 sin )1 sin

cos 1 sin

1 sincos cos

1 sincos

1 sin

θ θ θθ

θθ θθ

θ θθ

θ θθ

θθ

− = −

+= − ⋅+

−=

+

=+

=+


761


56.

2

2 2

1 cos1 cos sin

1 cos1 cossin

csc cot csc cot

csc cot csc cot(csc cot )

csc cot

θθ θ

θθθ

θ θ θ θθ θ θ θθ θθ θ

−− =

++

− −= ⋅+ −−=−

2

2

(csc cot )

1(csc cot )

θ θ

θ θ

−=

= −

57.

2 2

2 2

2

cos sincot tan

sin coscos sin

sin cos1 sin sin

sin cos1 2sin

sin cos

θ θθ θθ θ

θ θθ θ

θ θθ θ

θθ θ

− = −

−=

− −=

−=

58. ( ) ( )

( )( )[ ]

( )( )( )( )

( )

22 22

4 4 2 2 2 2

2

2 2

2

2

2

1 2sin2sin 1

sin cos sin cos sin cos

cos(2 )

1 cos sin 1

cos 2

cos 2

cos 2

1 2cos 1

1 2cos

θθθ θ θ θ θ θ

θθ θ

θθθ

θθ

− −− =− − +

−=

− − ⋅

=−

= −= − −

= −


cos sin cos sincos cos sin sin

cos sin cos sincos sin

sin coscot tan


α β α βα β α ββ αβ αβ α

+ −=

= −

= −

= −


sin cos sin cossin cos cos sin

sin cos sin cos1 cot tan



α β

− −=

= −

= −


cos cos cos coscos cos sin sin

cos cos cos cos1 tan tan



α β

− +=

= +

= +


sin cos sin coscos cos sin sin

sin cos sin coscos sin

sin coscot tan


α β α βα β α βα βα βα β

+ −=

= −

= −

= −

63. sin

(1 cos ) tan (1 cos ) sin2 1 cos

θ θθ θ θθ

+ = + ⋅ =+

64. 1 cos

sin tan sin 1 cos2 sin

θ θθ θ θθ

−= ⋅ = −

65. ( ) ( )( )

( )( )

2 2

2 2

2

2 2

2 2

2

cos 2cos2cot cot 2 2

sin sin 2

2cos cos sin

sin 2sin cos

cos sin

sincos sin

sin sincot 1

θθθ θθ θ

θ θ θθ θ θ

θ θθ

θ θθ θθ

= ⋅ ⋅

−=

−=

= −

= −

66. ( )( ) ( ) ( )( )( )

22sin 2 1 2sin 2sin 2 cos 2

sin 2 2

sin 4

θ θ θ θθ

θ

− == ⋅=

67. ( )( )

( )( )

22 2

2

1 8sin cos 1 2 2sin cos

1 2sin 2

cos 2 2

cos 4

θ θ θ θθ

θθ

− = −= −= ⋅=

68. ( ) ( )

( )( )

( )( )( )

sin 3 cos sin cos 3 sin 3

sin 2 sin 2

sin 2

sin 21


θθ

− −=

=

=


762


69. ( ) ( )( ) ( )

( ) ( )( ) ( )

( )( )( )

2 4 2 4

2 22 4 2 4

2 2

2sin cossin 2 sin 4

cos 2 cos 4 2cos cos

2sin 3 cos

2cos 3 cos

sin 3

cos 3

tan 3

θ θ θ θ


θ θθ θ

θθθ

+ −

+ +

+ =

+

−=

−

=

=

70. ( ) ( )( ) ( )

( )

( )

( ) ( )( ) ( )

( )

2 4 2 4cos

2 22 4 2 4

2 2

sin 2 sin 4 tan 3

sin 2 sin 4 tan

2sintan 3

tan2sin cos

2sin 3 cos tan 3

2sin cos 3 tan

θ θ θ θ

θ θ θ θ

θ θ θθ θ θ

θθ

θ θ θθ θ θ

+ −

− +

++

− = +

−= +

−

( ) ( ) ( )

( ) ( )

tan 3tan 3 cot

tantan 3 tan 3

tan tan0

θθ θ

θθ θθ θ

= − +

= − +

=

71. cos(2 ) cos(4 )

tan tan(3 )cos(2 ) cos(4 )

2sin(3 )sin( )tan tan(3 )

2cos(3 )cos( )

θ θ θ θθ θ

θ θ θ θθ θ

− −+

− −= −−

2sin(3 )sintan tan(3 )

2cos(3 )costan(3 ) tan tan tan(3 )0

θ θ θ θθ θ

θ θ θ θ

= −

= −=

72. 2 10 2 10

2 2

cos(2 ) cos(10 )

2sin sin

2sin(6 )sin( 4 )2sin(6 )sin(4 )

θ θ θ θθ θ

θ θθ θ

+ −− = −

= − −=

( )( ) ( ) ( )

( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

sin 42sin 6 cos 4

cos 4

sin 4 12 sin 6 4 sin 6 4

cos 4 2

tan 4 sin 10 sin 2

tan 4 sin 2 sin 10

θθ θ

θθ

θ θ θ θθθ θ θθ θ θ

= ⋅

= ⋅ ⋅ + −

= + = +

73. ( )sin165º sin 120º 45ºsin120º cos 45º cos120º sin 45º

= += ⋅ + ⋅

( )

3 2 1 2

2 2 2 2

6 2

4 41

6 24

= ⋅ + − ⋅

= −

= −

74. ( )tan105º tan 60º 45ºtan 60º tan 45º

1 tan 60º tan 45º3 1

1 3 13 1 1 3

1 3 1 3

= ++=

−+=

− ⋅+ += ⋅

− +

1 2 3 3

1 34 2 3

2

2 3

+ +=−

+=−

= − −

75.

( )

5 3 2cos cos

12 12 12


2 3 2 1

2 2 2 26 2

4 41

6 24

π π π = + π π π π= ⋅ − ⋅

= ⋅ − ⋅

= −

= −

76.

( )

2 3sin sin

12 12 12


1 2 3 2

2 2 2 22 6

4 41

2 64

π π π − = −

π π π π= ⋅ − ⋅

= ⋅ − ⋅

= −

= −


763


77. ( )cos80º cos 20º sin80º sin 20º cos 80º 20ºcos 60º1

2

⋅ + ⋅ = −=

=

78. ( )sin 70º cos 40º cos 70º sin 40º sin 70º 40ºsin 30º1

2

⋅ − ⋅ = −=

=

79.

21 cos 1

4 4 2tan tan8 2 21 cos 14 2

π π − − π = = = π + +

( )2

2 2

2 2

2 2 2 2

2 2 2 2

2 2

42 2 2

2 22 2 2

22 1

−=+− −= ⋅+ −

−=

−= ⋅

−=

= −

80.

