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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− −
We are finding the angle θ , 2 2
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−
We are finding the angle θ , 2 2
θπ π− < < , whose
tangent equals 0.
1
tan 0,2 2
0
tan 0 0
θ θ
θ−
π π= − < <
=
=
18. ( )1tan 1− −
We are finding the angle θ , 2 2
θπ π− < < , whose
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
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19. 1 2sin
2−
We are finding the angle θ , 2 2
θπ π− ≤ ≤ , whose
sine equals 2
2.
1
2sin ,
2 2 2
4
2sin
2 4
θ θ
θ
−
π π= − ≤ ≤
π=
π=
20. 1 3tan
3−
We are finding the angle θ , 2 2
θπ π− < < , whose
tangent equals 3
3.
1
3tan ,
3 2 2
6
3tan
3 6
θ θ
θ
−
π π= − < <
π=
π=
21. 1tan 3−
We are finding the angle θ , 2 2
θπ π− < < , whose
tangent equals 3 .
1
tan 3,2 2
3
tan 33
θ θ
θ
−
π π= − < <
π=
π=
22. 1 3sin
2−
−
We are finding the angle θ , 2 2
θπ π− ≤ ≤ , whose
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−
−
We are finding the angle θ , 2 2
θπ π− ≤ ≤ , whose
sine equals 2
2− .
1
2sin ,
2 2 2
4
2sin
2 4
θ θ
θ
−
π π= − − ≤ ≤
π= −
π− = −
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
π− − follows the form of the
equation ( )( ) ( )( )1 1tan tanf f x x x− −= = . Since
3
8
π− is in the interval ,2 2
π π −
, we can apply
the equation directly and get
1 3 3tan tan
8 8
π π− − = − .
40. 1 3sin sin
7
π− − follows the form of the
equation ( )( ) ( )( )1 1sin sinf f x x x− −= = . Since
3
7
π− is in the interval ,2 2
π π −
, we can apply
the equation directly and get
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− − = − = − .
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions
651
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42. 1 5cos cos
3
π− − follows the form of the
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
in the interval 0,π , we can apply the equation
above and get
1 15cos cos cos cos
3 3 3
π π π− − − = = .
43. 1 4tan tan
5
π−
follows the form of the
equation ( )( ) ( )( )1 1tan tanf f x x x− −= = , but
we cannot use the formula directly since 4
5
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
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
π− − follows the form of the
equation ( )( ) ( )( )1 1tan tanf f x x x− −= = . but we
cannot use the formula directly since 2
3
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
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 −
, we can apply the equation
above and get 1 12tan tan tan tan
3 3 3
π π π− − − = = .
45. 1 1sin sin
4−
follows the form of the equation
( )( ) ( )( )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− −
follows the form of the
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− − = −
.
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
real number, we can apply the equation directly
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
real number, we can apply the equation directly
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= −
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions
653
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( )( )
( )
1
1 1
2cos 3
cos 32
3 cos2
1cos
3 2
x y
xy
xy
xy f x
−
− −
= −
= −
= − = − =
The domain of ( )f x equals the range of
1( )f x− and is 03
xπ≤ ≤ , or 0,
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
2 2
x
x
x
− ≤ − ≤
≥ ≥ −− ≤ ≤
The domain of ( )1f x− is { }| 2 2x x− ≤ ≤ , or
2,2− 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
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
−
− −
=
=
=
= =
The domain of ( )f x equals the range of
1( )f x− and is 4 4
xπ π− ≤ ≤ , or ,
4 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
− ≤ ≤
− ≤ ≤
The domain of ( )1f x− is { }| 3 3x x− ≤ ≤ , or
3,3− 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
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).
The domain of ( )f x equals the range of
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
−
−
= + +
+ = −
+ = −
= − −
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
The domain of ( )f x equals the range of
1( )f x− and is 2 2x π− ≤ ≤ − , or 2, 2π− − in
interval notation. To find the domain of ( )1f x−
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
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
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
−
−
− −
= +
+ =
+ =
= − = − =
The domain of ( )f x equals the range of
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
− ≤ ≤
− ≤ ≤
The domain of ( )1f x− is { }| 3 3x x− ≤ ≤ , or
3,3− 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
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
−
−
− −
= +
+ =
+ = = − = − =
The domain of ( )f x equals the range of
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
− ≤ ≤
− ≤ ≤
The domain of ( )1f x− is { }| 2 2x x− ≤ ≤ , or
2,2− 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
2,2− .
61. 1
1
4sin
sin4
2sin
4 2
x
x
x
ππ
π
−
−
=
=
= =
The solution set is 2
2
.
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions
655
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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
−
−
=
=
=
= −
= −
The solution set is 1
4 −
.
64. ( )
( )
1
1
6sin 3
sin 36
3 sin6
13
21
6
x
x
x
x
x
ππ
π
−
−
− =
= −
= −
= −
= −
The solution set is 1
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
− −
−
−
− = −
= −
= −
= − = −
The solution set is 3
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
13.85 hours or 13 hours, 51 minutes
Dπ π− ⋅ ⋅
= ⋅ −π
≈
70. Note that 40 45 40.75θ ′= ° = ° .
a. ( ) ( )( )1
180 180cos tan 23.5 tan 40.7524 1
14.93 hours or 14 hours, 56 minutes
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
14.83 hours or 14 hours, 50 minutes
Dπ π− ⋅ ⋅
= ⋅ − π
≈
Chapter 7: Analytic Trigonometry
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71. Note that 21 18 21.3θ ′= ° = ° .
a. ( ) ( )( )1
180 180cos tan 23.5 tan 21.324 1
13.30 hours or 13 hours, 18 minutes
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
13.26 hours or 13 hours, 15 minutes
Dπ π− ⋅ ⋅
= ⋅ − π
≈
72. Note that 61 10 61.167θ ′= ° ≈ ° .
a. ( ) ( )( )1
180 180cos tan 23.5 tan 61.16724 1
18.96 hours or 18 hours, 57 minutes
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
18.64 hours or 18 hours, 38 minutes
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
Section 7.1: The Inverse Sine, Cosine, and Tangent Functions
657
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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 − − = − ≈ °
If you sit 15 feet from the screen, then the viewing angle is about 44.4° .
