Download Chapter 3 Analytic Trigonometry

Chapter 3
Analytic Trigonometry
15. sin −1 ( −1)
Section 3.1
1. Domain: { x x is any real number} ;
We are finding the angle θ , −
Range: { y − 1 ≤ y ≤ 1}
whose sine equals −1 .
π
π
sin θ = −1, − ≤ θ ≤
2
2
π
θ =−
2
π
sin −1 ( −1) = −
2
2. Answers may vary. One possibility is { x | x ≥ 1} .
3.
[3, ∞ )
4. True
5. 1;
3
2
16. cos −1 ( −1)
We are finding the angle θ , 0 ≤ θ ≤ π , whose
cosine equals −1 .
cos θ = −1, 0 ≤ θ ≤ π
θ =π
1
6. − ; −1
2
7. x = sin y
8.
9.
π
π
≤θ ≤ ,
2
2
cos −1 ( −1) = π
π
2
17. tan −1 0
π
We are finding the angle θ , −
5
π
π
< θ < , whose
2
2
tangent equals 0.
−1
10. False. The domain of y = sin x is −1 ≤ x ≤ 1 .
tan θ = 0,
−
θ =0
11. True
π
π
<θ <
2
2
−1
tan 0 = 0
12. True
18. tan −1 ( −1)
13. sin −1 0
We are finding the angle θ , −
π
π
≤ θ ≤ , whose
2
2
We are finding the angle θ , −
sine equals 0.
sin θ = 0,
θ =0
tangent equals −1 .
π
π
− ≤θ ≤
2
2
tan θ = −1,
−
π
π
< θ < , whose
2
2
π
π
<θ <
2
2
π
4
π
tan −1 (−1) = −
4
θ =−
−1
sin 0 = 0
14. cos −1 1
We are finding the angle θ , 0 ≤ θ ≤ π , whose
cosine equals 1.
cos θ = 1, 0 ≤ θ ≤ π
θ =0
cos −1 1 = 0
208
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Section 3.1: The Inverse Sine, Cosine, and Tangent Functions
19. sin −1
⎛
3⎞
22. sin −1 ⎜⎜ −
⎟⎟
⎝ 2 ⎠
2
2
We are finding the angle θ , −
2
.
2
2
sin θ =
,
2
π
θ=
4
2
π
sin −1
=
2
4
π
π
≤ θ ≤ , whose
2
2
We are finding the angle θ , −
sine equals
20. tan −1
sine equals −
−
π
π
≤θ ≤
2
2
We are finding the angle θ , −
tangent equals
3
,
3
π
θ=
6
tan θ =
tan −1
3
,
2
π
θ =−
3
⎛
⎞
3
π
sin −1 ⎜⎜ −
⎟⎟ = − 3
2
⎝
⎠
−
π
π
≤θ ≤
2
2
⎛
3⎞
23. cos −1 ⎜⎜ −
⎟⎟
⎝ 2 ⎠
We are finding the angle θ , 0 ≤ θ ≤ π , whose
π
π
< θ < , whose
2
2
3
.
2
3
cos θ = −
,
2
5π
θ=
6
⎛
⎞
3
5
π
cos −1 ⎜⎜ −
⎟⎟ =
⎝ 2 ⎠ 6
3
.
3
−
3
.
2
sin θ = −
3
3
π
π
≤ θ ≤ , whose
2
2
cosine equals −
π
π
<θ <
2
2
3 π
=
3
6
21. tan −1 3
0 ≤θ ≤ π
⎛
2⎞
24. sin −1 ⎜⎜ −
⎟⎟
⎝ 2 ⎠
π
π
We are finding the angle θ , − < θ < , whose
2
2
tangent equals 3 .
π
π
tan θ = 3, − < θ <
2
2
π
θ=
3
π
−1
tan
3=
3
We are finding the angle θ , −
sine equals −
π
π
≤ θ ≤ , whose
2
2
2
.
2
2
,
2
π
θ =−
4
⎛
2⎞
π
sin −1 ⎜⎜ −
⎟⎟ = −
4
⎝ 2 ⎠
sin θ = −
−
π
π
≤θ ≤
2
2
209
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
⎛
⎛ 3π
39. tan −1 ⎜⎜ tan ⎜ −
⎝ 8
⎝
25. sin −1 0.1 ≈ 0.10
26. cos −1 0.6 ≈ 0.93
(
)
(
)
equation f −1 f ( x ) = tan −1 tan ( x ) = x . Since
−1
27. tan 5 ≈ 1.37
3π
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply
8
⎣ 2 2⎦
the equation directly and get
⎛
3π
⎛ 3π ⎞ ⎞
.
tan −1 ⎜⎜ tan ⎜ − ⎟ ⎟⎟ = −
8
⎝ 8 ⎠⎠
⎝
−
28. tan −1 0.2 ≈ 0.20
29. cos −1
7
≈ 0.51
8
30. sin −1
1
≈ 0.13
8
⎛ ⎛ 3π
40. sin −1 ⎜⎜ sin ⎜ −
⎝ ⎝ 7
31. tan −1 (− 0.4) ≈ − 0.38
(
32. tan (− 3) ≈ −1.25
36. sin −1
3
≈ 0.35
5
⎛ ⎛ 9π
41. sin −1 ⎜⎜ sin ⎜
⎝ ⎝ 8
(
(
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
(
)
cannot use the formula directly since
)
4π
⎛
directly and get cos −1 ⎜ cos
5
⎝
⎞ 4π
.
⎟=
⎠ 5
⎛ ⎛ π ⎞⎞
38. sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ follows the form of the
⎝ ⎝ 10 ⎠ ⎠
)
(
9π
is not
8
⎡ π π⎤
in the interval ⎢ − , ⎥ . We need to find an
⎣ 2 2⎦
⎡ π π⎤
angle θ in the interval ⎢ − , ⎥ for which
⎣ 2 2⎦
9π
9π
sin
= sin θ . The angle
is in quadrant III
8
8
so sine is negative. Therefore we want θ to be in
quadrant IV so sine will still be negative and be
⎡ π π⎤
in the interval ⎢ − , ⎥ . From the unit circle,
⎣ 2 2⎦
in the interval ⎡⎣0, π ⎤⎦ , we can apply the equation
(
)
equation f −1 f ( x ) = sin −1 sin ( x ) = x , but we
4π ⎞
⎛
37. cos −1 ⎜ cos
⎟ follows the form of the equation
5 ⎠
⎝
4π
f −1 f ( x ) = cos −1 cos ( x ) = x . Since
is
5
)
(
3π
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply
7
⎣ 2 2⎦
the equation directly and get
⎛ ⎛ 3π ⎞ ⎞
3π
.
sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ = −
7
7
⎠⎠
⎝ ⎝
34. cos −1 (− 0.44) ≈ 2.03
2
≈ 1.08
3
)
−
33. sin −1 (− 0.12) ≈ − 0.12
35. cos −1
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
equation f −1 f ( x ) = sin −1 sin ( x ) = x . Since
−1
(
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
equation f −1 f ( x ) = sin −1 sin ( x ) = x . Since
π
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply
10
⎣ 2 2⎦
the equation directly and get
⎛ ⎛ π ⎞⎞
π
sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ = − .
10
10
⎠⎠
⎝ ⎝
−
we see sin
9π
π
⎛ π⎞
= sin ⎜ − ⎟ . Since − is in the
8
8
8
⎝
⎠
⎡ π π⎤
interval ⎢ − , ⎥ , we can apply the equation
⎣ 2 2⎦
above and get
π
⎛ 9π ⎞
⎛ π ⎞⎞
−1 ⎛
sin −1 ⎜ sin
⎟ = sin ⎜⎜ sin ⎜ − ⎟ ⎟⎟ = − .
8
8
8
⎝
⎠
⎠⎠
⎝ ⎝
210
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Section 3.1: The Inverse Sine, Cosine, and Tangent Functions
⎛
⎛ 5π
42. cos −1 ⎜⎜ cos ⎜ −
⎝ 3
⎝
(
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
(
⎛
⎛ 2π
44. tan −1 ⎜⎜ tan ⎜ −
⎝ 3
⎝
)
(
equation f −1 f ( x ) = cos −1 cos ( x ) = x , but
we cannot use the formula directly since −
5π
is
3
π⎞ π
⎞⎞
−1 ⎛
⎟ ⎟⎟ = cos ⎜ cos ⎟ = .