5 51 cos5 4 4sin sin

8 2 2

π π − π = =

21

2

2

2 2

4

2 2

2

− − =

+=

+=

81. 4 5

sin , 0 ; sin ,5 2 13 2

α α β βπ π= < < = < < π

3 4 12 5cos , tan , cos , tan ,

5 3 13 12α α β β= = = − = −

0 ,2 4 4 2 2

α βπ π π< < < <

a. sin( ) sin cos cos sin4 12 3 5

5 13 5 1348 15 33

65 65

α β α β α β+ = + = ⋅ − + ⋅ − += = −

b. cos( ) cos cos sin sin3 12 4 5

5 13 5 1336 20 56

65 65

α β α β α β+ = − = ⋅ − − ⋅ − −= = −

c. sin( ) sin cos cos sin4 12 3 5

5 13 5 1348 15 63

65 65

α β α β α β− = − = ⋅ − − ⋅ − −= = −

d. tan tan

tan( )1 tan tan

α βα βα β++ =

−

4 53 12

4 51

3 12

1111 9 3312

14 12 14 569

+ − =

− ⋅ −

= = ⋅ =

e. 4 3 24

sin(2 ) 2sin cos 25 5 25

α α α= = ⋅ ⋅ =

f. 2 2

2 2

cos(2 ) cos sin

12 5 144 25 119

13 13 169 169 169

β β β= − = − − = − =

g. 1 cos

sin2 2

121

132

2525 5 5 2613

2 26 2626

β β−=

− − =

= = = =

h. 1 cos

cos2 2

3 81 4 2 2 55 5

2 2 5 55

α α+=

+= = = = =


764


82. 4 5

cos , 0 ; cos , 05 2 13 2

α α β βπ π= < < = − < <

3 3 12 12sin , tan , sin , tan ,

5 4 13 5

0 , 02 4 4 2

α α β β

α β

= = = − = −

π π< < − < <


5 13 5 1315 48

6533

65

α β α β α β+ = + = ⋅ + ⋅ −

−=

= −


5 13 5 1320 36

6556

65

α β α β α β+ = − = ⋅ − ⋅ −

+=

=


3 5 4 12

5 13 5 1315 48

6563

65

α β α β α β− = −−= ⋅ − ⋅

+=

=

d. tan tan

tan( )1 tan tan

α βα βα β++ =

−

3 124 5

3 121

4 5

332020

56 202033

56

+ − =

− ⋅ −

−= ⋅

= −

e. 3 4 24

sin(2 ) 2sin cos 25 5 25

α α α = = ⋅ ⋅ =

f. 2 2cos(2 ) cos sinβ β β= −

2 2

5 12 25 144 119

13 13 169 169 169 = − − = − = −

g. 1 cos

sin2 2

51

132

84 2 2 1313

2 13 1313

β β−= −

−= −

= − = − = − = −

h. 1 cos

cos2 2

4 91 9 3 3 105 5

2 2 10 1010

α α+=

+= = = = =

83. 3 3 12 3

sin , ; cos , 25 2 13 2

α α β βπ π= − π < < = < < π

4 3 5 5cos , tan , sin , tan ,

5 4 13 123 3

,2 2 4 4 2

α α β β

α β

= − = = − = −

π π π< < < < π


5 13 5 1336 20

6516

65

α β α β α β+ = + = − ⋅ + − ⋅ − − +=

= −


5 13 5 1348 15

6563

65

α β α β α β+ = − = − ⋅ − − ⋅ − − −=

= −


5 13 5 1336 20

6556

65

α β α β α β− = − = − ⋅ − − ⋅ − − −=

= −


765


d. tan tan

tan( )1 tan tan

3 5 11 16 164 12 3

213 5 3 21 631

164 12

α βα βα β++ =

− + − = = = ⋅ = − −

e. sin(2 ) 2sin cos3 4 24

25 5 25

α α α= = ⋅ − ⋅ − =

f. 2 2

2 2

cos(2 ) cos sin

12 5

13 13144 25 119

169 169 169

β β β= − = − −

= − =

g. 1 cos

sin2 2

12 11 1 1 2613 13

2 2 26 2626

β β−=

−= = = = =

h. 1 cos

cos2 2

41

52

11 1 105

2 10 1010

α α+= −

+ − = −

= − = − = − = −

84. 4 5

sin , 0; cos ,5 2 13 2

α α β βπ π= − − < < = − < < π

3 4 12 12cos , tan , sin , tan ,

5 3 13 5

0,4 2 4 2 2

α α β β

α β

= = − = = −

π π π− < < < <


4 5 3 12

5 13 5 13

20 36

65 6556

65

α β α β α β+ = +

= − ⋅ − + ⋅

= +

=


5 13 5 1315 48

65 6533

65

α β α β α β+ = − = ⋅ − − − ⋅ −= +

=


5 13 5 1320 36

65 6516

65

α β α β α β− = − = − ⋅ − − ⋅

= −

= −

d. tan tan

tan( )1 tan tan

4 123 5

4 121

3 5

5656 15 5615

33 15 33 3315

α βα βα β++ =

−

− + − =

− − −

− = = − − = −

e. 4 3 24

sin(2 ) 2sin cos 25 5 25

α α α = = − = −

f. 2 2

2 2 5 12cos(2 ) cos sin

13 13β β β = − = − −

25 144

169 169119

169

= −

= −

g. 1 cos

sin2 2

51

132

189 3 3 1313

2 13 1313

β β−=

− − =

= = = =


766


h. 1 cos

cos2 2

31

52

84 2 2 55

2 5 55

α α+=

+=

= = = =

85. 3 3 12

tan , ; tan , 04 2 5 2

α α β βπ π= π < < = < <

3 4 12 5sin , cos , sin , cos ,

5 5 13 133

, 02 2 4 2 4

α α β β

α β

= − = − = =

π π π< < < <


5 13 5 1315 48

65 6563

65

α β α β α β+ = + = − ⋅ + − ⋅

= − −

= −


5 13 5 1320 36

65 6516

65

α β α β α β+ = − = − ⋅ − − ⋅

= − +

=


5 13 5 1315 48

65 6533

65

α β α β α β− = − = − ⋅ − − ⋅

= − +

=

d. tan tan

tan( )1 tan tan

3 124 53 12

14 5

6363 5 6320

4 20 4 165

α βα βα β++ =

−

+=

−

= = − = − −

e. 3 4 24

sin(2 ) 2sin cos 25 5 25

α α α = = − − =

f. 2 2

2 2

cos(2 ) cos sin

5 12 25 144 119

13 13 169 169 169

β β β= − = − = − = −

g. 1 cos

sin2 2

5 81 4 2 2 1313 13

2 2 13 1313

β β−=

−= = = = =

h. 1 cos

cos2 2

41

52

11 1 105

2 10 1010

α α+= −

+ − = −

= − = − = − = −

86. 4 12 3

tan , ; cot ,3 2 5 2

α α β βπ π= − < < π = π < <

4 3 5 12sin , cos , sin , cos ,

5 5 13 133

,4 2 2 2 2 4

α α β β

α β

= = − = − = −

π π π π< < < <


5 13 5 1348 15

6533

65

α β α β α β+ = + = ⋅ − + − ⋅ − − +=

= −


5 13 5 1336 20

6556

65

α β α β α β+ = − = − ⋅ − − ⋅ −

+=

=


767



5 13 5 1348 15

6563

65

α β α β α β− = − = ⋅ − − − ⋅ − − −=

= −

d. tan tan

tan( )1 tan tan

α βα βα β++ =

−

4 53 124 5

13 12

1112563611 36

12 5633

56

− +=

− − ⋅

−

=

= − ⋅

= −

e. 4 3 24

sin(2 ) 2sin cos 25 5 25

α α α = = − = −

f. 2 2

2 2

cos(2 ) cos sin

12 5

13 13144 25

169 169119

169

β β β= − = − − −

= −

=

g. 1 cos

sin2 2

121

132

2525 5 5 2613

2 26 2626

β β−=

− − =

= = = =

h. 1 cos

cos2 2

3 211 1 55 5

2 2 5 55

α α+=

+ − = = = = =

87. 3

sec 2, 0; sec 3, 22 2

α α β βπ π= − < < = < < π

3 1sin , cos , tan 3,

2 22 2 1

sin , cos , tan 2 2,3 3

30,

4 2 4 2

α α α

β β β

α β

= − = = −

= − = = −

π π− < < < < π


3 1 1 2 2

2 3 2 3

3 2 2

6

α β α β α β+ = + = − + −

− −=


1 1 3 2 2

2 3 2 3

1 2 6

6

α β α β α β+ = −

= ⋅ − − −

−=


3 1 1 2 2

2 3 2 3

3 2 2

6

α β α β α β− = −

= − ⋅ − ⋅ −

− +=

d.