( ) 1 134 620 tan tan 42.8
20 20θ − − = − ≈ °
If you sit 20 feet from the screen, then the viewing angle is about 42.8° .
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
− − − − − = −
= −
=
Chapter 7: Analytic Trigonometry
658
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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−
−
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
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
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
12. 1 1tan sin
2− −
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
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−
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
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− −
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
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−
Find the angle , ,2 2
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−
Find the angle , ,2 2
θ θπ π− < < whose tangent
equals 3 .
( )1
tan 3,2 2
3
sec tan 3 sec 23
θ θ
θ
−
π π= − < <
π=
π= =
17. 1sin tan ( 1)− −
Find the angle , ,2 2
θ θπ π− < < whose tangent
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−
−
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
equals 3
2− .
1
3sin ,
2 2 2
3
3 1cos sin cos
2 3 2
θ θ
θ
−
π π= − − ≤ ≤
π= −
π − = − =
Chapter 7: Analytic Trigonometry
660
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
19. 1 1sec sin
2− −
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
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−
−
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
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 = −
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
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 = −
Find the angle , ,2 2
θ θπ π− < < whose tangent
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 − = −
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
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 − = −
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
equals 1− .
1
cos 1, 0
3
cos tan3
θ θ
θ
π−
= − ≤ ≤ ππ=
π − =
Section 7.2: The Inverse Trigonometric Functions [Continued]
661
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
2 2θπ π− ≤ ≤ , θ is in quadrant I, and we let
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− − = = = = − −
Chapter 7: Analytic Trigonometry
662
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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 2θπ π− ≤ ≤ , θ is in quadrant I, and we let
2 5y = and 5r = .
Solve for x:
2
2
20 25
5
5
x
x
x
+ =
=
= ±
Since θ is in quadrant I, 5x = .
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
+ =
=
=
θ is in quadrant I.
1 1 5csc tan csc 5
2 1
r
yθ− = = = =
35. 1 13 2sin cos sin
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=
Section 7.2: The Inverse Trigonometric Functions [Continued]
663
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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−
We are finding the angle ,2 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−
−
We are finding the angle ,2 2
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− ≈ .
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
2θ = , so θ lies in
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− − − = −
We seek the angle , 0 ,θ θ π≤ ≤ whose cosine
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− − ≈ −
, which is an angle in
quadrant IV, so ( )1csc 3 0.34− − ≈ − .
50. 1 11cot tan ( 2)
2− − − = −
We seek the angle , 0 ,θ θ π≤ ≤ whose tangent
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− − − = −
We seek the angle , 0 ,θ θ π≤ ≤ whose tangent
equals 1
5− . Now
1tan
5θ = − , so θ lies in
quadrant II. The calculator yields
1 1tan 0.42
5− − ≈ −
, which is an angle in
quadrant IV. Since θ is in quadrant II, 0.42 2.72θ π≈ − + ≈ . Therefore,
( )1cot 5 2.72− − ≈ .
Section 7.2: The Inverse Trigonometric Functions [Continued]
665
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
52. ( )1 1 1cot 8.1 tan
8.1− − − = −
We seek the angle , 0 ,θ θ π≤ ≤ whose tangent
equals 1
8.1− . Now
1tan
8.1θ = − , so θ lies in
quadrant II. The calculator yields
1 1tan 0.12
8.1− − ≈ −
, which is an angle in
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− − ≈ −
, which is an angle in
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− − ≈
, which is an angle in
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− − ≈ −
, which is an angle in
quadrant IV. Since θ is in quadrant II,
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− − ≈ −
, which is an angle in
quadrant IV. Since θ is in quadrant II,
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
− = =
= − = −
Chapter 7: Analytic Trigonometry
666
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
− = = ⋅ =
−= = =
−=
64. Let 1sec uθ −= so that sec uθ = , 0 θ π≤ ≤ and
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− = = =
66. Let 1sec uθ −= so that sec uθ = , 0 θ π≤ ≤ and
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
+ =
+ =
=
= ± = ±
Since θ is in quadrant I, 5x = .
1 112 12 5cos sin cos
13 13 13
xg f
rθ− − = = = =
Section 7.2: The Inverse Trigonometric Functions [Continued]
667
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
+ − =
+ =
=
= ± = ±
Since θ is in quadrant IV, 4x = .
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
− + =
+ =
=
= ± = ±
Since θ is in quadrant II, 3y = .
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
π πθ− ≤ ≤ , θ is in quadrant I, and we let
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
π πθ− ≤ ≤ , θ is in quadrant I, and we let
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− − = = = =
Chapter 7: Analytic Trigonometry
668
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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
− + =
+ =
=
= ±
Since θ is in quadrant II, 15y = .
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
+ − =
+ =
=
= ±
Since θ is in quadrant IV, 21x = .
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
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
− = − +=
= =
The solution set is 3
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
− = + =
= = −
The solution set is 5
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
− +
.
Chapter 7: Analytic Trigonometry
670
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
= − =
= =
The solution set is 5
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
=
= ± = ±
5or , is any integer
6 6k k kθ θπ π= + π = + π
On the interval 0 2θ≤ < π , the solution set is 5 7 11
, , ,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
− =
=
=
= ± = ±
3or , is any integer
4 4k k k
π πθ π θ π= + = +
On the interval 0 2θ≤ < π , the solution set is 3 5 7
, , ,4 4 4 4
π π π π
.
Section 7.3: Trigonometric Equations
671
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
16. 2
2
2
4cos 3 0
4cos 3
3cos
4
3cos
2
θθ
θ
θ
− =
=
=
= ±
5or , is any integer
6 6k k kθ θπ π= + π = + π
On the interval 0 2θ≤ < π , the solution set is 5 7 11
, , ,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
=
= +
= +
On 0 2θ≤ < π , the solution set is 2
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
On the interval 0 2θ≤ < π , the solution set is 2 4 5
, , ,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
= +
= +
On 0 2θ≤ < π , the solution set is 5
4
π
.