3⎠ 3
⎠⎠
⎝
⎛ 2π ⎞
⎛π ⎞
unit circle, we see tan ⎜ −
⎟ = tan ⎜ ⎟ . Since
3
⎝
⎠
⎝3⎠
π
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply the
3
⎣ 2 2⎦
equation above and get
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
(
2π
is not
3
2π
⎛ 2π ⎞
is in
tan ⎜ −
⎟ = tan θ . The angle −
3
⎝ 3 ⎠
quadrant III so tangent is positive. Therefore we
want θ to be in quadrant I so tangent will still be
⎡ π π⎤
positive and in the interval ⎢ − , ⎥ . From the
⎣ 2 2⎦
⎡⎣ 0, π ⎤⎦ , we can apply the equation above and get
)
)
⎡ π π⎤
in the interval ⎢ − , ⎥ . We need to find an angle
⎣ 2 2⎦
⎡ π π⎤
θ in the interval ⎢ − , ⎥ for which
⎣ 2 2⎦
5π
⎛ 5π ⎞
is in
cos ⎜ −
⎟ = cos θ . The angle −
3
⎝ 3 ⎠
quadrant I. From the unit circle, we see
π
π
⎛ 5π ⎞
is in the interval
cos ⎜ − ⎟ = cos . Since
3
3
⎝ 3 ⎠
(
(
cannot use the formula directly since −
angle θ in the interval ⎡⎣0, π ⎤⎦ for which
⎛
⎛ 4π
43. tan −1 ⎜⎜ tan ⎜
⎝ 5
⎝
)
equation f −1 f ( x ) = tan −1 tan ( x ) = x . but we
not in the interval ⎡⎣ 0, π ⎤⎦ . We need to find an
⎛
⎛ 5π
cos −1 ⎜⎜ cos ⎜ −
⎝ 3
⎝
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
equation f −1 f ( x ) = tan −1 tan ( x ) = x , but
⎛
⎛ 2π
tan −1 ⎜⎜ tan ⎜ −
⎝ 3
⎝
4π
we cannot use the formula directly since
is
5
⎡ π π⎤
not in the interval ⎢ − , ⎥ . We need to find an
⎣ 2 2⎦
π⎞ π
⎞⎞
−1 ⎛
⎟ ⎟⎟ = tan ⎜ tan ⎟ = .
3⎠ 3
⎠⎠
⎝
1⎞
⎛
45. sin ⎜ sin −1 ⎟ follows the form of the equation
4⎠
⎝
1
f f −1 ( x ) = sin sin −1 ( x ) = x . Since
is in
4
the interval ⎡⎣ −1,1⎤⎦ , we can apply the equation
⎡ π π⎤
angle θ in the interval ⎢ − , ⎥ for which
⎣ 2 2⎦
4π
⎛ 4π ⎞
is in quadrant
tan ⎜
⎟ = tan θ . The angle
5
⎝ 5 ⎠
II so tangent is negative. Therefore we want θ to
be in quadrant IV so tangent will still be negative
⎡ π π⎤
and in the interval ⎢ − , ⎥ . From the unit
⎣ 2 2⎦
(
)
(
)
1⎞ 1
⎛
directly and get sin ⎜ sin −1 ⎟ = .
4⎠ 4
⎝
⎛
⎛ 2 ⎞⎞
46. cos ⎜⎜ cos −1 ⎜ − ⎟ ⎟⎟ follows the form of the
⎝ 3 ⎠⎠
⎝
π
⎛ 4π ⎞
⎛ π⎞
circle, we see tan ⎜
⎟ = tan ⎜ − ⎟ . Since −
5
5
5
⎝
⎠
⎝
⎠
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply the
⎣ 2 2⎦
equation above and get
⎛
π
⎛ 4π ⎞ ⎞
⎛ π ⎞⎞
−1 ⎛
tan −1 ⎜⎜ tan ⎜
⎟ ⎟⎟ = tan ⎜⎜ tan ⎜ − ⎟ ⎟⎟ = − .
5
5
5
⎝
⎠⎠
⎝
⎠⎠
⎝
⎝
(
)
(
)
equation f f −1 ( x ) = cos cos −1 ( x ) = x .
2
is in the interval ⎡⎣ −1,1⎤⎦ , we can
3
apply the equation directly and get
⎛
2
⎛ 2 ⎞⎞
cos ⎜⎜ cos −1 ⎜ − ⎟ ⎟⎟ = − .
3
⎝ 3 ⎠⎠
⎝
Since −
211
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
(
)
47. tan tan −1 4 follows the form of the equation
(
)
(
argument of the inverse sine function is
)
f f −1 ( x ) = tan tan −1 ( x ) = x . Since 4 is a
and that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That
is,
x−2
−1 ≤
≤1
5
−5 ≤ x − 2 ≤ 5
−3 ≤ x ≤ 7
The domain of f −1 ( x ) is { x | −3 ≤ x ≤ 7} , or
real number, we can apply the equation directly
(
)
and get tan tan −1 4 = 4 .
(
)
48. tan tan −1 ( −2 ) follows the form of the equation
(
)
(
)
f f −1 ( x ) = tan tan −1 ( x ) = x . Since −2 is a
real number, we can apply the equation directly
(
⎡⎣ −3, 7 ⎤⎦ in interval notation.
)
and get tan tan −1 ( −2 ) = −2 .
54.
49. Since there is no angle θ such that cos θ = 1.2 ,
)
cos cos −1 1.2 is not defined.
50. Since there is no angle θ such that sin θ = −2 ,
the quantity sin −1 ( −2 ) is not defined. Thus,
(
)
sin sin −1 ( −2 ) is not defined.
(
)
51. tan tan −1 π follows the form of the equation
(
f f
−1
( x ) ) = tan ( tan ( x ) ) = x .
−1
that x ≠
Since π is a
of the inverse tangent function can be any real
number. Thus, the domain of f −1 ( x ) is all real
)
and get tan tan −1 π = π .
numbers, or ( −∞, ∞ ) in interval notation.
52. Since there is no angle θ such that sin θ = −1.5 ,
the quantity sin −1 ( −1.5) is not defined. Thus,
(
sin sin
53.
−1
55.
( −1.5) ) is not defined.
f ( x ) = −2 cos ( 3 x )
y = −2 cos ( 3 x )
x = −2 cos ( 3 y )
f ( x ) = 5sin x + 2
cos ( 3 y ) = −
y = 5sin x + 2
x = 5sin y + 2
x−2
= f −1 ( x )
5
The domain of f ( x ) is the set of all real
y = sin −1
The domain of f ( x ) is the set of all real
numbers, or ( −∞, ∞ ) in interval notation. To
numbers, or ( −∞, ∞ ) in interval notation. To
−1
( x)
x
2
⎛ x⎞
3 y = cos −1 ⎜ − ⎟
⎝ 2⎠
1
⎛ x⎞
y = cos −1 ⎜ − ⎟ = f −1 ( x )
3
⎝ 2⎠
5sin y = x − 2
x−2
sin y =
5
find the domain of f
( 2k + 1) π
where k is an integer. To find
2
the domain of f −1 ( x ) we note that the argument
real number, we can apply the equation directly
(
f ( x ) = 2 tan x − 3
y = 2 tan x − 3
x = 2 tan y − 3
2 tan y = x + 3
x+3
tan y =
2
x+3
y = tan −1
= f −1 ( x )
2
The domain of f ( x ) is all real numbers such
the quantity cos −1 1.2 is not defined. Thus,
(
x−2
5
find the domain of f −1 ( x ) we note that the
we note that the
argument of the inverse cosine function is
−x
2
212
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.1: The Inverse Sine, Cosine, and Tangent Functions
The domain of f ( x ) is all real numbers such
and that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That
is,
x
−1 ≤ − ≤ 1
2
2 ≥ x ≥ −2
−2 ≤ x ≤ 2
The domain of f −1 ( x ) is { x | −2 ≤ x ≤ 2} , or
that x ≠
2
find the domain of f −1 ( x ) we note that the
argument of the inverse tangent function can be
any real number. Thus, the domain of f −1 ( x ) is
all real numbers, or ( −∞, ∞ ) in interval notation.
⎡⎣ −2, 2⎤⎦ in interval notation.
56.
58.
f ( x ) = 3sin ( 2 x )
x = cos ( y + 2 ) + 1
x = 3sin ( 2 y )
cos ( y + 2 ) = x − 1
x
3
2 y = sin −1
y + 2 = cos −1 ( x − 1)
x
3
y = cos −1 ( x − 1) − 2
The domain of f ( x ) is the set of all real
1
x
y = sin −1 = f −1 ( x )
2
3
The domain of f ( x ) is the set of all real
numbers, or ( −∞, ∞ ) in interval notation. To
find the domain of f −1 ( x ) we note that the
numbers, or ( −∞, ∞ ) in interval notation. To
argument of the inverse cosine function is x − 1
and that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That
find the domain of f −1 ( x ) we note that the
argument of the inverse sine function is
is, −1 ≤ x − 1 ≤ 1
0≤ x≤2
The domain of f −1 ( x ) is { x | 0 ≤ x ≤ 2} , or
x
and
3
that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That is,
x
−1 ≤ ≤ 1
3
−3 ≤ x ≤ 3
The domain of f −1 ( x ) is { x | −3 ≤ x ≤ 3} , or
⎡⎣ 0, 2 ⎤⎦ in interval notation.
59.
x = 3sin ( 2 y + 1)
f ( x ) = − tan ( x + 1) − 3
sin ( 2 y + 1) =
y = − tan ( x + 1) − 3
x
3
x
3
⎛x⎞
2 y = sin −1 ⎜ ⎟ − 1
⎝3⎠
2 y + 1 = sin −1
x = − tan ( y + 1) − 3
tan ( y + 1) = − x − 3
y + 1 = tan −1 ( − x − 3)
1
⎛ x⎞ 1
y = sin −1 ⎜ ⎟ − = f −1 ( x )
2
⎝3⎠ 2
y = −1 + tan −1 ( − x − 3)
= −1 − tan −1 ( x + 3) = f −1 ( x )
(note here we used the fact that y = tan
odd function).
f ( x ) = 3sin ( 2 x + 1)
y = 3sin ( 2 x + 1)
⎣⎡ −3,3⎦⎤ in interval notation.