( )( )( )

tan tantan( )

1 tan tan

3 2 2

1 3 2 2

3 2 2 1 2 6

1 2 6 1 2 6

α βα βα β++ =

−− + −

=− − −

− − += ⋅ − +

9 3 8 2

23

8 2 9 3

23

− −=−+=

e. 3 1 3

sin(2 ) 2sin cos 22 2 2

α α α = = − = −

f. 2 2

22

cos(2 ) cos sin

1 2 2 1 8 7

3 3 9 9 9

β β β= − = − − = − = −


768


g. 1 cos

sin2 2

1 21 1 1 33 3

2 2 3 33

β β−=

−= = = = =

h. 1 cos

cos2 2

1 31 3 32 2

2 2 4 2

α α+=

+= = = =

88. csc 2, ; sec 3,2 2

α α β βπ π= < < π = − < < π

1 3 3sin , cos , tan ,

2 2 32 2 1

sin , cos , tan 2 2,3 3

,4 2 2 4 2 2

α α α

β β β

α β

= = − = −

= = − = −

π π π π< < < <


1 1 3 2 2

2 3 2 3

1 2 6

6

α β α β α β+ = + = ⋅ − + − ⋅

− −=


3 1 1 2 2

2 3 2 3

3 2 2

6

α β α β α β+ = − = − − −

−=


1 1 3 2 2

2 3 2 3

1 2 6

6

α β α β α β− = − = − − −

− +=

d.

( )( )

tan tantan( )

1 tan tan

32 2

33

1 2 23

3 6 23

3 2 63

α βα βα β++ =

−

− + −=

− − −

− −

=−

3 6 2 3 2 6

3 2 6 3 2 627 3 24 2

159 3 8 2

5

− − += ⋅− +

− −=−

+=

e. sin(2 ) 2sin cos

1 3 32

2 2 2

α α α= = − = −

f. 2 2

22

cos(2 ) cos sin

1 2 2

3 3

1 8 7

9 9 9

β β β= − = − −

= − = −

g. 1 cos

sin2 2

11

32

42 2 3 63

2 3 33 3

β β−=

− − =

= = = ⋅ =

h. 1 cos

cos2 2

31

2

2

2 32 3 2 32

2 4 2

α α+=

+ − =

−− −= = =


769


89. 2 3 2 3

sin , ; cos ,3 2 3 2

α α β βπ π= − π < < = − π < <

5 2 5 5cos , tan , sin ,

3 5 35 3 3

tan , ,2 2 2 4 2 2 4

α α β

α ββ

= − = = −

π π π π= < < < <


2 2 5 5

3 3 3 3

4 5

9 91

α β α β α β+ = + = − − + − −

= +

=


5 2 2 5

3 3 3 3

2 5 2 5

9 90

α β α β α β+ = − = − − − − −

= −

=


2 2 5 5

3 3 3 3

4 5

9 91

9

α β α β α β− = − = − − − − −

= −

= −

d. tan tan

tan( )1 tan tan

α βα βα β++ =

−

2 5 55 22 5 5

15 2

4 5 5 510

1 1

9 510 ; Undefined0

+=

− ⋅

+

=−

=

e. sin(2 ) 2sin cos

2 5 4 52

3 3 9

α α α= = − − =

f. 2 2

22

cos(2 ) cos sin

2 5 4 5 1

3 3 9 9 9

β β β= − = − − − = − = −

g. 1 cos

sin2 2

2 515 303 3

2 2 6 6

β β−=

− − = = = =

h.

51

31 coscos

2 2 2

α α

+ − + = − = −

( )

3 532

3 5

6

6 3 5

6

6 3 5

6

−

= −

−= −

−= −

−= −

90. tan 2, ; cot 2,2 2

α α β βπ π= − < < π = − < < π

1 2 1 1tan , sin , cos ,sin ,

2 5 5 52

cos , , 4 2 2 4 2 25

β α α β

α ββ

= − = = − =

π π π π= − < < < <


2 2 1 1

5 5 5 54 1

5 55

51

α β α β α β+ = + = − + −

= − −

= −

= −


1 2 2 1

5 5 5 52 2

5 50

α β α β α β+ = − = − − −

= −

=


770



2 2 1 1

5 5 5 54 1

5 53

5

α β α β α β− = − = ⋅ − − − ⋅

= − +

= −

d. tan tan

tan( )1 tan tan

12

21

1 ( 2)2

52 ; undefined

0

α βα βα β++ =

− − + − = − − ⋅ −

−=

e. 2 1 4

sin(2 ) 2sin cos 255 5

α α α = = − = −

f. 2 2

2 2

cos(2 ) cos sin

2 1 4 1 3

5 5 55 5

β β β= − = − − = − =

g.

21

1 cos 5sin

2 2 2

β β − − − = =

5 2

52

5 2

2 5

5 2 5

10

10 5 2 5

10

+

=

+=

+=

+=

h.

11

1 cos 5cos

2 2 2

α α + − + = =

5 1

52

5 1

2 5

5 5

10

10 5 5

10

−

=

−=

−=

−=

91. 1 13 1cos sin cos

5 2− − −

Let 1 3sin

5α −= and 1 1

cos2

β −= . α is in


sin5

α = ,

02

α π≤ ≤ , and 1

cos , 02 2

β β π= ≤ ≤ .

2

2

cos 1 sin

3 9 16 41 1

5 25 25 5

α α= −

= − = − = =

2

2

sin 1 cos

1 1 3 31 1

2 4 4 2

β β= −

= − = − = =

( )1 13 1cos sin cos cos

5 2cos cos sin sin

4 1 3 3

5 2 5 24 3 3 4 3 3

10 10 10

α β

α β α β

− − − = −

= +

= ⋅ + ⋅

+= + =

92. 1 15 4sin cos cos

13 5− − −

Let 1 15 4cos and cos .

13 5α β− −= = α is in


cos13

α = ,

02

α π≤ ≤ , and 4

cos , 05 2

β β π= ≤ ≤ .


771


2

2

sin 1 cos

5 25 144 121 1

13 169 169 13

α α= −

= − = − = =

2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

β β= −

= − = − = =

( )1 15 4sin cos cos sin

13 5sin cos cos sin12 4 5 3

13 5 13 548 15 33

65 65 65

α β

α β α β

− − − = −

= −

= ⋅ − ⋅

= − =

93. 1 11 3tan sin tan

2 4− − − −

Let 1 1sin

2α − = −

and 1 3

tan4

β −= . α is in

quadrant IV; β is in quadrant I. Then,

1sin

2α = − , 0

2α π≤ ≤ , and

3tan

4β = ,

02

β π< < .

2

2

cos 1 sin

1 1 3 31 1

2 4 4 2

α α= −

= − − = − = =

1 3tan

33α = − = −

( )1 11 3tan sin tan tan

2 4α β− − − − = −

tan tan

1 tan tan

3 33 4

3 31

3 4

4 3 9123 3

112

9 4 3 12 3 3

12 3 3 12 3 3

144 75 3

117

48 25 3

39

48 25 3

39

α βα β−=

+

− −=

+ −

− −

=−

− − += ⋅− +

− −=

− −=

+= −

94. 1 1 4cos tan ( 1) cos

5− − − + −

Let 1 1 4tan ( 1) and cos .

5α β− − = − = −

α is in

quadrant IV; β is in quadrant II. Then

tan 1, 02

α απ= − − < < , and 4

cos5

β = − ,

2

π β≤ ≤ π .

2 2sec 1 tan 1 ( 1) 2

1 2cos

22

α α

α

= + = + − =

= =

2

2

sin 1 cos

2 1 1 21 1

2 2 2 2

α α= − −

= − − = − − = − = −

2

2

sin 1 cos

4 16 9 31 1

5 25 25 5

β β= −

= − − = − = =


772


( )1 1 4cos tan ( 1) cos cos

5α β− − − + − = +

cos cos sin sin

2 4 2 3

2 5 2 5

4 2 3 2

10 102

10

α β α β= − = − − − −= +

= −

95. 1 3sin 2cos

5− −

Let 1 3cos .

5α − = −

α is in quadrant II. Then

3cos ,

5 2

πα α= − ≤ ≤ π .

2

2

sin 1 cos

3 9 16 41 1

5 25 25 5

α α= −

= − − = − = =

1 3sin 2cos sin 2

52sin cos

4 3 242

5 5 25

α

α α

− − = =

= − = −

96. 1 4cos 2 tan

3−

Let 1 4tan .