23. 2sin 1 0
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
.
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
On 0 2θ≤ < π , the solution set is 3 7
,4 4
π π
.
Chapter 7: Analytic Trigonometry
672
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
26. 3 cot 1 0
3 cot 1
1 3cot
33
θθ
θ
+ =
= −
= − = −
2, is any integer
3k kθ π= + π
On 0 2θ≤ < π , the solution set is 2 5
,3 3
π π
.
27. 4sec 6 2
4sec 8
sec 2
θθθ
+ = −= −= −
2 42 or 2 , is any integer
3 3k k kθ θπ π= + π = + π
On 0 2θ≤ < π , the solution set is 2 4
,3 3
π π
.
28. 5csc 3 2
5csc 5
csc 1
θθθ
− ===
2 , is any integer2
k kθ π= + π
On 0 2θ≤ < π , the solution set is 2
π
.
29. 3 2 cos 2 1
3 2 cos 3
1 2cos
22
θθ
θ
+ = −
= −
= − = −
3 52 or 2 , is any integer
4 4k k kθ θπ π= + π = + π
On 0 2θ≤ < π , the solution set is 3 5
,4 4
π π
.
30. 4sin 3 3 3
4sin 2 3
2 3 3sin
4 2
θθ
θ
+ =
= −
= − = −
4 52 or 2 , is any integer
3 3k k kθ θπ π= + π = + π
On 0 2θ≤ < π , the solution set is 4 5
,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
On 0 2θ≤ < π , the solution set is 3 7
,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= +
On the interval 0 2θ≤ < π , the solution set is 4 22 40
, ,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
On 0 2θ≤ < π , the solution set is 11
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.
On 0 2θ≤ < π , the solution set is 7
4
π
.
Section 7.3: Trigonometric Equations
673
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
θ π π π π π π= .
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
θ π π π π π π= .
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
θ π π π π π π= .
39. cos 0θ =
32 or = 2
2 2k kθ θ θ π π = + π + π
, k is any
integer
Six solutions are 3 5 7 9 11
, , , , ,2 2 2 2 2 2
θ π π π π π π= .
40. 2
sin2
θ =
32 or 2
4 4k k
πθ θ θ π π = + π = +
, k is any
integer
Six solutions are 3 9 11 17 19
, , , , ,4 4 4 4 4 4
θ π π π π π π= .
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
Six solutions are 2 4 5 7 8
, , , , ,3 3 3 3 3 3
θ π π π π π π= .
42. ( )sin 2 1θ = −
32 2 , is any integer
2k kθ π= + π
3, is any integer
4k kθ θ π = + π
Six solutions are 3 7 11 15 19 23
, , , , ,4 4 4 4 4 4
θ π π π π π π= .
43. 3
sin2 2
θ = −
4 52 or 2 , is any integer
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
θ π π π π π π= .
44. tan 12
θ = −
3, is any integer
2 4k k
θ π π= +
32 , is any integer
2k k
πθ θ π = +
Six solutions are 3 7 11 15 19 23
, , , , ,2 2 2 2 2 2
θ π π π π π π= .
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 .
Chapter 7: Analytic Trigonometry
674
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
46.
( )1
cos 0.6
cos 0.6 0.93
θθ −
=
= ≈
0.93θ ≈ or 2 0.93 5.36θ π≈ − ≈ .
The solution set is { }0.93, 5.36 .
47.
( )1
tan 5
tan 5 1.37
θθ −
=
= ≈
1.37θ ≈ or 1.37 4.51θ π≈ + ≈ .
The solution set is { }1.37, 4.51 .
48.
1
cot 2
1tan
21
tan 0.462
θ
θ
θ −
=
=
= ≈
0.46θ ≈ or 0.46 3.61θ π≈ + ≈ .
The solution set is { }0.46, 3.61 .
49.
( )1
cos 0.9
cos 0.9 2.69
θθ −
= −
= − ≈
2.69θ ≈ or 2 2.69 3.59θ π≈ − ≈ .
The solution set is { }2.69, 3.59 .
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≈ − −≈
.
The solution set is { }3.34, 6.08 .
51.
1
sec 4
1cos
41
cos 1.824
θ
θ
θ −
= −
= −
= − ≈
1.82θ ≈ or 2 1.82 4.46θ π≈ − ≈ .
The solution set is { }1.82, 4.46 .
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≈ − −≈
.
The solution set is { }3.48, 5.94 .
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≈ − +≈
The solution set is { }2.08, 5.22 .
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≈ − +≈
.
The solution set is { }2.47, 5.61 .
55.
1
3sin 2 0
3sin 2
2sin
32
sin 0.733
θθ
θ
θ −
− ==
=
= ≈
0.73θ ≈ or 0.73 2.41θ π≈ − ≈ .
The solution set is { }0.73, 2.41 .
56.
1
4cos 3 0
4cos 3
3cos
43
cos 2.424
θθ
θ
θ −
+ == −
= −
= − ≈
2.42θ ≈ or 2 2.42 3.86θ π≈ − ≈ .
The solution set is { }2.42, 3.86 .
Section 7.3: Trigonometric Equations
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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= =
The solution set is 3
, 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=
The solution set is 5
, , 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 ==
The solution set is 5
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=
The solution set is 3 7
, 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
= + =
= = −
=
The solution set is 2 4 3
, , , 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
+ = − === −=
=
The solution set is 7 11
, , 2 6 6
π π π
.
Chapter 7: Analytic Trigonometry
676
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
− = − ====
=
The solution set is 5
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=
The solution set is 5
, 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=
The solution set is 3 7
, 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=
The solution set is 5
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=
The solution set is 3 5 7
, , , 4 4 4 4
π π π π
.
Section 7.3: Trigonometric Equations
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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=
The solution set is 5 3
, , 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= =
The solution set is 2
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=
The solution set is 2 4
, 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
+ = + == − = −
=
The solution set is 3
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=
The solution set is 5
, 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= =
The solution set is 5 3
, , , 4 2 4 2
π π π π
.