57.
f ( x ) = cos ( x + 2 ) + 1
y = cos ( x + 2 ) + 1
y = 3sin ( 2 x )
sin ( 2 y ) =
( 2k + 1) π − 1 where k is an integer. To
The domain of f ( x ) is the set of all real
−1
numbers, or ( −∞, ∞ ) in interval notation. To
x is an
find the domain of f −1 ( x ) we note that the
213
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
argument of the inverse sine function is
62. 2 cos −1 x = π
x
and
3
cos −1 x =
that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That is,
x
−1 ≤ ≤ 1
3
−3 ≤ x ≤ 3
The domain of f −1 ( x ) is { x | −3 ≤ x ≤ 3} , or
63. 3cos −1 ( 2 x ) = 2π
cos −1 ( 2 x ) =
y = 2 cos ( 3x + 2 )
2π
3
1
2
1
x=−
4
2x = −
x
2
⎛x⎞
3 y + 2 = cos −1 ⎜ ⎟
⎝2⎠
⎛x⎞
3 y = cos −1 ⎜ ⎟ − 2
⎝2⎠
⎧ 1⎫
The solution set is ⎨− ⎬ .
⎩ 4⎭
64. −6sin −1 ( 3x ) = π
1
⎛ x⎞ 2
y = cos −1 ⎜ ⎟ − = f −1 ( x )
3
⎝2⎠ 3
sin −1 ( 3x ) = −
π
6
⎛ π⎞
3 x = sin ⎜ − ⎟
⎝ 6⎠
1
3x = −
2
1
x=−
6
⎧ 1⎫
The solution set is ⎨− ⎬ .
⎩ 6⎭
The domain of f ( x ) is the set of all real
numbers, or ( −∞, ∞ ) in interval notation. To
find the domain of f −1 ( x ) we note that the
x
2
and that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That
argument of the inverse cosine function is
is,
x
≤1
2
−2 ≤ x ≤ 2
The domain of f −1 ( x ) is { x | −2 ≤ x ≤ 2} , or
−1 ≤
65. 3 tan −1 x = π
tan −1 x =
π
3
x = tan
⎡⎣ −2, 2 ⎤⎦ in interval notation.
π
3
= 3
The solution set is
61. 4sin −1 x = π
π
{ 3} .
66. −4 tan −1 x = π
4
π
2
4
2
⎧⎪ 2 ⎫⎪
The solution set is ⎨ ⎬ .
⎩⎪ 2 ⎭⎪
x = sin
2π
3
2 x = cos
x = 2 cos ( 3 y + 2 )
sin −1 x =
π
=0
2
The solution set is {0} .
f ( x ) = 2 cos ( 3 x + 2 )
cos ( 3 y + 2 ) =
2
x = cos
⎡⎣ −3,3⎤⎦ in interval notation.
60.
π
tan −1 x = −
=
π
4
⎛ π⎞
x = tan ⎜ − ⎟ = −1
⎝ 4⎠
The solution set is {−1} .
214
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Section 3.1: The Inverse Sine, Cosine, and Tangent Functions
71. Note that θ = 21°18′ = 21.3° .
67. 4 cos −1 x − 2π = 2 cos −1 x
2 cos −1 x − 2π = 0
a.
2 cos −1 x = 2π
cos −1 x = π
x = cos π = −1
The solution set is {−1} .
b.
68. 5sin −1 x − 2π = 2sin −1 x − 3π
3sin −1 x = −π
sin −1 x = −
c.
π
3
3
⎛ π⎞
x = sin ⎜ − ⎟ = −
2
⎝ 3⎠
⎪⎧ 3 ⎪⎫
The solution set is ⎨−
⎬.
⎩⎪ 2 ⎭⎪
b.
c.
a.
π
⎡ cos −1 ( tan ( 23.5 ⋅ 180
) tan ( 29.75 ⋅ 180π ) ) ⎤
⎥
D = 24 ⋅ ⎢1 −
π
⎢⎣
⎥⎦
≈ 13.92 hours or 13 hours, 55 minutes
b.
π
⎡ cos −1 ( tan ( 0 ⋅ 180
) tan ( 29.75 ⋅ 180π ) ) ⎤
⎥
D = 24 ⋅ ⎢1 −
π
⎢⎣
⎥⎦
≈ 12 hours
c.
π
⎡ cos −1 ( tan ( 22.8 ⋅ 180
) tan ( 29.75 ⋅ 180π ) ) ⎤
⎥
D = 24 ⋅ ⎢1 −
π
⎢⎣
⎥⎦
≈ 13.85 hours or 13 hours, 51 minutes
73. a.
70. Note that θ = 40°45′ = 40.75° .
a.
b.
c.
π
⎛ cos −1 ( tan ( 0 ⋅ 180
) tan ( 21.3 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
π
⎛ cos −1 ( tan ( 22.8 ⋅ 180
) tan ( 21.3 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 13.26 hours or 13 hours, 15 minutes
72. Note that θ = 61°10′ ≈ 61.167° .
69. Note that θ = 29°45′ = 29.75° .
a.
π
⎛ cos −1 ( tan ( 23.5 ⋅ 180
) tan ( 21.3 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 13.30 hours or 13 hours, 18 minutes
b.
π
⎛ cos −1 ( tan ( 23.5 ⋅ 180
) tan ( 40.75 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 14.93 hours or 14 hours, 56 minutes
c.
π
⎛ cos −1 ( tan ( 0 ⋅ 180
) tan ( 40.75 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
π
⎛ cos −1 ( tan ( 23.5 ⋅ 180
) tan ( 61.167 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 18.96 hours or 18 hours, 57 minutes
π
⎛ cos −1 ( tan ( 0 ⋅ 180
) tan ( 61.167 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
π
⎛ cos −1 ( tan ( 22.8 ⋅ 180
) tan ( 61.167 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 18.64 hours or 18 hours, 38 minutes
π
⎛ cos −1 ( tan ( 23.5 ⋅ 180
) tan ( 0 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
π
⎛ cos −1 ( tan ( 0 ⋅ 180
) tan ( 0 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
π
⎛ cos −1 ( tan ( 22.8 ⋅ 180
) tan ( 0 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
d. There are approximately 12 hours of
daylight every day at the equator.
π
⎛ cos −1 ( tan ( 22.8 ⋅ 180
) tan ( 40.75 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 14.83 hours or 14 hours, 50 minutes
215
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Chapter 3: Analytic Trigonometry
74. Note that θ = 66°30′ = 66.5° .
a.
b.
c.
2π (2710) 39.65
=
, so
24
t
24(39.65)
t=
≈ 0.05589 hours ≈ 3.35 minutes
2π (2710)
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.
and
π
⎛ cos −1 ( tan ( 23.5 ⋅ 180
) tan ( 66.5 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 24 hours
π
⎛ cos −1 ( tan ( 0 ⋅ 180
) tan ( 66.5 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜ 1 −
⎜
⎟
π
⎝
⎠
≈ 12 hours
⎛ 34 ⎞
⎛6⎞
76. θ ( x ) = tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ .
⎝ x ⎠
⎝x⎠
⎛ 34 ⎞
⎛ 6⎞
a. θ (10 ) = tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ ≈ 42.6°
10
⎝ ⎠
⎝ 10 ⎠
If you sit 10 feet from the screen, then the
viewing angle is about 42.6° .
⎛ 34 ⎞
⎛ 6⎞
θ (15 ) = tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ ≈ 44.4°
⎝ 15 ⎠
⎝ 15 ⎠
If you sit 15 feet from the screen, then the
viewing angle is about 44.4° .
⎛ 34 ⎞
⎛ 6 ⎞
θ ( 20 ) = tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ ≈ 42.8°
⎝ 20 ⎠
⎝ 20 ⎠
If you sit 20 feet from the screen, then the
viewing angle is about 42.8° .
π
⎛ cos −1 ( tan ( 22.8 ⋅ 180
) tan ( 66.5 ⋅ 180π ) ) ⎞⎟
D = 24 ⋅ ⎜1 −
⎜
⎟
π
⎝
⎠
≈ 22.02 hours or 22 hours, 1 minute
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 center of Earth, and
arrange the figure so that segment CQ lies along
the x-axis (see figure).
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(r − 1) = 14.3
5 + 3r − 3 = 14.3
y
P
D (x,y)
s
C
θ
2710 mi
x
Q (2710,0)
3r = 12.3
r = 4.1
Sitting in the 4th row should provide the
largest viewing angle.
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 mile
1530 ft ⋅
≈ 0.29 mile . Thus, point
5280 feet
D( x, y ) = (2700, y ) lies on a circle with radius
r = 2710.29 miles. Thus,
x
2710
cos θ = =
r 2710.29
⎛ 2710 ⎞
θ = cos −1 ⎜
⎟ ≈ 0.01463 radians
⎝ 2710.29 ⎠
Finally, s = rθ = 2710(0.01463) ≈ 39.65 miles ,
216
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Section 3.2: The Inverse Trigonometric Functions [Continued]
c.