3α −= α is in quadrant I. Then

4tan

3α = , 0

2α π< < .

2

2

sec tan 1

4 16 25 51 1

3 9 9 3

α α= +

= + = + = =

3cos

5α =

( )1

2

2

4cos 2 tan cos 2

32cos 1

3 9 72 1 2 1

5 25 25

α

α

− =

= − = − = − = −

97. 1

cos2

θ =

52 or 2

3 3k k

π πθ π θ π= + = + , k is any integer

On 0 2θ π≤ < , the solution set is 5

, 3 3

π π

.

98. 3

sin2

θ = −

4 52 or 2

3 3k k


On 0 2θ π≤ < , the solution set is 4 5

, 3 3

π π

.

99. 2cos 2 0

2cos 2

2cos

2

θθ

θ

+ =

= −

= −

3 52 or 2

4 4k k



, 4 4

π π

.

100. tan 3 0

tan 3

θθ

+ =

= −

2, is any integer

3k kθ π= + π

On the interval 0 2θ π≤ < , the solution set is 2 5

, 3 3

π π

.

101. sin(2 ) 1 0

sin(2 ) 1

θθ+ =

= −

32 2

23

,4

k

k

πθ π

πθ π

= +

= + k is any integer


, 4 4

π π


773


102. ( )cos 2 0θ =

32 2 or 2 2

2 23

,4 4

k k

k k

θ θ

θ θ

π π= + π = + π

π π= + π = + π

where k is any integer On the interval 0 2θ π≤ < , the solution set is

3 5 7, , ,

4 4 4 4

π π π π

.

103. ( )tan 2 0θ =

2 0

, where is any integer2

k

kk

θ

θ

= + ππ=

On the interval 0 2θ π≤ < , the solution set is 3

0, , ,2 2

π ππ

.

104. ( )sin 3 1θ =

3 22

2

6 3

k

k

θ

θ

π= + π

π π= + , where k is any integer


, ,6 6 2

π π π

.

105. 2sec 4θ = sec 2

1cos

2

θ

θ

= ±

= ±

2+ or + ,

3 3where is any integer

k k

k

π πθ π θ π= =

On the interval 0 2θ π≤ < , the solution set is 2 4 5

, , ,3 3 3 3

π π π π

.

106. 2csc 1θ = csc 1

sin 1

+ , where is any integer2

k k

θθ

πθ π

= ±= ±

=

On the interval 0 2θ π≤ < , the solution set is 3

,2 2

π π

.

107. 0.2sin 0.05θ = Find the intersection of 1 0.2sinY θ= and

2 0.05Y = :

On the interval 0 2θ π≤ < , 0.25x ≈ or 2.89x ≈


108. 0.9cos(2 ) 0.7θ =

Find the intersection of 1 0.9cos(2 )Y θ= and

2 0.7Y = :

On the interval 0 2θ π≤ < , 0.34x ≈ or 2.80x ≈ or 3.48x ≈ or 5.94x ≈

The solution set is { }0.34,2.80,3.48,5.94 .

109. sin sin(2 ) 0

sin 2sin cos 0

sin (1 2cos ) 0

θ θθ θ θ

θ θ

+ =+ =

+ =

1 2cos 0 or sin 0

1 0,cos

22 4

,3 3

θ θθθ

θ

+ = == π= −

π π=


0, , ,3 3

π ππ

.


774


110. ( )2

2

cos 2 sin

1 2sin sin

2sin sin 1 0

(2sin 1)(sin 1) 0

θ θ

θ θθ θ

θ θ

=

− =

+ − =− + =

2sin 1 0 or sin 1 0

1 sin 1sin

2 35 2,

6 6

θ θθθ

πθπ πθ

− = + == −==

=


, ,6 6 2

π π π

.

111. sin(2 ) cos 2sin 1 0

2sin cos cos 2sin 1 0

cos (2sin 1) 1(2sin 1) 0

(2sin 1)(cos 1) 0

θ θ θθ θ θ θ

θ θ θθ θ

− − + =− − + =− − − =

− − =

1sin or cos 1

205

, 6 6

θ θθπ πθ

= ==

=


0, ,6 6

π π

.

112. sin(2 ) sin 2cos 1 0

2sin cos sin 2cos 1 0

sin (2cos 1) 1(2cos 1) 0

(2cos 1)(sin 1) 0

θ θ θθ θ θ θ

θ θ θθ θ

− − + =− − + =− − − =

− − =

1cos or sin 1

25

, 23 3

θ θππ π θθ

= =

==


, ,3 2 3

π π π

.

113. 22sin 3sin 1 0

(2sin 1)(sin 1) 0

θ θθ θ

− + =− − =

2sin 1 0 or sin 1 0

1 sin 1sin

25 2,

6 6

θ θθθ

πθπ πθ

− = − ====

=


, ,6 2 6

π π π

.

114. 22cos cos 1 0

(2cos 1)(cos 1) 0

θ θθ θ

+ − =− + =

2cos 1 0 or cos 1 0

1 cos 1cos

25

,3 3

θ θθθθ π

π πθ

− = + == −==

=


, ,3 3

π ππ

.

115.

( )

( )( )

2

2

2

2

4sin 1 4cos

4 1 cos 1 4cos

4 4cos 1 4cos

4cos 4cos 3 0

2cos 1 2cos 3 0

θ θ

θ θ

θ θθ θ

θ θ

= +

− = +

− = +

+ − =− + =

2cos 1 0 or 2cos 3 0

1 3cos cos

2 25 (not possible)

,3 3

θ θ

θ θ

π πθ

− = + =

= = −

=


,3 3

π π

.

116. 2 28 12sin 4cosθ θ− =

( )2 2

2 2

2

2

8 12sin 4 1 sin

8 12sin 4 4sin

4 8sin

4 1sin

8 2

1 2sin

2 23 5 7

, , ,4 4 4 4

θ θ

θ θθ

θ

θ

π π π πθ

− = −

− = −

=

= =

= ± = ±

=

On 0 2θ π≤ < , the solution set is 3 5 7

, , ,4 4 4 4

π π π π

.


775


117. ( )

( )

sin 2 2 cos

2sin cos 2 cos

2sin cos 2 cos 0

cos 2sin 2 0

θ θ

θ θ θθ θ θ

θ θ

=

=

− =

− =

cos 0 or 2sin 2 0

3 2, sin2 2 2

3,

4 4

θ θπ πθ θ

π πθ

= − =

= =

=


, , ,4 2 4 2

π π π π

.

118. ( )( )

( )( )

2

2

1 3 cos cos 2 0

1 3 cos 2cos 1 0

2cos 3 cos 0

cos 2cos 3 0

θ θ

θ θ

θ θ

θ θ

+ + =

+ + − =

+ =

+ =

cos 0 or 2cos 3 0

3 3, cos2 2 2

5 7= ,

6 6

θ θπ πθ θ

π πθ

= + =

= = −

On 0 2θ π≤ < , the solution set is 5 7 3

, , ,2 6 6 2

π π π π

.

119. sin cos 1θ θ− =

Divide each side by 2 : 1 1 1

sin cos2 2 2

θ θ− =


where 1

cos2

φ = , 1

sin2

φ = , and 4

φ π= :

1sin cos cos sin

2

2sin( )

2

θ φ θ φ

θ φ

− =

− =

3 or

4 43

or 4 4 4 4

or 2

θ φ θ φ

θ θ

θ θ

π π− = − =

π π π π− = − =

π= = π

On 0 2θ π≤ < , the solution set is ,2

π π

.

120. sin 3 cos 2θ θ− =

Divide each side by 2 :

1 3sin cos 1

2 2θ θ− =

Rewrite using the difference of two angles

formula where 3

φ π= , so 1

cos2

φ = and

3sin

2φ = .

( )sin cos cos sin 1

sin 1θ φ θ φ

θ φ− =

− =

2

3 25

6

θ φ

θ

θ

π− =

π π− =

π=


6

π

.