Chapter 7: Analytic Trigonometry
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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 ≈
Section 7.3: Trigonometric Equations
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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
Chapter 7: Analytic Trigonometry
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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
Section 7.3: Trigonometric Equations
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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 =
Chapter 7: Analytic Trigonometry
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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
.
c. From the graph in part (a) and the results of part (b), the solution of ( ) ( )f x g x> on
[ ]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
.
c. From the graph in part (a) and the results of part (b), the solution of ( ) ( )f x g x> on
[ ]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≤ ≤ .
Section 7.3: Trigonometric Equations
683
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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 = − + =
on the interval [ ]0,80 .
125sin 0.157 125 2502
125sin 0.157 1252
sin 0.157 12
t
t
t
π
π
π
− + =
− = − =
Chapter 7: Analytic Trigonometry
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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 = − + >
on the interval [ ]0,40 .
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
Section 7.3: Trigonometric Equations
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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
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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.
Section 7.3: Trigonometric Equations
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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° .
Chapter 7: Analytic Trigonometry
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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
θ θ 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
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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
+ = ⋅
= ⋅
=
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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
θ θ θ θθ θθ 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
+ = −
+=−
Section 7.4: Trigonometric Identities
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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
= ⋅− −
=−
=−
Chapter 7: Analytic Trigonometry
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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 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 θ θ θθ θ θ θ
θ θθ θ θ θθ θ θ θ θ θ
θ θ θ θθ 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
+ +
= + +
= + +
Section 7.4: Trigonometric Identities
693
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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
θ θ θ θθ θ θ 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
θ θ θ θ θθ θ θ
θ θ θ θ θθ θ θ
+ − + − +=− − +
− + − + − +=− − + −
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
θ θ θ θθ θ θ θ
− ++ −
− + + += ⋅+ − + +
2 2
2
sin cos sin cos sin cos 1
(sin cos ) 1
θ θ θ θ θ θθ θ
− + + + − +=+ −
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 θθ θθ 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
−
=+
−=+
=+
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63.
2 2
2 2
2 2
sin costan cot cos sin1 1
sin costan cotcos sin
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 costan cotcos sin
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
− = −
= −
= −
Section 7.4: Trigonometric Identities
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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
θ θ θ θθ θ θ θ
θ θ θ θθ θ θ θ
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
+=−+=
+ =
= +
= +
Chapter 7: Analytic Trigonometry
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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 θ θ θθ θ θ 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 sin cos cos )
sin cos
sin cos sin cos
1 sin cos
θ θθ θ
θ θ θ θ θ θθ θ
θ θ θ θθ θ
++
+ − +=+
= + −= −
82. 3 3
2
2 2
2 2
sin cos
1 2cos
(sin cos )(sin sin cos cos )
1 cos cos
θ θθ
θ θ θ θ θ θθ θ
+−
+ − +=− −
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 θ θ θ θθ θ
θ θ θ θθ θ θ θ
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
− −=− −
−=−
= − ⋅−
=
Section 7.4: Trigonometric Identities
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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
θ θ θ θθ θ θ 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 θ θ θθ θ θ
θ θ θ θ θθ θ θ
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
θ θ θ θθ θ θ 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 )
θ θ θ θ θθ θ θ
θ θ θ θ θθ θ θ
+ + + + +=+ + −
+ + + + + −=+ + − −
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
θ θ θ θθ θ θ θ
θ θ θ θθ θ θ θ
+ + −
= + +
+ − +
= + + +
= +
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
=
+ − +
= + − +
= + +
= +
=
=
Chapter 7: Analytic Trigonometry
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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
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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 θ θ θθ θ θ θ
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
Chapter 7: Analytic Trigonometry
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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
sin cos cos sin4 6 4 6
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
− + −=−
− +=−
= −
Section 7.5: Sum and Difference Formulas
701
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
cos cos sin sin4 3 4 3
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
⋅ + ⋅ = −=
=
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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+ = −
= ⋅ − ⋅ −
+=
=
Section 7.5: Sum and Difference Formulas
703
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
+ − = >
= − = >=
−= = = −
a. sin( ) sin cos cos sin
2 5 3 5 4
5 5 5 5
6 5 4 5
25
2 5
25
α β α β α β+ = +
= ⋅ + ⋅ −
−=
=
b. cos( ) cos cos sin sin
5 3 2 5 4
5 5 5 5
3 5 8 5
25
11 5
25
α β α β α β+ = −
= ⋅ − ⋅ −
+=
=
c. sin( ) sin cos cos sin
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
α βα βα β−− =
+
42
34
1 23
103532
− − = + ⋅ −
=−
= −
35. 4
tan ,3 2
1cos , 0
2 2
α α
β β
π= − < < π
π= < <
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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
+ = >
= − = >
=
= = =
a. sin( ) sin cos cos sin
4 1 3 3
5 2 5 2
4 3 3
10
α β α β α β+ = +
= ⋅ + − ⋅
−=
b. cos( ) cos cos sin sin
3 1 4 3
5 2 5 2
3 4 3
10
α β α β α β+ = −
= − ⋅ − ⋅
− −=
c. sin( ) sin cos cos sin
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
α βα βα β−− =
+
− −=
+ − ⋅
− −
=−
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
+ − = <
= − = <
= −
−= − = =−
a. sin( ) sin cos cos sin
5 3 12 1
13 2 13 2
5 3 12 12 5 3
26 26
α β α β α β+ = +
= − ⋅ − + − ⋅ −
+ += =
Section 7.5: Sum and Difference Formulas
705
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
b. cos( ) cos cos sin sin
12 3 5 1
13 2 13 2
12 3 5
26