Set the graphing calculator in degree mode
⎛ 34 ⎞
⎛6⎞
and let Y1 = tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ :
⎝ x ⎠
⎝ x⎠
90°
1
1
; b = ; The area is:
2
2
⎛1⎞
⎛ 1⎞
sin −1 b − sin −1 a = sin −1 ⎜ ⎟ − sin −1 ⎜ − ⎟
⎝2⎠
⎝ 2⎠
a=−
b.
=
=
0
6
⎛ π⎞
−⎜− ⎟
⎝ 6⎠
π
square units
3
79. Here we have α1 = 41°50 ' , β1 = −87°31' ,
50
0°
Use MAXIMUM:
90°
π
α 2 = 21°18' , and β 2 = −157°51' .
Converting minutes to degrees gives
( )
(
)
31
α1 = 41 56 ° , β1 = −87 60
° , α 2 = 21.3° , and
β 2 = −157.85° . Substituting these values, and
50
r = 3960 , into our equation gives d ≈ 4251
miles. The distance from Chicago to Honolulu is
about 4251 miles.
(remember that S and W angles are negative)
a = 0 ; b = 3 ; The area is:
80. Here we have α1 = 21°18' , β1 = −157°51' ,
α 2 = −37°47 ' , and β 2 = 144°58' .
Converting minutes to degrees gives α1 = 21.3° ,
0
0°
The maximum viewing angle will occur
when x ≈ 14.3 feet.
77. a.
tan −1 b − tan −1 a = tan −1 3 − tan −1 0
=
=
b.
a=−
π
3
π
3
(
Section 3.2
π
⎛ π⎞
= −⎜− ⎟
4 ⎝ 6⎠
5π
=
square units
12
⎧
π⎫
1. Domain: ⎨ x x ≠ odd multiples of ⎬ ,
2⎭
⎩
Range: { y y ≤ −1 or y ≥ 1}
3
; The area is:
2
⎛ 3⎞
−1
sin −1 b − sin −1 a = sin −1 ⎜⎜
⎟⎟ − sin 0
⎝ 2 ⎠
2. True
a = 0; b =
=
π
3
π
3
)
r = 3960 , into our equation gives d ≈ 5518
miles. The distance from Honolulu to
Melbourne is about 5518 miles.
(remember that S and W angles are negative)
3
; b = 1 ; The area is:
3
=
)
29
β 2 = 144 30
° . Substituting these values, and
square units
⎛
3⎞
tan −1 b − tan −1 a = tan −1 1 − tan −1 ⎜⎜ −
⎟⎟
⎝ 3 ⎠
78. a.
(
β1 = −157.85° , α 2 = −37 47
° , and
60
−0
3. −
2
5
=−
2 5
5
4. x = sec y , ≥ 1 , 0 , π
−0
square units
5.
2
2
⎛
π
2⎞
−1
⎜ cos ( tan 1) = cos =
⎟
4
2 ⎠
⎝
6. False
217
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Chapter 3: Analytic Trigonometry
7. True
⎡
⎛ 1 ⎞⎤
12. tan ⎢sin −1 ⎜ − ⎟ ⎥
⎝ 2 ⎠⎦
⎣
8. True
Find the angle θ , −
⎛
2⎞
9. cos ⎜⎜ sin −1
⎟⎟
2
⎝
⎠
Find the angle θ , −
equals
1
equals − .
2
π
π
≤ θ ≤ , whose sine
2
2
1
sin θ = − ,
2
π
θ =−
6
2
.
2
2
π
π
, − ≤θ ≤
2
2
2
π
θ=
4
π
2⎞
2
⎟⎟ = cos =
2 ⎠
4
2
sin θ =
⎛
cos ⎜⎜ sin −1
⎝
1
,
2
π
θ=
3
equals
1
,
2
π
θ=
3
⎡
⎛ 1 ⎞⎤
14. cot ⎢sin −1 ⎜ − ⎟ ⎥
⎝ 2 ⎠⎦
⎣
Find the angle θ , −
⎡
⎛
3 ⎞⎤
11. tan ⎢cos −1 ⎜⎜ −
⎟⎟ ⎥
⎝ 2 ⎠ ⎦⎥
⎣⎢
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
θ=
π
π
≤ θ ≤ , whose sine
2
2
1
equals − .
2
1
π
π
sin θ = − , − ≤ θ ≤
2
2
2
π
θ =−
6
⎡ −1 ⎛ 1 ⎞ ⎤
⎛ π⎞
cot ⎢sin ⎜ − ⎟ ⎥ = cot ⎜ − ⎟ = − 3
⎝ 2 ⎠⎦
⎝ 6⎠
⎣
3
.
2
3
,
2
0 ≤θ ≤ π
1⎞
π
⎛
sec ⎜ cos −1 ⎟ = sec = 2
2
3
⎝
⎠
0 ≤θ ≤ π
cos θ = −
1
.
2
cos θ =
π
1⎞
3
⎛
sin ⎜ cos −1 ⎟ = sin =
2⎠
3
2
⎝
equals −
π
π
≤θ ≤
2
2
1⎞
⎛
13. sec ⎜ cos −1 ⎟
2⎠
⎝
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
1
.
2
cos θ =
−
⎡
3
⎛ 1 ⎞⎤
⎛ π⎞
tan ⎢sin −1 ⎜ − ⎟ ⎥ = tan ⎜ − ⎟ = −
2
6
3
⎝
⎠⎦
⎝
⎠
⎣
1⎞
⎛
10. sin ⎜ cos −1 ⎟
2⎠
⎝
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
equals
π
π
≤ θ ≤ , whose sine
2
2
0 ≤θ ≤ π
5π
6
⎡
⎛
3 ⎞⎤
5π
3
tan ⎢cos −1 ⎜⎜ −
=−
⎟⎟ ⎥ = tan
2
6
3
⎝
⎠ ⎥⎦
⎣⎢
218
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Section 3.2: The Inverse Trigonometric Functions [Continued]
(
)
15. csc tan −1 1
Find the angle θ , −
π
π
< θ < , whose tangent
2
2
equals 1.
tan θ = 1,
θ=
(
−
π
4
)
csc tan −1 1 = csc
(
16. sec tan −1 3
π
Find the angle θ , −
π
π
<θ <
2
2
1
sin θ = − ,
2
π
θ =−
6
π
π
< θ < , whose tangent
2
2
θ=
(
π
π
π
<θ <
2
2
equals −
3
)
sec tan −1 3 = sec
3
.
2
π
=2
3
cos θ = −
θ=
17. sin ⎡⎣ tan −1 (−1) ⎤⎦
Find the angle θ , −
equals −1 .
θ =−
−
π
π
<θ <
2
2
2
⎛ π⎞
sin ⎣⎡ tan −1 (−1) ⎦⎤ = sin ⎜ − ⎟ = −
2
⎝ 4⎠
equals −
5π
6
2
.
2
cos θ = −
⎡
⎛
3 ⎞⎤
18. cos ⎢sin −1 ⎜⎜ −
⎟⎟ ⎥
⎢⎣
⎝ 2 ⎠ ⎥⎦
equals −
0 ≤θ ≤ π
⎛
5π ⎞
2⎞
⎛
21. cos −1 ⎜ cos ⎟ = cos −1 ⎜⎜ −
⎟⎟
4 ⎠
⎝
⎝ 2 ⎠
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
π
4
Find the angle θ , −
3
2
⎡
⎛
3 ⎞⎤
5π
=2
csc ⎢ cos −1 ⎜⎜ −
⎟⎟ ⎥ = csc
6
⎢⎣
⎝ 2 ⎠ ⎥⎦
π
π
< θ < , whose tangent
2
2
tan θ = −1,
π
π
≤θ ≤
2
2
⎡
⎛
3 ⎞⎤
20. csc ⎢ cos −1 ⎜⎜ −
⎟⎟ ⎥
⎝ 2 ⎠ ⎦⎥
⎣⎢
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
3.
−
−
⎡
⎛ 1 ⎞⎤
⎛ π⎞ 2 3
sec ⎢sin −1 ⎜ − ⎟ ⎥ = sec ⎜ − ⎟ =
2
3
⎝
⎠⎦
⎝ 6⎠
⎣
)
tan θ = 3,
π
π
≤ θ ≤ , whose sine
2
2
1
equals − .
2
= 2
4
Find the angle θ , −
equals
⎡
⎛ 1 ⎞⎤
19. sec ⎢sin −1 ⎜ − ⎟ ⎥
⎝ 2 ⎠⎦
⎣
2
,
2
0 ≤θ ≤ π
3π
4
5π ⎞ 3π
−1 ⎛
cos ⎜ cos ⎟ =
4 ⎠ 4
⎝
θ=
π
π
≤ θ ≤ , whose sine
2
2
3
.