121. ( )1sin 0.7 0.78− ≈

122. 1 4cos 0.64

5− ≈

123. ( )1tan 2 1.11− − ≈ −

124. ( )1cos 0.2 1.77− − ≈


776


125. ( )1 1 1sec 3 cos

3− − =


equals 1

3. Now

1cos


quadrant I. The calculator yields 1 1cos 1.23

3− ≈ ,

which is an angle in quadrant I, so

( )1sec 3 1.23− ≈ .

126. ( )1 1 1cot 4 tan

4− − − = −


equals 1

4− . Now

1tan



1 1tan 0.24

4− − ≈ −


quadrant IV. Since θ lies in quadrant II, 0.24 2.90θ π≈ − + ≈ . Therefore,

( )1cot 4 2.90− − ≈ .

127. 2 5cosx x= Find the intersection of 1 2Y x= and

2 5cosY x= :

−6

6

−2π 2π

1.11x ≈

The solution set is { }1.11 .

128. 2 5sinx x= Find the intersection of 1 2Y x= and 2 5sinY x= :

−6

6

−2π 2π

−6

6

−2π 2π

On the interval 0 2x π≤ ≤ , 0x ≈ or 2.13x ≈ .

The solution set is { }0,2.13 .

129. 2sin 3cos 4x x x+ = Find the intersection of 1 2sin 3cosY x x= + and

2 4Y x= :

−6

6

−2π 2π

0.87x ≈ .


130. 3cos sinx x x+ = Find the intersection of 1 3cosY x x= + and

2 sinY x= :

−6

6

−2π 2π

−6

6

−2π 2π

On the interval 0 2x π≤ ≤ , 1.89x ≈ or 3.07x ≈ .

The solution set is { }1.89,3.07 .

131. sin lnx x= Find the intersection of 1 sinY x= and 2 lnY x= :

−2

2

−2π 2π

2.22x ≈



777


132. sin xx e−=

Find the intersection of 1 sinY x= and 2xY e−= :

−6

6

−2π 2π

−6

6

−2π 2π

On the interval 0 2x π≤ ≤ , 0.59x ≈ or 3.10x ≈ .


133. 1

1

3sin

sin3

sin3

3

2

x

x

x

ππ

π

−

−

− =

= −

= −

= −


2

−

.

134. 1 1

1

1

1

2cos 4cos

2cos 0

2cos

cos2

cos 02

x x

x

x

x

x

ππ

ππ

π

− −

−

−

−

+ =− + =

− = −

=

= =

The solution set is {0}.

135. Using a half-angle formula: 30

sin15 sin2

1 cos30

2

31 2 3 2 32

2 4 2

° ° =

− °=

− − −= = =

Note: since 15º lies in quadrant I, we have sin15 0° > .

Using a difference formula:

( )

sin15 sin(45 30 )

sin(45 )cos(30 ) cos(45 )sin(30 )

2 3 2 1

2 2 2 2

6 2 6 2 16 2

4 4 4 4

° = ° − °= ° ° − ° °

= ⋅ − ⋅

−= − = = −

Verifying equality:

( )

( )

( )

( )

2

1 6 26 2

4 4

2 3 2

4

2 3 1

4

2 3 1

4

2 3 2 3 1

16

−− =

⋅ −=

−=

− =

− +=

( )

( )

2 4 2 3

16

2 2 2 3

16

2 3

4

2 3

2

−=

⋅ −=

−=

−=

136. Given the value of cosθ , the most efficient Double-angle Formula to use is

( ) 2cos 2 2cos 1θ θ= − .


778


Chapter 7 Test

1. Let 1 2sec

3θ − =

. We seek the angle θ , such

that 0 θ π≤ ≤ and 2πθ ≠ , whose secant equals

2

3. The only value in the restricted range with

a secant of 2

3 is

6π

. Thus, 1 2sec

63

π− =

.

2. Let 1 2sin

2θ −

= −

. We seek the angle θ , such

that 2 2π πθ− ≤ ≤ , whose sine equals

22

− . The

only value in the restricted range with a sine of

2

2− is

4

π− . Thus, 1 2sin

2 4

π− − = −

.

3. 1 11sin sin

5

π−


( )( ) ( )( )1 1sin sinf f x x x− −= = , but because

11

5

π is not in the interval ,

2 2

π π −

, we cannot

directly use the equation. We need to find an angle θ in the interval

,2 2

π π −

for which 11

sin sin5

π θ= . The angle

11

5

π is in quadrant I. The reference angle of

11

5

π is

5

π and

11sin sin

5 5

π π= . Since 5

π is in

the interval ,2 2

π π −


above and get 1 11sin sin

5 5

π π− =

.

4. 1 7tan tan

3−

follows the form

( )( ) ( )1 1tan tanf f x x x− −= = . Since the

domain of the inverse tangent is all real numbers,

we can directly apply this equation to get

1 7 7tan tan

3 3− =

.

5. ( )1cot csc 10−

Since 1csc 10ry

θ− = = , 2 2π πθ− ≤ ≤ , let

10r = and 1y = . Solve for x:

( )22 2

2

2

1 10

1 10

9

3

x

x

x

x

+ =

+ =

==


Thus, ( )1 3cot csc 10 cot 3

1xy

θ− = = = = .

6. Let 1 3cos

4θ − = −

.

1

1

34

3sec cos sec

41

cos1

3cos cos

41

43

θ

θ

−

−

− =

=

= −

=−

= −

7. ( )1sin 0.382 0.39 radian− ≈

8. 1 1 1sec 1.4 cos 0.78 radian

1.4− − = ≈

Chapter 7 Test

779


9. 1tan 3 1.25 radians− ≈

10. 1 1 1cot 5 tan 0.20 radian

5− − = ≈

11. csc cotsec tan

θ θθ θ

++

( )( )

( )( )

( )( )

( ) ( )

2 2

2 2

csc cot csc cotsec tan csc cot

csc cotsec tan csc cot

1sec tan csc cot

1 sec tansec tansec tan csc cot

sec tan

sec tan csc cot

sec tancsc cot


θ θθ θ θ θ

θ θ θ θθ θθ θθ θ θ θ

θ θθ θ θ θ

θ θθ θ

+ −= ⋅+ −

−=+ −

=+ −

−= ⋅−+ −

−=− −

−=−

12. 2 2

2 2

sinsin tan cos sin cos

cossin coscos cossin cos

cos1

cossec

θθ θ θ θ θθ

θ θθ θθ θ

θ

θθ

+ = ⋅ +

= +

+=

=

=

13. sin cos

tan cotcos sin

θ θθ θ θ θ+ = +

( )( )

2 2

2 2

sin cossin cos sin cos

sin cossin cos

1sin cos

22sin cos

2sin 2

2csc 2

θ θθ θ θ θθ θθ θ

θ θ

θ θ

θθ

= +

+=

=

=

=

=

14. ( )sin

tan tanα β

α β+

+

sin cos cos sin cos cos

1 sin cos cos sin

sin cos cos sinsinsin

cos cos

sin cos cos sinsin cos cos sincos cos cos cos

sin cos cos sinsin cos cos sin

cos cos

cos cos


α β α ββα

α βα β α βα β α βα β α βα β α βα β α β

α β

α β

+= ⋅

+

+=+

+=+

+=+

=

15. ( )( )

( ) ( )( )

( )

2 2

2 3 2

2 3

2 3

3 3

3

sin 3sin 2sin cos 2 cos sin 2

sin cos sin cos 2sin cos

sin cos sin 2sin cos

3sin cos sin

3sin 1 sin sin

3sin 3sin sin3sin 4sin

θθ θ

θ θ θ θθ θ θ θ θ θθ θ θ θ θ

θ θ θθ θ θθ θ θθ θ

= += += − + ⋅

= − += −= − −

= − −= −


780


16.

sin costan cot cos sin


θ θθ θ θ θ

θ θθ θθ θ

−− =+ +

( )