α β α β α β+ = −
= − ⋅ − − − ⋅ −
−=
c. sin( ) sin cos cos sin
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
α βα βα β−− =
+ ⋅
−−= =
++ ⋅
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
= − + ==
−= = = −
a. sin( ) sin cos cos sin
5 1 12 3
13 2 13 2
5 12 3 5 12 3 or
26 26
α β α β α β+ = +
= ⋅ − + − ⋅
− − += −
b. cos( ) cos cos sin sin
12 1 5 3
13 2 13 2
12 5 3
26
α β α β α β+ = −
= − ⋅ − − ⋅
−=
c. sin( ) sin cos cos sin
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
α βα βα β−− =
+
− − −=
+ − ⋅ −
− +
=+
5 12 3 12 5 3
12 5 3 12 5 3
240 169 3
69
− + −= ⋅ + −
− +=
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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
+ = >
= − = >
= =
= = =
a. sin( ) sin cos cos sin
3 2 2 1 1
2 3 2 3
1 2 6
6
α β α β α β+ = +
= − ⋅ + ⋅
−=
b. cos( ) cos cos sin sin
1 2 2 3 1
2 3 2 3
3 2 2
6
α β α β α β+ = −
= ⋅ − − ⋅
+=
c. sin( ) sin cos cos sin
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
α βα βα β−− =
+
− −=
+ − ⋅
− −
=−
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 − = ⋅ + ⋅
= − +
− +=
Section 7.5: Sum and Difference Formulas
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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
+ = += ⋅ + ⋅
= + −
−= − =
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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
+ = += ⋅ − ⋅
= − −
+= + =
43. 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
− = −= ⋅ + ⋅
= + −
−= − =
44. From the solution to Problem 41, we have
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
− = −= ⋅ − ⋅
= − −
+= + =
45. From the solution to Problem 41, we have
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++ = + =
−
+ −=
− −
−= ⋅
+
− −= ⋅+ −
− − +=− + −− −= =
−
46. From the solution to Problem 45, we have
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−− = − =
+
− −=
+ −
+= ⋅
−
+ += ⋅− +
+ + +=+ − −+ += = −
−
Section 7.5: Sum and Difference Formulas
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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 α β
α β α βα β
+ + −= +
+ −=
58. cos( ) cos( )
cos cos sin sin
cos cos sin sin
2cos cos
α β α βα β α β
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 α βα β α β
α β
+ +=
= +
= +
60. sin( ) sin cos cos sin
cos cos cos cos
sin cos cos sin
cos cos cos cos
tan tan
α β α β α βα β α β
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 α β α βα β α β
α β α βα β α β
α β
+ −=
= −
= −
62. cos( ) cos cos sin sin
sin cos sin cos
cos cos sin sin
sin cos sin cos
cot tan
α β α β α βα β α β
α β α βα β α βα β
− +=
= +
= +
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63. sin( ) sin cos cos sin
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
+
=−
+=
−
+=−
64. cos( ) cos cos sin sin
cos( ) cos cos sin sin
α β α β α βα β α β α β
+ −=− +
cos cos sin sincos cos
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
−− =−
+=−
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
+
=−
+=
−
+=−
67. 1
sec( )cos( )
1
cos cos sin sin
α βα β
α β α β
+ =+
=−
1sin sin
cos cos sin sinsin 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
cos cos sin sincos 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
=+
⋅=
+
=+
Section 7.5: Sum and Difference Formulas
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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
− += + −
= −
= − − −
= − − +
= −
71. sin( ) sin cos cos sin
(sin )( 1) (cos )(0)
( 1) sin , any integer
k
k
k k k
k
θ θ θθ θ
θ
+ π = ⋅ π + ⋅ π
= − +
= −
72. cos( ) cos cos sin sin
(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 =
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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
Section 7.5: Sum and Difference Formulas
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
quadrant I; β is in quadrant I. Then 5
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
quadrant I; β is in quadrant I. Then 4
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
= = ⋅ =
Chapter 7: Analytic Trigonometry
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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
5α −= . α is in quadrant I. Then
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:
Section 7.5: Sum and Difference Formulas
715
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )( )
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
α β α βα β α β
+=−
+ = −
= = −−
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
uα απ π= − < < , and
tan ,2 2
vβ βπ π= − < < .
u−∞ < < ∞ , v−∞ < < ∞
2 2sec tan 1 1uα α= + = +
2
1cos
1uα =
+
Chapter 7: Analytic Trigonometry
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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
uα απ π= − < < , and
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= − = −
Section 7.5: Sum and Difference Formulas
717
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )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= =
The solution set is 7
, 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= =
The solution set is 2
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=
The solution set is 4
π
.
Chapter 7: Analytic Trigonometry
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94. 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 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+ =
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 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+ = −
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
+ = −
+ = −
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+ = .
Section 7.5: Sum and Difference Formulas
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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
+ −
− −=
− + −=
− − −=
−= − ⋅ − ⋅
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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 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( )
sin sin sin sin 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 sin sin sin sin
sin sin sinsin sin sin sin
sin sin sin 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
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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+=
+= = = = =
Chapter 7: Analytic Trigonometry
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8. 3
cos , 05 2
θ θ π= < < . Thus, 02 4
θ π< < , which
means 2
θ lies in quadrant I.
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
−− =
+
=
+=
Section 7.6: Double-angle and Half-angle Formulas
723
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
θ lies in quadrant I.
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
θ lies in quadrant II.
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
−=
−
=
−=
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
θ lies in quadrant I.
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
θ lies in quadrant II.
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
+ − = −
−
= −
−= −
Section 7.6: Double-angle and Half-angle Formulas
725
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
15. cot 2, sec 0θ θ= − < , so 2
θπ < < π . Thus,
4 2 2
θπ π< < , which means 2
θ lies in quadrant I.
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
θ lies in quadrant II.
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
θ lies in quadrant II.