2
3
π
π
, − ≤θ ≤
2
2
2
π
θ =−
3
⎡
⎛
3 ⎞⎤
⎛ π⎞ 1
cos ⎢sin −1 ⎜⎜ −
⎟⎟ ⎥ = cos ⎜ − ⎟ =
⎝ 3⎠ 2
⎢⎣
⎝ 2 ⎠ ⎥⎦
sin θ = −
219
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Chapter 3: Analytic Trigonometry
2π ⎞
⎛
22. tan −1 ⎜ tan ⎟ = tan −1 − 3
3 ⎠
⎝
π
π
Find the angle θ , − < θ < , whose tangent
2
2
equals − 3 .
π
π
tan θ = − 3, − < θ <
2
2
π
θ =−
3
π
2
π
⎛
⎞
tan −1 ⎜ tan
⎟=−
3 ⎠
3
⎝
(
)
1⎞
⎛
25. tan ⎜ sin −1 ⎟
3⎠
⎝
1
1
Let θ = sin −1 . Since sin θ = and
3
3
π
π
− ≤ θ ≤ , θ is in quadrant I, and we let
2
2
y = 1 and r = 3 .
Solve for x:
x2 + 1 = 9
x2 = 8
x = ± 8 = ±2 2
Since θ is in quadrant I, x = 2 2 .
⎡ ⎛ 7π ⎞⎤
1
23. sin −1 ⎢sin ⎜ − ⎟ ⎥ = sin −1
2
⎣ ⎝ 6 ⎠⎦
π
π
Find the angle θ , − ≤ θ ≤ , whose sine
2
2
1
equals .
2
1
π
π
sin θ = , − ≤ θ ≤
2
2
2
π
θ=
6
⎡ ⎛ 7π ⎞ ⎤ π
sin −1 ⎢sin ⎜ −
⎟⎥ =
⎣ ⎝ 6 ⎠⎦ 6
1⎞
y
1
2
2
⎛
tan ⎜ sin −1 ⎟ = tan θ = =
⋅
=
x 2 2 2
3⎠
4
⎝
1⎞
⎛
26. tan ⎜ cos −1 ⎟
3⎠
⎝
1
1
. Since cos θ = and 0 ≤ θ ≤ π ,
3
3
θ is in quadrant I, and we let x = 1 and r = 3 .
Solve for y:
1 + y2 = 9
Let θ = cos −1
y2 = 8
y = ± 8 = ±2 2
Since θ is in quadrant I, y = 2 2 .
⎡ ⎛ π ⎞⎤
1
24. cos −1 ⎢ cos ⎜ − ⎟ ⎥ = cos −1
2
⎣ ⎝ 3 ⎠⎦
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
equals
1⎞
y 2 2
⎛
=2 2
tan ⎜ cos −1 ⎟ = tan θ = =
x
3
1
⎝
⎠
1
.
2
1
cos θ = ,
2
π
θ=
3
⎡ ⎛ π ⎞⎤ π
cos −1 ⎢ cos ⎜ − ⎟ ⎥ =
⎣ ⎝ 3 ⎠⎦ 3
1⎞
⎛
27. sec ⎜ tan −1 ⎟
2⎠
⎝
0 ≤θ ≤ π
Let θ = tan −1
1
1
. Since tan θ = and
2
2
π
π
< θ < , θ is in quadrant I, and we let
2
2
x = 2 and y = 1 .
Solve for r:
22 + 1 = r 2
−
r2 = 5
r= 5
θ is in quadrant I.
1⎞
r
5
⎛
sec ⎜ tan −1 ⎟ = sec θ = =
x
2⎠
2
⎝
220
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Section 3.2: The Inverse Trigonometric Functions [Continued]
⎛
2⎞
28. cos ⎜⎜ sin −1
⎟
3 ⎟⎠
⎝
Let θ = sin −1
31. sin ⎡⎣ tan −1 (−3) ⎤⎦
Let θ = tan −1 (−3) . Since tan θ = −3 and
π
π
< θ < , θ is in quadrant IV, and we let
2
2
x = 1 and y = −3 .
Solve for r:
1+ 9 = r2
2
2
. Since sin θ =
and
3
3
−
π
π
≤ θ ≤ , θ is in quadrant I, and we let
2
2
y = 2 and r = 3 .
Solve for x:
x2 + 2 = 9
−
r 2 = 10
r = ± 10
x =7
2
Since θ is in quadrant IV, r = 10 .
y
sin ⎡⎣ tan −1 (−3) ⎤⎦ = sin θ =
r
10
3 10
−3
=
⋅
=−
10
10 10
x=± 7
Since θ is in quadrant I, x = 7 .
⎛
2⎞
x
7
cos ⎜⎜ sin −1
⎟⎟ = cos θ = =
3
r
3
⎝
⎠
⎡
⎛
2 ⎞⎤
29. cot ⎢sin −1 ⎜⎜ −
⎟⎟ ⎥
⎢⎣
⎝ 3 ⎠ ⎥⎦
⎛
2⎞
2
Let θ = sin −1 ⎜⎜ −
and
⎟⎟ . Since sin θ = −
3
3
⎝
⎠
π
π
− ≤ θ ≤ , θ is in quadrant IV, and we let
2
2
y = − 2 and r = 3 .
Solve for x:
x2 + 2 = 9
⎡
⎛
3 ⎞⎤
32. cot ⎢ cos −1 ⎜⎜ −
⎟⎟ ⎥
⎝ 3 ⎠ ⎥⎦
⎣⎢
⎛
3⎞
3
Let θ = cos −1 ⎜⎜ −
and
⎟⎟ . Since cos θ = −
3
⎝ 3 ⎠
0 ≤ θ ≤ π , θ is in quadrant II, and we let
x = − 3 and r = 3 .
Solve for y:
3 + y2 = 9
y2 = 6
x2 = 7
y=± 6
x=± 7
Since θ is in quadrant II, y = 6 .
Since θ is in quadrant IV, x = 7 .
⎡
⎛
x
2 ⎞⎤
7
2
14
⋅
=−
cot ⎢sin −1 ⎜⎜ −
⎟⎟ ⎥ = cot θ = =
y
3
2
2
2
−
⎝
⎠ ⎦⎥
⎣⎢
⎡
⎛
3 ⎞⎤
x
cot ⎢cos −1 ⎜⎜ −
⎟⎟ ⎥ = cot θ =
y
⎢⎣
⎝ 3 ⎠ ⎥⎦
=
30. csc ⎡⎣ tan −1 (− 2) ⎤⎦
Let θ = tan −1 ( − 2) . Since tan θ = − 2 and
6
=
−1
2
⋅
2
2
=−
2
2
⎛
2 5⎞
33. sec ⎜⎜ sin −1
⎟
5 ⎟⎠
⎝
π
π
− < θ < , θ is in quadrant IV, and we let
2
2
x = 1 and y = − 2 .
Solve for r:
1+ 4 = r2
Let θ = sin −1
2 5
2 5
. Since sin θ =
and
5
5
π
π
≤ θ ≤ , θ is in quadrant I, and we let
2
2
y = 2 5 and r = 5 .
Solve for x:
−
r =5
2
r=± 5
Since θ is in quadrant IV, r = 5 .
csc ⎡⎣ tan −1 (− 2) ⎤⎦ = csc θ =
− 3
r
5
5
=
=−
y −2
2
221
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Chapter 3: Analytic Trigonometry
39. csc −1 (−1)
x 2 + 20 = 25
π
π
≤θ ≤ ,
2
2
θ ≠ 0 , whose cosecant equals −1 .
π
π
csc θ = −1, − ≤ θ ≤ , θ ≠ 0
2
2
π
θ =−
2
π
−1
csc (−1) = −
2
x2 = 5
We are finding the angle θ , −
x=± 5
Since θ is in quadrant I, x = 5 .
⎛
2 5⎞
r
5
sec ⎜⎜ sin −1
= 5
⎟⎟ = sec θ = =
5 ⎠
x
5
⎝
1⎞
⎛
34. csc ⎜ tan −1 ⎟
2⎠
⎝
Let θ = tan −1
1
1
. Since tan θ = and
2
2
40. csc −1 2
π
π
< θ < , θ is in quadrant I, and we let
2
2
x = 2 and y = 1 .
Solve for r:
22 + 1 = r 2
π
π
≤θ ≤ ,
2
2
θ ≠ 0 , whose cosecant equals 2 .
π
π
csc θ = 2, − ≤ θ ≤ , θ ≠ 0
2
2
π
θ=
4
π
−1
csc
2=
4
We are finding the angle θ , −
−
r2 = 5
r= 5
θ is in quadrant I.
1⎞
r
5
⎛
csc ⎜ tan −1 ⎟ = csc θ = =
= 5
2
y
1
⎝
⎠
41. sec −1
2 3
3
⎛
3π ⎞
2⎞
π
⎛
35. sin −1 ⎜ cos ⎟ = sin −1 ⎜⎜ −
⎟⎟ = −
4 ⎠
4
⎝
⎝ 2 ⎠
We are finding the angle θ , 0 ≤ θ ≤ π , θ ≠
7π ⎞
⎛
⎛ 1 ⎞ 2π
36. cos −1 ⎜ sin ⎟ = cos −1 ⎜ − ⎟ =
6
⎝
⎠
⎝ 2⎠ 3
whose secant equals
2 3
,
3
π
θ=
6
sec θ =
37. cot −1 3
We are finding the angle θ , 0 < θ < π, whose
cotangent equals
cot θ = 3,
3.