( )

2 2

2 2

2 2

2 2

2

2

sin cossin cos

sin cossin cos

sin cos

sin coscos 2

12cos 1

1 2cos

θ θθ θ

θ θθ θ

θ θθ θ

θ

θ

θ

−

=+

−=+

−=

= − −

= −

17. ( )

( )( )

cos15 cos 45 30cos 45 cos30 sin 45 sin 30

2 3 2 12 2 2 22

3 146 2 1

or 6 24 4

° = ° − °= ° ° + ° °

= ⋅ + ⋅

= +

+= +

18. ( )tan 75 tan 45 30° = ° + °

2

tan 45 tan 301 tan 45 tan 30

31

33

1 13

3 3

3 33 3 3 3

3 3 3 39 6 3 3

3 312 6 3

62 3

° + °=− ° °

+=

− ⋅

+=−+ += ⋅− ++ +=

−+=

= +

19. 11 3sin cos

2 5−

Let 1 3cos

5θ −= . Since 0

2πθ< < (from the

range of 1cos x− ),

135

1 1 cossin

2 2

31 cos cos 15

2 21 55 5

θθ

−

− =

− − = =

= =

20. 1 6tan 2sin

11−

Let 1 6sin

11θ −= . Then

6sin

11θ = and θ lies in

quadrant I. Since 6

sin11

yr

θ = = , let 6y = and

11r = , and solve for x: 2 2 2

2

6 11

85

85

x

x

x

+ ===

6 6 85tan

8585

yx

θ = = =

( ) 2 2

6 852

852 tantan 2

1 tan 6 851

85

12 8512 85 8585

36 85 49185

12 8549

θθθ

= =

− −

= = ⋅−

=

21. 1 12 3cos sin tan

3 2− − +

Let 1 2sin

3α −= and 1 3

tan2

β −= . Then

2sin

3α = and

3tan

2β = , and both α and β

lie in quadrant I. Since 1

1

2sin

3yr

α = = , let

1 2y = and 1 3r = . Solve for 1x : 2 2 21

21

21

1

2 3

4 9

5

5

x

x

x

x

+ =+ =

==

Thus, 1

1

5cos

3xr

α = = .

Chapter 7 Test

781


Since 2

2

3tan

2yx

β = = , let 2 2x = and 2 3y = .

Solve for 2x : 2 2 22

22

22

2

2 3

4 9

13

13

r

r

r

r

+ =+ =

==

Thus, 2

2

3sin

13

yr

β = = .

Therefore, ( )

( )

cos cos cos sin sin

5 2 2 33 313 13

2 5 6

3 13

2 13 5 3

39

α β α β α β+ = −

= ⋅ − ⋅

−=

−=

22. Let 75α = ° , 15β = ° .

Since ( ) ( )1sin cos sin sin

2α β α β α β= + + − ,

( ) ( )

( )

1sin 5 cos15 sin 90 sin 60

21 3 1 2 3

1 2 32 2 4 4

7 ° ° = ° + ° += + = + =

23.

( ) ( )

sin 75 sin15

75 15 75 152sin cos

2 2

2 3 62sin 45 cos 30 2

2 2 2

° + °° + ° ° − ° =

= ° ° = =

24.

( ) ( )

cos 65 cos 20 sin 65 sin 20

2cos 65 20 cos 45

2

° ° + ° °

= ° − ° = ° =

25. 2

2

2

4sin 3 0

4sin 3

3sin

4

3sin

2

θθ

θ

θ

− =

=

=

= ±

On the interval [ )0,2π , the sine function takes

on a value of 3

2 when

3πθ = or

23πθ = . The

sine takes on a value of 3

2− when

43πθ = and

53πθ = . The solution set is { }2 4 5

, , ,3 3 3 3π π π π

.

26. 3cos tan2

3sin tan

sin0 3sin

cos1

0 sin 3cos

π θ θ

θ θθ θθ

θ θ

− − = − =

= +

= +

sin 0θ = or 1

3 0cos

1cos

3

θ

θ

+ =

= −

On the interval [ )0,2π , the sine function takes on

a value of 0 when 0θ = or θ π= . The cosine

function takes on a value of 13

− in the second and

third quadrants when 1 1cos

3θ π −= − and

1 1cos

3θ π −= + . That is 1.911θ ≈ and 4.373θ ≈ .

The solution set is { }0,1.911, , 4.373π .

27.

( )( ) ( )

( ) ( )( )

2 2

2 2

cos 2sin cos sin 0

cos sin 2sin cos 0

cos 2 sin 2 0

sin 2 cos 2

tan 2 1

θ θ θ θ

θ θ θ θ

θ θθ θθ

+ − =

− + =

+ =

= −

= −

The tangent function takes on the value 1−

when its argument is 34

kπ π+ . Thus, we need


782


( )

32

438 2

3 48

k

k

k

πθ π

π πθ

πθ

= +

= +

= +

On the interval [ )0,2π , the solution set is

{ }3 7 11 15, , ,

8 8 8π π π π

8.

28. ( )sin 1 cos

sin cos1 cos sin1 cos

sin cos1 cos sin1 coscos cos

tan cos1 sin1 1

tan cos1 1 sin1

1 sin1tan

cos1

θ θθ θ θθ θ θ

θ θθθ

θ

+ =+ =+ =

+ == −

−=

Therefore, 1 1 sin1tan 0.285

cos1θ − − = ≈

or

1 1 sin1tan 3.427

cos1θ π − − = + ≈


29. 2

2

4sin 7sin 2

4sin 7sin 2 0

θ θθ θ

+ =

+ − =

Let sinu θ= . Then,

( )( )24 7 2 0

4 1 2 0

u u

u u

+ − =− + =

4 1 0

4 1

14

u

u

u

− ==

=

or 2 0

2

u

u

+ == −

Substituting back in terms of θ , we have 1

sin4

θ = or sin 2θ = −

The second equation has no solution since 1 sin 1θ− ≤ ≤ for all values of θ .

Therefore, we only need to find values of θ

between 0 and 2π such that 1

sin4

θ = . These will

occur in the first and second quadrants. Thus,

1 1sin 0.253

4θ −= ≈ and 1 1

sin 2.8894

θ π −= − ≈ .


Chapter 7 Cumulative Review

1. 23 1 0x x+ − =

( )( )( )

2

2

42

1 1 4 3 1

2 3

1 1 126

1 136

b b acx

a− ± −=

− ± − −=

− ± +=

− ±=

The solution set is 1 13 1 13

,6 6

− − − +

.

2. Line containing points ( 2,5)− and (4, 1)− :

( )2 1

2 1

1 5 61

64 2y y

mx x

− − − −= = = = −− − −

Using 1 1( )y y m x x− = − with point (4, 1)− ,

( )( )

( 1) 1 4

1 1 4

1 4

3 or 3

y x

y x

y x

y x x y

− − = − −

+ = − −+ = − +

= − + + =

Distance between points ( 2,5)− and (4, 1)− :

( ) ( )

( )( ) ( )

( )

2 2

2 1 2 1

2 2

22

4 2 1 5

6 6 72 36 2 6 2

d x x y y= − + −

= − − + − −

= + − = = ⋅ =

Midpoint of segment with endpoints ( 2,5)− and

(4, 1)− :

( ) ( )1 2 1 25 12 4

, , 1,22 2 2 2

x x y y + − + + − + = =

3. 23 9x y+ =

x-intercept: 23 0 9

3 9

3

x

x

x

+ ===

; ( )3,0

y-intercepts: ( ) 2

2

3 0 9

9

3

y

y

y

+ =

== ±

; ( ) ( )0, 3 , 0,3−


783


Tests for symmetry:

x-axis: Replace y with y− : ( )2

2

3 9

3 9

x y

x y

+ − =

+ =

Since we obtain the original equation, the graph is symmetric with respect to the x-axis.

y-axis: Replace x with x− : ( ) 2

2

3 9

3 9

x y

x y

− + =

− + =

Since we do not obtain the original equation, the graph is not symmetric with respect to the y-axis.