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=
= ⋅ − ⋅
= − = −
Chapter 7: Analytic Trigonometry
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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
+ °=
+ + += = =
Section 7.6: Double-angle and Half-angle Formulas
727
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
− +=
= − +
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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= =
= ⋅ ⋅ − = − = −
30. From the solution to Problem 29, we have
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
−+ =
−
=
−=
−=
Section 7.6: Double-angle and Half-angle Formulas
729
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
=
=−
− − −= = = =− −− −
34. From the solution to Problem 29, we have
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= = −
= − −
= − = − = −
Chapter 7: Analytic Trigonometry
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36. From the solution to Problem 35, we have
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
+ − = −
= − = − = − ⋅ = − = −
39. From the solution to Problem 35, we have
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
=
=−
= = = = −− −−
Section 7.6: Double-angle and Half-angle Formulas
731
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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
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 θ θθ θ θ θ
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
− = + −
= ⋅
=
Chapter 7: Analytic Trigonometry
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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 )(cos sin )
cos sin 2sin cos
θ θ θθ θ θ
θ θ θ θθ θ θ θ
−=+ +
− +=+ +
(cos sin )(cos sin )
(cos sin )(cos sin )
cos sin
cos sincos sin
sincos sin
sincos sinsin sincos sinsin sincot 1
cot 1
θ θ θ θθ θ θ θ
θ θθ θθ θ
θθ θ
θθ θθ θθ θθ θθθ
− +=+ +
−=+−
=+
−=
+
−=+
Section 7.6: Double-angle and Half-angle Formulas
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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 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
−− =
−=
=
=
=
Chapter 7: Analytic Trigonometry
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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
− + − − + + +=
−− + − − + + + + +=
−− +=
−−
=−
=
Section 7.6: Double-angle and Half-angle Formulas
735
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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
=
= ±
=
The solution set is 2 4 5
, , , 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=
The solution set is 2 4
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=
The solution set is 5 3
, , , 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
.
Chapter 7: Analytic Trigonometry
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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=
The solution set is 2 4
, 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
The solution set is 5
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=
The solution set is 7 3 11
, , , 2 6 2 6
π π π π
.
Section 7.6: Double-angle and Half-angle Formulas
737
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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= ≤ ≤ .
Chapter 7: Analytic Trigonometry
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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
5α −= . α is in quadrant I. Then
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= − − ≤ ≤ .
Section 7.6: Double-angle and Half-angle Formulas
739
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
− = + == −= =
=
The zeros on 0 2x π≤ < are 5
, ,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
=+ =
− + ===
=
The zeros on 0 2x π≤ < are 3
,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˚.
Chapter 7: Analytic Trigonometry
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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+ =
Divide each side by 2 : 1 1
sin(2 ) cos(2 ) 02 2
θ θ+ =
Rewrite in the sum of two angles form using
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.
Section 7.6: Double-angle and Half-angle Formulas
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
= − ⋅ − ⋅
= − ⋅ +
= − ⋅ − +
= − ⋅
= −
Chapter 7: Analytic Trigonometry
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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.
Section 7.6: Double-angle and Half-angle Formulas
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
= ⋅ − ⋅ = ⋅ − − = ⋅ − +
= ⋅ − = − ⋅
Chapter 7: Analytic Trigonometry
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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
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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 )
sin(135 )cos(120 ) cos(135 )sin(120 ) sin(135 )cos(120 ) cos(135 )sin(120 )
sin(135 )cos(120 ) cos(135 )sin(120 ) sin(135 )cos(120 ) cos(135 )sin(120 )
° − ° = ° + ° − ° − °= ° ° + ° ° − ° ° − ° °= ° ° + ° ° − ° ° + ° °
2cos(135 )sin(120 )
2 3 62
2 2 2
= ° °
= − = −
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
+ − + =
= −
=
Section 7.7: Product-to-Sum and Sum-to-Product Formulas
747
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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
+ − − − =+ −+
− −=−
− −=
=
Chapter 7: Analytic Trigonometry
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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
+ − + =
− +−
+ − = ⋅
+ −
+ − =
Section 7.7: Product-to-Sum and Sum-to-Product Formulas
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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
= − − −
= −= −
− + = ==
Chapter 7: Analytic Trigonometry
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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
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
.
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
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
.
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.
Section 7.7: Product-to-Sum and Sum-to-Product Formulas
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
= + +
− + = + +
= − + + +
+ −= − +
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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
−
= − −
==
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753
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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 )
cos cos cos cos cos cos
sin cos ( ) cos cos sin cos( ) cos cos
cos cos cos 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 cos cos sin
sin( ) sin cos cos sin
sin( ) sin( ) 2sin cos
1sin cos sin( ) sin( )
2
α β α β α βα β α β α β
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 α β α β
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 α β α β
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 + − − = −
.
Chapter 7: Analytic Trigonometry
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Chapter 7 Review Exercises
1. 1sin 1−
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
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−
Find the angle , ,2 2
θ θπ π− < < whose tangent
equals 1.
tan 1,2 2
4
θ θ
θ
π π= − < <
π=
Thus, ( )1tan 14
− π= .
4. 1 1sin
2− −
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
equals 1
2− .
1sin ,
2 2 2
6
θ θ
θ
π π= − − ≤ ≤
π= −
Thus, 1 1sin
2 6− π − = −
.
5. 1 3cos
2−
−
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
equals 3
2− .
3cos , 0
25
6
θ θ
θ
= − ≤ ≤ π
π=
Thus, 1 3 5cos
2 6− π− =
.
6. ( )1tan 3− −
Find the angle , ,2 2
θ θπ π− < < whose tangent
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− − = .
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9. 1 3sin sin
8
π−
follows the form of the
equation ( )( ) ( )( )1 1sin sinf f x x x− −= = . Since
3
8
π is in the interval ,
2 2
π π −
, we can apply
the equation directly and get
1 3 3sin sin
8 8
π π− = .
10. 1 3cos cos
4
π−
follows the form of the equation
( )( ) ( )( )1 1cos cosf f x x x− −= = . Since 3
4
π is
in the interval 0,π , we can apply the equation
directly and get 1 3 3cos cos
4 4
π π− =
.
11. 1 2tan tan
3
π−
follows the form of the
equation ( )( ) ( )( )1 1tan tanf f x x x− −= = but we
cannot use the formula directly since 2
3
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
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 −
, we can apply the equation
above and get
1 12tan tan tan tan
3 3 3
π π π− − = − = − .