0 <θ < π
sec −1
π
6
π
3=
6
θ=
cot
−1
π
,
2
2 3
.
3
0 ≤ θ ≤ π, θ ≠
π
2
2 3 π
=
3
6
42. sec −1 ( − 2 )
We are finding the angle θ , 0 ≤ θ ≤ π , θ ≠
whose secant equals −2 .
38. cot −1 1
We are finding the angle θ , 0 < θ < π, whose
cotangent equals 1.
cot θ = 1, 0 < θ < π
π
θ=
4
π
−1
cot 1 =
4
sec θ = − 2,
0 ≤ θ ≤ π, θ ≠
π
,
2
π
2
2π
3
2
π
sec −1 ( − 2 ) =
3
θ=
222
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Section 3.2: The Inverse Trigonometric Functions [Continued]
⎛
3⎞
43. cot −1 ⎜⎜ −
⎟⎟
⎝ 3 ⎠
We are finding the angle θ , 0 < θ < π, whose
cotangent equals −
46. csc −1 5 = sin −1
We seek the angle θ , −
3
,
3
2π
θ=
3
⎛
⎞
3
2π
cot −1 ⎜⎜ −
⎟⎟ =
3
⎝
⎠ 3
0 <θ < π
⎛ 2 3⎞
44. csc −1 ⎜⎜ −
⎟
3 ⎟⎠
⎝
2 3
,
3
π
θ =−
3
⎛ 2 3⎞
π
csc −1 ⎜⎜ −
⎟⎟ = −
3 ⎠
3
⎝
−
π
1
2
We seek the angle θ , 0 ≤ θ ≤ π , whose tangent
47. cot −1 2 = tan −1
π
π
We are finding the angle θ , − ≤ θ ≤ ,
2
2
2 3
θ ≠ 0 , whose cosecant equals −
.
3
csc θ = −
π
≤ θ ≤ , whose sine
2
2
1
1
equals . Now sin θ = , so θ lies in
5
5
1
quadrant I. The calculator yields sin −1 ≈ 0.20 ,
5
which is an angle in quadrant I, so
csc−1 5 ≈ 0.20 .
3
.
3
cot θ = −
1
5
1
1
. Now tan θ = , so θ lies in
2
2
1
quadrant I. The calculator yields an −1 ≈ 0.46 ,
2
which is an angle in quadrant I, so
cot −1 ( 2 ) ≈ 0.46 .
equals
π
π
≤θ ≤ , θ ≠ 0
2
2
1
4
We seek the angle θ , 0 ≤ θ ≤ π , whose cosine
45. sec −1 4 = cos −1
⎛ 1⎞
48. sec −1 (−3) = cos −1 ⎜ − ⎟
⎝ 3⎠
We seek the angle θ , 0 ≤ θ ≤ π , whose cosine
1
1
. Now cos θ = , so θ lies in quadrant
4
4
1
I. The calculator yields cos −1 ≈ 1.32 , which is
4
an angle in quadrant I, so sec−1 ( 4 ) ≈ 1.32 .
equals
1
1
equals − . Now cos θ = − , θ lies in
3
3
quadrant II. The calculator yields
⎛ 1⎞
cos −1 ⎜ − ⎟ ≈ 1.91 , which is an angle in
⎝ 3⎠
quadrant II, so sec−1 ( − 3) ≈ 1.91 .
223
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Chapter 3: Analytic Trigonometry
⎛ 1⎞
49. csc −1 ( −3) = sin −1 ⎜ − ⎟
⎝ 3⎠
We seek the angle θ , −
π
2
≤θ ≤
π
2
⎛ 1 ⎞
52. cot −1 ( −8.1) = tan −1 ⎜ −
⎟
⎝ 8.1 ⎠
We seek the angle θ , 0 ≤ θ ≤ π , whose tangent
, whose sine
1
1
, so θ lies in
. Now tan θ = −
8.1
8.1
quadrant II. The calculator yields
⎛ 1 ⎞
tan −1 ⎜ −
⎟ ≈ −0.12 , which is an angle in
⎝ 8.1 ⎠
quadrant IV. Since θ is in quadrant II,
equals −
1
1
equals − . Now sin θ = − , so θ lies in
3
3
quadrant IV. The calculator yields
⎛ 1⎞
sin −1 ⎜ − ⎟ ≈ −0.34 , which is an angle in
⎝ 3⎠
quadrant IV, so csc−1 ( −3) ≈ −0.34 .
θ ≈ −0.12 + π ≈ 3.02 . Thus, cot −1 ( −8.1) ≈ 3.02 .
⎛ 3⎞
⎛ 2⎞
53. csc −1 ⎜ − ⎟ = sin −1 ⎜ − ⎟
⎝ 2⎠
⎝ 3⎠
⎛ 1⎞
50. cot −1 ⎜ − ⎟ = tan −1 (− 2)
⎝ 2⎠
We seek the angle θ , 0 ≤ θ ≤ π , whose tangent
equals −2 . Now tan θ = −2 , so θ lies in
quadrant II. The calculator yields
tan −1 ( − 2 ) ≈ −1.11 , which is an angle in
We seek the angle θ , −
)
1
. Now tan θ = −
1
2
, θ ≠0,
⎛ 4⎞
⎛ 3⎞
54. sec −1 ⎜ − ⎟ = cos −1 ⎜ − ⎟
⎝ 3⎠
⎝ 4⎠
We are finding the angle θ , 0 ≤ θ ≤ π , θ ≠
π
,
2
3
3
whose cosine equals − . Now cos θ = − , so
4
4
θ lies in quadrant II. The calculator yields
⎛ 3⎞
cos −1 ⎜ − ⎟ ≈ 2.42 , which is an angle in
⎝ 4⎠
⎛ 4⎞
quadrant II, so sec −1 ⎜ − ⎟ ≈ 2.42 .
⎝ 3⎠
, so θ lies in
5
5
quadrant II. The calculator yields
⎛ 1 ⎞
tan −1 ⎜ −
⎟ ≈ −0.42 , which is an angle in
5⎠
⎝
quadrant IV. Since θ is in quadrant II,
θ ≈ −0.42 + π ≈ 2.72 . Therefore,
(
π
2
2
. Now sin θ = − , so θ
3
3
lies in quadrant IV. The calculator yields
⎛ 2⎞
sin −1 ⎜ − ⎟ ≈ −0.73 , which is an angle in
⎝ 3⎠
⎛ 3⎞
quadrant IV, so csc−1 ⎜ − ⎟ ≈ −0.73 .
⎝ 2⎠
⎛ 1 ⎞
51. cot −1 − 5 = tan −1 ⎜ −
⎟
5⎠
⎝
We seek the angle θ , 0 ≤ θ ≤ π , whose tangent
equals −
2
≤θ ≤
whose sine equals −
quadrant IV. Since θ lies in quadrant II,
θ ≈ −1.11 + π ≈ 2.03 . Therefore,
⎛ 1⎞
cot −1 ⎜ − ⎟ ≈ 2.03 .
⎝ 2⎠
(
π
)
cot −1 − 5 ≈ 2.72 .
224
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Section 3.2: The Inverse Trigonometric Functions [Continued]
⎛ 3⎞
⎛ 2⎞
55. cot −1 ⎜ − ⎟ = tan −1 ⎜ − ⎟
2
⎝
⎠
⎝ 3⎠
We are finding the angle θ , 0 ≤ θ ≤ π , whose
59. Let θ = sin −1 u so that sin θ = u , −
−1 ≤ u ≤ 1 . Then,
(
2
2
. Now tan θ = − , so θ
3
3
lies in quadrant II. The calculator yields
⎛ 2⎞
tan −1 ⎜ − ⎟ ≈ −0.59 , which is an angle in
⎝ 3⎠
quadrant IV. Since θ is in quadrant II,
⎛ 3⎞
θ ≈ −0.59 + π ≈ 2.55 . Thus, cot −1 ⎜ − ⎟ ≈ 2.55 .
⎝ 2⎠
)
=
=
)
1
1
. Now tan θ = −
, so θ
10
10
lies in quadrant II. The calculator yields
⎛ 1 ⎞
tan −1 ⎜ −
⎟ ≈ −0.306 , which is an angle in
⎝ 10 ⎠
quadrant IV. Since θ is in quadrant II,
(
θ≠
(
π
2
57. Let θ = tan u so that tan θ = u , −
−∞ < u < ∞ . Then,
(
2
<θ <
2
1 − sin 2 θ
1− u2
sin 2 θ
1 − cos 2 θ
=
cos θ
cos θ
=
1− u2
u
)
sin sec −1 u = sin θ = sin 2 θ = 1 − cos 2 θ
= 1−
1
sec θ
2
=
sec2 θ − 1
sec 2 θ
u2 −1
u
62. Let θ = cot −1 u so that cot θ = u , 0 < θ < π ,
−∞ < u < ∞ . Then,
1
sin cot −1 u = sin θ = sin 2 θ =
csc2 θ
1
1
=
=
2
1 + cot θ
1+ u2
,
(
)
1
1
cos tan u = cos θ =
=
sec θ
sec 2 θ
1
1
=
=
2
1 + tan θ
1+ u2
−1
)
58. Let θ = cos −1 u so that cos θ = u , 0 ≤ θ ≤ π ,
−1 ≤ u ≤ 1 . Then,
(
,
sin θ
=
=
π
2
, u ≥ 1 . Then,
)
π
cos θ
u
π
61. Let θ = sec −1 u so that sec θ = u , 0 ≤ θ ≤ π and
θ ≈ −0.306 + π ≈ 2.84 . So, cot −1 − 10 ≈ 2.84 .