Origin: Replace x with x− and y with y− :

( ) ( )2

2

3 9

3 9

x y

x y

− + − =

− + =

Since we do not obtain the original equation, the graph is not symmetric with respect to the origin.

4. 3 2y x= − +

Using the graph of y x= , shift horizontally to

the right 3 units and vertically up 2 units.

5. 3 2xy e= −

Using the graph of xy e= , stretch vertically by a

factor of 3, and shift down 2 units.

6. cos 12

y xπ = − −

Using the graph of cosy x= , horizontally shift

to the right 2π

units, and vertically shift down 1

unit.

7. a. 3y x= y

x−5 5

5

(1, 1)

(−1,−1) (0, 0)

−5

Inverse function: 3y x=

y

x−5 5

5

(1, 1)

(−1,−1)

(0, 0)

−5


784


b. xy e=

y

x−5 5

5

(0, 1)

−5

(1, )e11,

e −

_

Inverse function: lny x=

y

x−5 5

5

(1, 0)

−5

( , 1)e

1, 1

e − _

c. siny x= , 2 2

xπ π− ≤ ≤

y

x−2 2

2

(0, 0)

−2, 1

2π − −

_

, 12π _

Inverse function: 1siny x−=

y

x−2 2

2

(0, 0)

−21,

2π − −

_

1,2π

_

d. cosy x= , 0 x π≤ ≤

y

x3

3

(0, 1)

−1

, 02π _

(π,−1)−1

Inverse function: 1cosy x−=

y

x3

3

(1, 0)−1

0,2π

_

(−1, π)

−1


785


8. 1 3

sin ,3 2

θ π θ π= − < < , so θ lies in Quadrant III.

a. In Quadrant III, cos 0θ < 2

2 1cos 1 sin 1

3

1 81

9 9

2 23

θ θ = − − = − − −

= − − = −

= −

b.

1sin 3tancos 2 2

3

1 3 1 23 42 2 2 2

θθ θ

−= =

−

= − − = =

c. 1 2 2

sin(2 ) 2sin cos 23 3

4 29

θ θ θ = = − −

=

d. 2 2

2 2

cos(2 ) cos sin

2 2 1 8 1 73 3 9 9 9

θ θ θ= −

= − − = − =

e. Since 32

π θ π< < , we have that 3

2 2 4π θ π< < .

Thus, 12

θ lies in Quadrant II and 1

2sin 0θ >

.

2 21

31 1 cossin

2 2 2

3 2 23 2 23

2 6

θθ

− −

− = =

++= =

f. Since 12

θ lies in Quadrant II, 1

cos 02

θ <

.

2 21

31 1 coscos

2 2 2

3 2 23 2 23

2 6

θθ

+ −

+ = − = −

−−= − = −

9. ( )1cos tan 2−

Let 1tan 2θ −= . Then 2

tan1

yx

θ = = ,

2 2θπ π− ≤ ≤ . Let 1x = and 2y = .

Solve for r: 2 2 2

2 2 2

2

1 2

5

5

r x y

r

r

r

= +

= +

=

=


( )1 1 1 5 5cos tan 2 cos

55 5 5

xr

θ− = = = = ⋅ =

10. 1 1 3

sin , ; cos ,3 2 3 2

α α π β βπ π= < < = − π < <

a. Since 2

α ππ < < , we know that α lies in

Quadrant II and cos 0α < . 2

2

cos 1 sin

1 1 81 1

3 9 9

2 23

α α= − −

= − − = − − = −

= −

b. 32

π β π< < , we know that β lies in

Quadrant III and sin 0β < .

2

2

sin 1 cos

11

3

1 8 2 21

9 9 3

β β= − −

= − − −

= − − = − = −

c. 2 2

2 2

cos(2 ) cos sin

2 2 1 8 1 73 3 9 9 9

α α α= −

= − − = − =

d. cos( ) cos cos sin sin

2 2 1 1 2 23 3 3 3

2 2 2 2 4 29 9 9

α β α β α β+ = −

= − − − −

= + =


786


e. Since 32

π β π< < , we have that 3

2 2 4βπ π< < .

Thus, 2β

lies in Quadrant II and sin 02β > .

11

31 cossin

2 2 2

44 2 2 6 63

2 6 6 36

β β − − − = =

= = = = =

11. 5 4 3 2( ) 2 4 2 2 1f x x x x x x= − − + + −

a. ( )f x has at most 5 real zeros.

Possible rational zeros: 1

1; 1, 2; 1,2

pp q

q= ± = ± ± = ± ±

Using the Bounds on Zeros Theorem:

( )

{ }{ }{ }

5 4 3 2

4 3 2 1 0

( ) 2 0.5 2 0.5

0.5, 2, 1, 1, 0.5

Max 1, 0.5 1 1 2 0.5

Max 1, 5 5

1 Max 0.5 , 1 , 1 , 2 , 0.5

1 2 3

f x x x x x x

a a a a a

= − − + + −

= − = − = = = −

− + + + − + −

= =

+ − − −

= + =

The smaller of the two numbers is 3. Thus, every zero of f must lie between 3− and 3.

Use synthetic division with –1:

1 2 1 4 2 2 1

2 3 1 3 1

2 3 1 3 1 0

− − − −− −

− − −

Since the remainder is 0, ( )1 1x x− − = + is

a factor. The other factor is the quotient: 4 3 22 3 3 1x x x x− − + − .

Use synthetic division with 1 on the quotient:

1 2 3 1 3 1

2 1 2 1

2 1 2 1 0

− − −− −

− −

Since the remainder is 0, 1x − is a factor. The other factor is the quotient:

3 22 2 1x x x− − + .

Factoring:

( ) ( )( )( )( )( ) ( )

3 2 2

2

2 2 1 2 1 1 2 1

2 1 1

2 1 1 1

x x x x x x

x x

x x x

− − + = − − −

= − −

= − − +

Therefore,

( ) ( ) ( ) ( )

( ) ( )

2 2

2 2

2 1 1 1

12 1 1

2

f x x x x

x x x

= − − +

= − − +

The real zeros are 1− and 1 (both with

multiplicity 2) and 12

(multiplicity 1).

b. x-intercepts: 1, 1

2, 1−

y-intercept: 1−

The intercepts are (0, 1)− , (1,0) , 1

,02

,

and ( 1,0)−

c. f resembles the graph of 52y x= for large

x .

d. Let 5 4 3 21 2 4 2 2 1Y x x x x x= − − + + −

e. Four turning points exist. Use the

MAXIMUM and MINIMUM features to locate local maxima at ( )1,0− , ( )0.69,0.10

and local minima at ( )1,0 , ( )0.29, 1.33− − .

f. To graph by hand, we determine some additional

information about the intervals between the x-intercepts:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

Interval , 1 1,0.5 0.5,1 1,Test 2 0 0.7 2numberValue 45 1 0.1 27of

Below Below Above AboveLocation -axis -axis -axis -axisPoint 2, 45 0, 1 0.7,0.1 2, 27

f

x x x x

−∞ − − ∞

−

− − ≈

− − −

Chapter 7 Projects

787


f is above the x-axis for ( )0.5,1 and

( )1,∞ , and below the x-axis for ( ), 1−∞ −

and ( )1,0.5− .

g. f is increasing on ( ), 1−∞ − , ( )0.29,0.69− ,

and ( )1,∞ . f is decreasing on ( )1, 0.29− −

and ( )0.69,1 .