12. 1sin sin8
π− − follows the form of the
equation ( )( ) ( )( )1 1sin sinf f x x x− −= = . Since
8
π− is in the interval ,2 2
π π −
, we can apply
the equation directly and get
1sin sin8 8
π π− − = − .
13. 1 15cos cos
7
π−
follows the form of the
equation ( )( ) ( )( )1 1cos cosf f x x x− −= = , but
we cannot use the formula directly since 15
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
in the interval 0,π , we can apply the equation
above and get
1 115cos cos cos cos
7 7 7
π π π− − = = .
14. 1 8sin sin
9
π− − follows the form of the
equation ( )( ) ( )( )1 1sin sinf f x x x− −= = , but we
cannot use the formula directly since 8
9
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
8sin sin
9
π θ − =
. The angle 8
9
π− is in
quadrant III so sine is negative. The reference
Chapter 7: Analytic Trigonometry
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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
in the interval ,2 2
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
real number, we can apply the equation directly
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.
21. 1 12 1sin cos sin
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−
−
Find the angle , ,2 2
θ θπ π− ≤ ≤ whose sine
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− −
Find the angle , 0 ,θ θ≤ ≤ π whose cosine
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 − = = −
.
Chapter 7 Review Exercises
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27. 1 3sec tan
3−
Find the angle , ,2 2
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−
Find the angle , ,2 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
+ =
== ±
Since θ is in quadrant I, 4x = .
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
+ === ±
Since θ is in quadrant IV, 3x = .
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
+ === ±
Since θ is in quadrant II, 4y = .
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
−
− −
=
=
=
= = =
The domain of ( )f x equals the range of
Chapter 7: Analytic Trigonometry
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( )1f x− and is 6 6
xπ π− ≤ ≤ , or ,
6 6
π π −
in
interval notation. To find the domain of ( )1f x−
we note that the argument of the inverse sine
function is 2
x and that it must lie in the interval
1,1− . That is,
1 12
2 2
x
x
− ≤ ≤
− ≤ ≤
The domain of ( )1f x− is { }| 2 2x x− ≤ ≤ , or
2,2− 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
[ ]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
−
−
− −
= + −
= + −
+ = +
+ = +
= + −
= + − =
The domain of ( )f x equals the range of
( )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− −
= − += − +
− = −− =
= − =
The domain of ( )f x equals the range of
( )1f x− and is 0 x π≤ ≤ , or 0,π in interval
notation. To find the domain of ( )1f x− we note
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
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
2,4 .
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
interval notation. To find the domain of ( )1f x−
we note that the argument of the inverse sine
function is 2
x and that it must lie in the interval
Chapter 7 Review Exercises
759
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1,1− . That is,
1 12
2 2
x
x
− ≤ ≤
− ≤ ≤
The domain of ( )1f x− is { }| 2 2x x− ≤ ≤ , or
2,2− 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
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
− = =
= − = −
38. 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
− = = ⋅ =
−= = =
−=
39. Let 1csc uθ −= so that csc uθ = , 2 2
π πθ− ≤ ≤
and 0θ ≠ , 1u ≥ . Then,
( )1 1 1sin csc sin
cscu
uθ
θ− = = =
40. Let 1csc uθ −= so that csc uθ = , 2 2
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 θ θθ θ θ θ
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
− +=−
−=−
−=−
=
=
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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 θ θθ θ θ θ
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
− = −
+= − ⋅+
−=
+
=+
=+
Chapter 7 Review Exercises
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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
− −− =− − +
−=
− − ⋅
=−
= −= − −
= −
59. cos( ) cos cos sin sin
cos sin cos sincos cos sin sin
cos sin cos sincos sin
sin coscot tan
α β α β α βα β α β
α β α βα β α ββ αβ αβ α
+ −=
= −
= −
= −
60. sin( ) sin cos cos sin
sin cos sin cossin cos cos sin
sin cos sin cos1 cot tan
α β α β α βα β α β
α β α βα β α β
α β
− −=
= −
= −
61. cos( ) cos cos sin sin
cos cos cos coscos cos sin sin
cos cos cos cos1 tan tan
α β α β α βα β α β
α β α βα β α β
α β
− +=
= +
= +
62. cos( ) cos cos sin sin
sin cos sin coscos cos sin sin
sin cos sin coscos sin
sin coscot tan
α β α β α βα β α β
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
θ θ θ θ θ θθ θ
θθ
− −=
=
=
Chapter 7: Analytic Trigonometry
762
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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 θ θ θθ θθ 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
cos cos sin sin4 6 4 6
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
sin cos cos sin6 4 6 4
1 2 3 2
2 2 2 22 6
4 41
2 64
π π π − = −
π π π π= ⋅ − ⋅
= ⋅ − ⋅
= −
= −
Chapter 7 Review Exercises
763
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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+=
+= = = = =
Chapter 7: Analytic Trigonometry
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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< < − < <
a. sin( ) sin cos cos sin3 5 4 12
5 13 5 1315 48
6533
65
α β α β α β+ = + = ⋅ + ⋅ −
−=
= −
b. cos( ) cos cos sin sin4 5 3 12
5 13 5 1320 36
6556
65
α β α β α β+ = − = ⋅ − ⋅ −
+=
=
c. sin( ) sin cos cos sin
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
a. sin( ) sin cos cos sin3 12 4 5
5 13 5 1336 20
6516
65
α β α β α β+ = + = − ⋅ + − ⋅ − − +=
= −
b. cos( ) cos cos sin sin4 12 3 5
5 13 5 1348 15
6563
65
α β α β α β+ = − = − ⋅ − − ⋅ − − −=
= −
c. sin( ) sin cos cos sin3 12 4 5
5 13 5 1336 20
6556
65
α β α β α β− = − = − ⋅ − − ⋅ − − −=
= −
Chapter 7 Review Exercises
765
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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− < < < <
a. sin( ) sin cos cos sin
4 5 3 12
5 13 5 13
20 36
65 6556
65
α β α β α β+ = +
= − ⋅ − + ⋅
= +
=
b. cos( ) cos cos sin sin3 5 4 12
5 13 5 1315 48
65 6533
65
α β α β α β+ = − = ⋅ − − − ⋅ −= +
=
c. sin( ) sin cos cos sin4 5 3 12
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−=
− − =
= = = =
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
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< < < <
a. sin( ) sin cos cos sin3 5 4 12
5 13 5 1315 48
65 6563
65
α β α β α β+ = + = − ⋅ + − ⋅
= − −
= −
b. cos( ) cos cos sin sin4 5 3 12
5 13 5 1320 36
65 6516
65
α β α β α β+ = − = − ⋅ − − ⋅
= − +
=
c. sin( ) sin cos cos sin3 5 4 12
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< < < <
a. sin( ) sin cos cos sin4 12 3 5
5 13 5 1348 15
6533
65
α β α β α β+ = + = ⋅ − + − ⋅ − − +=
= −
b. cos( ) cos cos sin sin3 12 4 5
5 13 5 1336 20
6556
65
α β α β α β+ = − = − ⋅ − − ⋅ −
+=
=
Chapter 7 Review Exercises
767
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
c. sin( ) sin cos cos sin4 12 3 5