−1
=
2
≤θ ≤
)
⎛ 1 ⎞
56. cot −1 − 10 = tan −1 ⎜ −
⎟
⎝ 10 ⎠
We are finding the angle θ , 0 ≤ θ ≤ π , whose
tangent equals −
sin θ
2
60. Let θ = cos −1 u so that cos θ = u , 0 ≤ θ ≤ π ,
−1 ≤ u ≤ 1 . Then,
sin θ
tan cos −1 u = tan θ =
cos θ
(
(
sin θ
cos θ
tan sin −1 u = tan θ =
tangent equals −
π
)
sin cos −1 u = sin θ = sin 2 θ
= 1 − cos 2 θ = 1 − u 2
225
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Chapter 3: Analytic Trigonometry
63. Let θ = csc−1 u so that csc θ = u , −
π
2
≤θ ≤
π
2
,
68.
u ≥ 1 . Then,
(
)
cos csc−1 u = cos θ = cos θ ⋅
=
sin θ
= cot θ sin θ
sin θ
cot θ
cot 2 θ
csc 2 θ − 1
=
=
csc θ
csc θ
csc θ
52 + y 2 = 132
25 + y 2 = 169
u2 −1
=
u
y 2 = 144
y = ± 144 = ±12
Since θ is in quadrant I, y = 12 .
64. Let θ = sec−1 u so that sec θ = u , 0 ≤ θ ≤ π and
θ≠
π
(
2
, u ≥ 1 . Then,
)
cos sec −1 u = cos θ =
⎛
5⎞
y 12
⎛ 5 ⎞⎞
⎛
f ⎜ g −1 ⎜ ⎟ ⎟ = sin ⎜ cos −1 ⎟ = sin θ = =
13 ⎠
r 13
⎝ 13 ⎠ ⎠
⎝
⎝
1
1
=
sec θ u
⎛
69. g −1 ⎜
⎝
−1
65. Let θ = cot u so that cot θ = u , 0 < θ < π ,
−∞ < u < ∞ . Then,
1
1
tan cot −1 u = tan θ =
=
cot θ u
(
)
π
(
⎛ 7π
f⎜
⎝ 4
2
70.
, u ≥ 1 . Then,
−1
)
⎛ ⎛ 5π ⎞ ⎞
5π ⎞
−1 ⎛
f −1 ⎜ g ⎜
⎟ ⎟ = sin ⎜ cos
⎟
6 ⎠
⎝
⎝ ⎝ 6 ⎠⎠
⎛
π
3⎞
= sin −1 ⎜⎜ −
⎟⎟ = −
3
⎝ 2 ⎠
tan sec u = tan θ = tan θ
2
= sec2 θ − 1 = u 2 − 1
⎛
⎛
⎛ 3 ⎞⎞
⎛ 3 ⎞⎞
71. h ⎜ f −1 ⎜ − ⎟ ⎟ = tan ⎜ sin −1 ⎜ − ⎟ ⎟
5
⎝
⎠⎠
⎝ 5 ⎠⎠
⎝
⎝
3
⎛ 3⎞
Let θ = sin −1 ⎜ − ⎟ . Since sin θ = − and
5
⎝ 5⎠
⎛
12 ⎞
⎛ 12 ⎞ ⎞
⎛
67. g ⎜ f −1 ⎜ ⎟ ⎟ = cos ⎜ sin −1 ⎟
13 ⎠
⎝ 13 ⎠ ⎠
⎝
⎝
12
12
and
Let θ = sin −1 . Since sin θ =
13
13
π
≤θ ≤
π
π
≤ θ ≤ , θ is in quadrant IV, and we let
2
2
y = −3 and r = 5 . Solve for x:
−
π
, θ is in quadrant I, and we let
2
2
y = 12 and r = 13 . Solve for x:
−
7π ⎞
⎞⎞
−1 ⎛
⎟ ⎟ = cos ⎜ sin
⎟
4 ⎠
⎠⎠
⎝
⎛
2 ⎞ 3π
= cos −1 ⎜⎜ −
⎟⎟ =
⎝ 2 ⎠ 4
66. Let θ = sec−1 u so that sec θ = u , 0 ≤ θ ≤ π and
θ≠
⎛
5⎞
⎛ 5 ⎞⎞
⎛
f ⎜ g −1 ⎜ ⎟ ⎟ = sin ⎜ cos −1 ⎟
13
13
⎝ ⎠⎠
⎝
⎠
⎝
5
5
and
Let θ = cos −1 . Since cos θ =
13
13
0 ≤ θ ≤ π , θ is in quadrant I, and we let x = 5
and r = 13 . Solve for y:
x 2 + (−3) 2 = 52
x 2 + 122 = 132
x 2 + 9 = 25
x 2 + 144 = 169
x 2 = 16
x = ± 16 = ±4
Since θ is in quadrant IV, x = 4 .
⎛
⎛
⎛ 3 ⎞⎞
⎛ 3 ⎞⎞
h ⎜ f −1 ⎜ − ⎟ ⎟ = tan ⎜ sin −1 ⎜ − ⎟ ⎟
5
⎝
⎠
⎝ 5 ⎠⎠
⎝
⎠
⎝
y −3
3
= tan θ = =
=−
x
4
4
x 2 = 25
x = ± 25 = ±5
Since θ is in quadrant I, x = 5 .
⎛
12 ⎞
x 5
⎛ 12 ⎞ ⎞
⎛
g ⎜ f −1 ⎜ ⎟ ⎟ = cos ⎜ sin −1 ⎟ = cos θ = =
13 ⎠
r 13
⎝ 13 ⎠ ⎠
⎝
⎝
226
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Section 3.2: The Inverse Trigonometric Functions [Continued]
⎛
75. g −1 ⎜
⎝
⎛
⎛
⎛ 4 ⎞⎞
⎛ 4 ⎞⎞
72. h ⎜ g −1 ⎜ − ⎟ ⎟ = tan ⎜ cos −1 ⎜ − ⎟ ⎟
5
⎝
⎠⎠
⎝ 5 ⎠⎠
⎝
⎝
4
⎛ 4⎞
Let θ = cos −1 ⎜ − ⎟ . Since cos θ = − and
5
⎝ 5⎠
0 ≤ θ ≤ π , θ is in quadrant II, and we let
x = −4 and r = 5 . Solve for y:
(−4) 2 + y 2 = 52
⎛ 4π ⎞ ⎞
⎛ 4π ⎞ ⎞
−1 ⎛
f ⎜−
⎟ ⎟ = cos ⎜ sin ⎜ −
⎟⎟
3
⎝
⎠⎠
⎝ ⎝ 3 ⎠⎠
= cos −1
⎛
76. g −1 ⎜
⎝
⎛ 5π
f ⎜−
⎝ 6
⎞⎞
⎛ 5π
−1 ⎛
⎟ ⎟ = cos ⎜ sin ⎜ −
⎠⎠
⎝ ⎝ 6
y =9
2
y = ± 9 = ±3
Since θ is in quadrant II, y = 3 .
⎛
⎛
⎛ 1 ⎞⎞
⎛ 1 ⎞⎞
77. h ⎜ g −1 ⎜ − ⎟ ⎟ = tan ⎜ cos −1 ⎜ − ⎟ ⎟
⎝ 4 ⎠⎠
⎝ 4 ⎠⎠
⎝
⎝
1
⎛ 1⎞
Let θ = cos −1 ⎜ − ⎟ . Since cos θ = − and
4
4
⎝
⎠
0 ≤ θ ≤ π , θ is in quadrant II, and we let
x = −1 and r = 4 . Solve for y:
(−1) 2 + y 2 = 42
⎛
⎛
⎛ 4 ⎞⎞
⎛ 4 ⎞⎞
h ⎜ g −1 ⎜ − ⎟ ⎟ = tan ⎜ cos −1 ⎜ − ⎟ ⎟
⎝ 5 ⎠⎠
⎝ 5 ⎠⎠
⎝
⎝
y −3
3
= tan θ = =
=−
x
4
4
⎛ ⎛ 12 ⎞ ⎞
12 ⎞
⎛
73. g ⎜ h −1 ⎜ ⎟ ⎟ = cos ⎜ tan −1 ⎟
5⎠
⎝
⎝ ⎝ 5 ⎠⎠
12
12
and
Let θ = tan −1 . Since tan θ =
5
5
π
1 + y 2 = 16
y 2 = 15
y = ± 15
π
Since θ is in quadrant II, y = 15 .