12. 2( ) 2 3 1f x x x= + + ; 2( ) 3 2g x x x= + +

a. 2

( ) 0

2 3 1 0

(2 1)( 1) 0

f x

x x

x x

=

+ + =+ + =

1

or 12

x x= − = −

The solution set is { }11,

2− − .

b. 2 2

2

( ) ( )

2 3 1 3 2

1 0

( 1)( 1) 0

f x g x

x x x x

x

x x

=

+ + = + +

− =+ − =

1 or 1x x= − =

The solution set is { }1, 1− .

c. 2

( ) 0

2 3 1 0

(2 1)( 1) 0

f x

x x

x x

>

+ + >+ + >

( ) ( )( ) 2 1 1f x x x= + +

The zeros of f are 1

and 12

x x= − = −

( ) 1 1Interval , 1 1, ,

2 2

Test number 2 0.75 0

Value of 3 0.125 1

Conclusion Positive Negative Positive

f

−∞ − − − − ∞

− −−

The solution set is ( ) 1, 1 ,

2 −∞ − ∪ − ∞

.

d. 2 2

2

( ) ( )

2 3 1 3 2

1 0

( 1)( 1) 0

f x g x

x x x x

x

x x

≥

+ + ≥ + +

− ≥+ − ≥

( ) ( )( )1 1p x x x= − +

The zeros of p are 1 and 1x x= − = .

( ) ( ) ( )Interval , 1 1,1 1,

Test number 2 0 2

Value of 3 1 3

Conclusion Positive Negative Positive

p

−∞ − − ∞−

−

The solution set is ( ] [ ), 1 1,−∞ − ∪ ∞ .

Chapter 7 Projects

Project I – Internet Based Project

Project II

a. Amplitude = 0.00421 m

b. 2.68ω = radians/sec

c. 2.68

0.42652 2

fωπ π= = ≈ vibrations/sec

d. 2 2

0.0919968.3k

π πλ = = ≈ m

e. If 1x = , the resulting equation is 0.00421sin(68.3 2.68 )y t= − . To graph, let

1 0.00421sin(68.3 2.68 )Y x= − .

−0.005

0.005

−4 4


788


f. Note: ( ) ( ) 2 2kx t kx t kx tω ω φ ω φ− + − + = − + and

( ) ( )kx t kx tω ω φ φ− − − + = − .

1 2 sin( ) sin( )

[sin( ) sin( )]

2 22sin cos

2 2

2 22 sin cos

2 2

m m

m

m

m

y y y kx t y kx t

y kx t kx t

kx wty

kx wty

ω ω φω ω φ

φ φ

φ φ

+ = − + − += − + − +

− + − = − + =

g. 0.0045my = , 2.5φ = , 0.09λ = , 2.3f =

Let 1x = : 2

0.09

20069.8

9

k

k

πλ

π

= =

= ≈

2.32

4.6 14.45

fωπ

ω π

= =

= ≈

1 sin( )

0.0045sin(69.8 1 14.45 )

0.0045sin(69.8 14.45 )

my y kx t

t

t

ω= −= ⋅ −= −

2 sin( )

0.0045sin(69.8 1 14.45 2.5)

0.0045sin(72.3 14.45 )

my y kx t

t

t

ω φ= − += ⋅ − += −

1 2

2 22 sin cos

2 2m

kx ty y y

ω φ φ− + + =

2 69.8 1 2 14.45 2.5 2.52 0.0045sin cos

2 2t⋅ ⋅ − ⋅ + = ⋅

( )

142.1 28.90.009sin cos(1.25)

2

0.009sin 71.05 14.45 cos(1.25)

t

t

− =

= −

h. Let 1 0.0045sin(69.8 14.45 )Y x= − ,

2 0.0045sin(72.3 14.45 )Y x= − , and

( )3 0.009sin 71.05 14.45 cos(1.25)Y x= − .

−0.01

0.01

−0.4 0.41 2y y+1y

2y

i. 0.0045my = , 0.4φ = , 0.09λ = , 2.3f =

Let 1x = :

20.09

20069.8

9

k

k

πλ

π

= =

= =

2.32

4.6 14.45

fωπ

ω π

= =

= =

1 0.0045sin(69.8 14.45 )y t= −

2 sin( )

0.0045sin(69.8 1 14.45 0.4)

0.0045sin(70.2 14.45 )

my y kx t

t

t

ω φ= − += ⋅ − += −

1 2

2 22 sin cos

2 2m

kx ty y y

ω φ φ− + + =

2 69.8 1 2 14.45 0.4 0.42 0.0045sin cos

2 2t⋅ ⋅ − ⋅ + = ⋅

( )

140 28.90.009sin cos(0.2)

2

0.009sin 70 14.45 cos(0.2)

t

t

− =

= −

Let 1 0.0045sin(69.8 14.45 )Y x= − ,

2 0.0045sin(70.2 14.45 )Y x= − , and

( )3 0.009sin 70 14.45 cos(0.2)Y x= − .

−0.4 0.4

−0.01

0.01 1 2y y+1y

2y

j. The phase shift causes the amplitude of 1 2y y+

to increase from 0.009cos(1.25) 0.003≈ to

0.009cos(0.2) 0.009≈ .

Project III

a. y

x5ππ 2π 3π 4π

−h

h

b. Let 1

sin(3 ) sin(5 ) sin(7 )4 sin1

1 3 5 7x x xx

Y π = + + + +

−3

3

0 5π

Chapter 7 Projects

789


c. Let 1

sin(3 ) sin(17 )4 sin1 ...

1 3 17x xx

Y π = + + + +

−3

3

0 5π

d. Let 1

sin(3 ) sin(37 )4 sin1 ...

1 3 37x xx

Y π = + + + +

−3

3

0 5π

e. The best one is the one with the most terms.

Project IV

a. ( ) sinf x x= (see table column 2)

( ) ( ) ( ) ( ) ( )0 0 0.954 0.311 0.749 6.085

10.791 0.703 2.437 4.011

6 2

20.607 1.341 1.387 3.052

4 23

0.256 0.978 0.588 1.2433 2

1 0.256 0.670 0.063 0.4132

2 30.607 0.703 0.153 8.507

3 23 2

0.791 0.623 2.380 6.4 2

x f x g x h x k x m x

π

π

π

π

π

π

− −

−

− −

− −

− − −

− −

− − − 822

5 10.954 0 0.594 2.695

6 20 0.954 0.311 0.817 1.536

7 10.791 0.117 0.013 5.248

6 2

5 20.607 1.341 1.387 3.052

4 24 3

0.256 0.978 0.588 1.2433 2

31 0.256 0.670 0.063 0.705

25 3

0.607 0.703 0.3063 2

7 20.791 0.623

4 2

π

ππ

π

π

π

π

π

− −

− −

− − − − −

− − −

− − −

− −

− −

−

11 10.954

6 22 1

π

π

−

b. ( ) ( )1

1

( ) i i

i i

f x f xg x

x x+

+

−=

− (see table column 3)

c.

−1.2

1.2

−0.5 6.5

The shape looks like a sinusoidal graph.


790


Rounding a, b, c, and d to the nearest tenth, we have that sin( 1.8)y x= + .

Barring error due to rounding and approximation, this looks like cosy x=

d. ( ) ( )1

1

( ) i i

i i

g x g xh x

x x+

+

−=


−1.8

1.8

−0.5 6.5

The shape is sinusoidal. It looks like an upside-

down sine wave.

Rounding a, b, c, and d to the nearest tenth, we

have that 0.5sin(6.4 )y x= .

e. ( ) ( )1

1

( ) i i

i i

h x h xk x

x x+

+

−=


−2

3

−0.5 6

This curve is losing its sinusoidal features, although it still looks like one. It takes on the features of an upside-down cosine curve

. Rounding a, b, c, and d to the nearest tenth, we have that 0.8sin(1.1 ) 0.3y x= + .

Note: The rounding error is getting greater and greater.

f. ( ) ( )1

1

( ) i i

i i

k x k xm x

x x+

+

−=


−10

12

−0.5 5.5

The sinusoidal features are gone.

Rounding a, b, c, and d to the nearest tenth, we have that 2.1sin(5.1 1.5) 0.6y x= − + .

g. It would seem that the curves would be less “involved”, but the rounding error has become incredibly great that the points are nowhere near accurate at this point in calculating the differences.

Chapter 7 Analytic Trigonometry - bsdweb.bsdvt.orgshubbard/HPCanswers/PR_9e_ISM_07all.pdf · tan...

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