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
a. sin( ) sin cos cos sin
3 1 1 2 2
2 3 2 3
3 2 2
6
α β α β α β+ = + = − + −
− −=
b. cos( ) cos cos sin sin
1 1 3 2 2
2 3 2 3
1 2 6
6
α β α β α β+ = −
= ⋅ − − −
−=
c. sin( ) sin cos cos sin
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= − = − − = − = −
Chapter 7: Analytic Trigonometry
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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< < < <
a. sin( ) sin cos cos sin
1 1 3 2 2
2 3 2 3
1 2 6
6
α β α β α β+ = + = ⋅ − + − ⋅
− −=
b. cos( ) cos cos sin sin
3 1 1 2 2
2 3 2 3
3 2 2
6
α β α β α β+ = − = − − −
−=
c. sin( ) sin cos cos sin
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+=
+ − =
−− −= = =
Chapter 7 Review Exercises
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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= < < < <
a. sin( ) sin cos cos sin
2 2 5 5
3 3 3 3
4 5
9 91
α β α β α β+ = + = − − + − −
= +
=
b. cos( ) cos cos sin sin
5 2 2 5
3 3 3 3
2 5 2 5
9 90
α β α β α β+ = − = − − − − −
= −
=
c. sin( ) sin cos cos sin
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= − < < < <
a. sin( ) sin cos cos sin
2 2 1 1
5 5 5 54 1
5 55
51
α β α β α β+ = + = − + −
= − −
= −
= −
b. cos( ) cos cos sin sin
1 2 2 1
5 5 5 52 2
5 50
α β α β α β+ = − = − − −
= −
=
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
c. sin( ) sin cos cos sin
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
quadrant I; β is in quadrant I. Then 3
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
quadrant I; β is in quadrant I. Then 5
cos13
α = ,
02
α π≤ ≤ , and 4
cos , 05 2
β β π= ≤ ≤ .
Chapter 7 Review Exercises
771
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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= −
= − − = − = =
Chapter 7: Analytic Trigonometry
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( )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
π πθ π θ π= + = + , k is any integer
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
π πθ π θ π= + = + , k is any integer
On 0 2θ π≤ < , the solution set is 3 5
, 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
On the interval 0 2θ π≤ < , the solution set is 3 7
, 4 4
π π
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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
On the interval 0 2θ π≤ < , the solution set is 5 3
, ,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 ≈
The solution set is { }0.25, 2.89 .
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=
On 0 2θ π≤ < , the solution set is 2 4
0, , ,3 3
π ππ
.
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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
− = + == −==
=
On 0 2θ π≤ < , the solution set is 5 3
, ,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
= ==
=
On 0 2θ π≤ < , the solution set is 5
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
= =
==
On 0 2θ π≤ < , the solution set is 5
, ,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
− = − ====
=
On 0 2θ π≤ < , the solution set is 5
, ,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
− = + == −==
=
On 0 2θ π≤ < , the solution set is 5
, ,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
− = + =
= = −
=
On 0 2θ π≤ < , the solution set is 5
,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
.
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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
= − =
= =
=
On 0 2θ π≤ < , the solution set is 3 3
, , ,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− =
Rewrite in the difference of two angles form
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=
On 0 2θ π≤ < , the solution set is 5
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− − ≈
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125. ( )1 1 1sec 3 cos
3− − =
We seek the angle , 0 ,θ θ π≤ ≤ whose cosine
equals 1
3. Now
1cos
3θ = , so θ lies in
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− − − = −
We seek the angle , 0 ,θ θ π≤ ≤ whose tangent
equals 1
4− . Now
1tan
4θ = − , so θ lies in
quadrant II. The calculator yields
1 1tan 0.24
4− − ≈ −
, which is an angle in
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 ≈ .
The solution set is { }0.87 .
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 ≈
The solution set is { }2.22 .
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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 ≈ .
The solution set is { }0.59,3.10 .
133. 1
1
3sin
sin3
sin3
3
2
x
x
x
ππ
π
−
−
− =
= −
= −
= −
The solution set is 3
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= − .
Chapter 7: Analytic Trigonometry
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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−
follows the form of the equation
( )( ) ( )( )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 −
, we can apply the equation
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
+ =
+ =
==
θ is in quadrant I.
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− − = ≈
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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 θ θ θθ θ θ θ
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 β α βα β α β
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
= += += − + ⋅
= − += −= − −
= − −= −
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16.
sin costan cot cos sin
sin costan cotcos 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 = = .
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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
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( )
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 − − = + ≈
The solution set is { }0.285,3.427 .
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 −= − ≈ .
The solution set is { }0.253,2.889 .
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−
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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
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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
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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
= +
= +
=
=
θ is in quadrant I.
( )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+ = −
= − − − −
= + =
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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
−∞ − − ∞
−
− − ≈
− − −
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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
Chapter 7: Analytic Trigonometry
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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
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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.
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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+
+
−=
− (see table column 4)
−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+
+
−=
− (see table column 5)
−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+
+
−=
− (see table column 6)
−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.