≤ θ ≤ , θ is in quadrant I, and we let
2
2
x = 5 and y = 12 . Solve for r:
⎛
⎛
⎛ 1 ⎞⎞
⎛ 1 ⎞⎞
h ⎜ g −1 ⎜ − ⎟ ⎟ = tan ⎜ cos −1 ⎜ − ⎟ ⎟
⎝ 4 ⎠⎠
⎝ 4 ⎠⎠
⎝
⎝
r 2 = 52 + 122
= tan θ =
r 2 = 25 + 144 = 169
r = ± 169 = ±13
Now, r must be positive, so r = 13 .
⎛ ⎛ 12 ⎞ ⎞
12 ⎞
x 5
⎛
g ⎜ h −1 ⎜ ⎟ ⎟ = cos ⎜ tan −1 ⎟ = cos θ = =
5
5
13
r
⎝
⎠
⎝ ⎝ ⎠⎠
74.
π
≤θ ≤
π
y
15
=
= − 15
x
−1
⎛
⎛
⎛ 2 ⎞⎞
⎛ 2 ⎞⎞
78. h ⎜ f −1 ⎜ − ⎟ ⎟ = tan ⎜ sin −1 ⎜ − ⎟ ⎟
5
⎝
⎠⎠
⎝ 5 ⎠⎠
⎝
⎝
2
⎛ 2⎞
Let θ = sin −1 ⎜ − ⎟ . Since sin θ = − and
5
⎝ 5⎠
⎛ ⎛ 5 ⎞⎞
5⎞
⎛
f ⎜ h −1 ⎜ ⎟ ⎟ = sin ⎜ tan −1 ⎟
12 ⎠
⎝
⎝ ⎝ 12 ⎠ ⎠
5
5
and
Let θ = tan −1 . Since tan θ =
12
12
−
⎞⎞
⎟⎟
⎠⎠
⎛ 1 ⎞ 2π
= cos −1 ⎜ − ⎟ =
⎝ 2⎠ 3
16 + y 2 = 25
−
3 π
=
2
6
π
π
≤ θ ≤ , θ is in quadrant IV, and we let
2
2
y = −2 and r = 5 . Solve for x:
−
x 2 + (−2)2 = 52
x 2 + 4 = 25
, θ is in quadrant I, and we let
2
2
x = 12 and y = 5 . Solve for r:
x 2 = 21
x = ± 21
Since θ is in quadrant IV, x = 21 .
⎛
⎛
⎛ 2 ⎞⎞
⎛ 2 ⎞⎞
h ⎜ f −1 ⎜ − ⎟ ⎟ = tan ⎜ sin −1 ⎜ − ⎟ ⎟
⎝ 5 ⎠⎠
⎝ 5 ⎠⎠
⎝
⎝
r 2 = 122 + 52
r 2 = 144 + 25 = 169
r = ± 169 = ±13
Now, r must be positive, so r = 13 .
⎛ ⎛ 5 ⎞⎞
5⎞
y 5
⎛
f ⎜ h −1 ⎜ ⎟ ⎟ = sin ⎜ tan −1 ⎟ = sin θ = =
12 ⎠
r 13
⎝
⎝ ⎝ 12 ⎠ ⎠
= tan θ =
y
−2
2 21
=
=−
x
21
21
227
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Chapter 3: Analytic Trigonometry
79. a.
b.
Since the diameter of the base is 45 feet, we
45
= 22.5 feet. Thus,
have r =
2
⎛ 22.5 ⎞
θ = cot −1 ⎜
⎟ = 31.89° .
⎝ 14 ⎠
θ = cot −1
81. a.
⎛ 2x ⎞
⎜ 2 y + gt 2 ⎟⎟
⎝
⎠
The artillery shell begins at the origin and
lands at the coordinates ( 6175, 2450 ) . Thus,
r
h
⎛
θ = cot −1 ⎜
the diameter is 2 ( 27.32 ) = 54.64 feet.
The artilleryman used an angle of elevation
of 22.3° .
From part (b), we get h =
≈ cot −1 ( 2.437858 ) ≈ 22.3°
r
.
cot θ
b.
sec θ =
θ = cot
x sec θ ( 6175 ) sec ( 22.3° )
=
t
2.27
= 2940.23 ft/sec
82. Let. y = cot −1 x = cos −1
−10
r
h
x +1
10
−
r
r
→ h=
h
cot θ
Here we have θ = 50.14° and r = 4 feet.
4
= 4.79 feet. The
Thus, h =
cot ( 50.14° )
π
2
Note that the range of y = cot −1 x is ( 0, π ) , so
tan −1
1
will not work.
x
83. y = sec −1 x = cos −1
bunker will be 4.79 feet high.
θTG
x
2
3π
2
cot θ =
c.
v0 t
x
v0 =
Since the diameter of the base is 6.68 feet,
6.68
= 3.34 feet. Thus,
we have r =
2
⎛ 3.34 ⎞
θ = cot −1 ⎜
⎟ = 50.14°
⎝ 4 ⎠
−1
2 ⋅ 6175
⎜ 2 ⋅ 2450 + 32.2 2.27 2
( )
⎝
122
= 61 feet.
2
r
61
h=
=
≈ 37.96 feet.
cot θ 22.5 /14
Thus, the height is 37.96 feet.
b.
⎞
⎟
⎟
⎠
r
→ r = h cot θ
h
Here we have θ = 31.89° and h = 17 feet.
Thus, r = 17 cot ( 31.89° ) = 27.32 feet and
The radius is
80. a.
2x
2 y + gt 2
θ = cot −1 ⎜
cot θ =
c.
cot θ =
π
−1 ⎛
4.22 ⎞
= cot ⎜
⎟ = 54.88°
⎝ 6 ⎠
From part (a) we have θUSGA = 50.14° . 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.
−5
0
1
x
5
228
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Section 3.3: Trigonometric Identities
84. y = csc −1 x = sin −1
_π
1
x
13.
2
−10
10
−
_π
2
85 – 86. Answers will vary.
sin θ + cos θ cos θ − sin θ
+
cos θ
sin θ
2
sin θ + sin θ cos θ + cos θ ( cos θ − sin θ )
=
sin θ cos θ
sin 2 θ + sin θ cos θ + cos 2 θ − cos θ sin θ
=
sin θ cos θ
sin 2 θ + cos 2 θ + sin θ cos θ − cos θ sin θ
=
sin θ cos θ
1
=
sin θ cos θ
Section 3.3
1
1
1 + cos v + 1 − cos v
+
=
1 − cos v 1 + cos v (1 − cos v )(1 + cos v )
1. True
=
14.
2
1 − cos 2 v
2
=
sin 2 v
2. True
3. identity; conditional
4. −1
15.
5. 0
7. False
8. True
9. tan θ ⋅ csc θ =
sin θ
1
1
⋅
=
cos θ sin θ cos θ
10. cot θ ⋅ sec θ =
cos θ
1
1
⋅
=
sin θ cos θ sin θ
12.
sin θ cos θ
sin θ + 2sin θ cos θ + cos 2 θ − 1
=
sin θ cos θ
2
sin θ + cos 2 θ + 2sin θ cos θ − 1
=
sin θ cos θ
1 + 2sin θ cos θ − 1
=
sin θ cos θ
2sin θ cos θ
=
sin θ cos θ
=2
2
6. True
11.
( sin θ + cos θ )( sin θ + cos θ ) − 1
16.
cos θ 1 + sin θ cos θ (1 + sin θ )
⋅
=
1 − sin θ 1 + sin θ
1 − sin 2 θ
cos θ (1 + sin θ )
=
cos 2 θ
1 + sin θ
=
cos θ
( tan θ + 1)( tan θ + 1) − sec2 θ
tan θ
tan θ + 2 tan θ + 1 − sec2 θ
=
tan θ
2
tan θ + 1 + 2 tan θ − sec 2 θ
=
tan θ
2
sec θ + 2 tan θ − sec 2 θ
=
tan θ
2 tan θ
=
tan θ
=2
2
sin θ 1 − cos θ sin θ (1 − cos θ )
⋅
=
1 + cos θ 1 − cos θ
1 − cos 2 θ
sin θ (1 − cos θ )
=
sin 2 θ
1 − cos θ
=
sin θ
229
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
17.
3sin 2 θ + 4sin θ + 1 ( 3sin θ + 1)( sin θ + 1)
=
sin 2 θ + 2sin θ + 1
( sin θ + 1)( sin θ + 1)
=
18.
1
− cos 2 u
sin u
= 1 − cos 2 u
26. sin u csc u − cos 2 u = sin u ⋅
3sin θ + 1
sin θ + 1
= sin 2 u
( cos θ + 1)( cos θ − 1)
cos 2 θ − 1
=
2
cos θ ( cos θ − 1)
cos θ − cos θ
=
27. (sec θ − 1)(sec θ + 1) = sec 2 θ − 1 = tan 2 θ
28. (csc θ − 1)(csc θ + 1) = csc 2 θ − 1 = cot 2 θ
cos θ + 1
cos θ
19. csc θ ⋅ cos θ =
1
cos θ
⋅ cos θ =
= cot θ
sin θ
sin θ
20. sec θ ⋅ sin θ =
1
sin θ
⋅ sin θ =
= tan θ
cos θ
cos θ
29. (sec θ + tan θ )(sec θ − tan θ ) = sec 2 θ − tan 2 θ = 1
30. (csc θ + cot θ )(csc θ − cot θ ) = csc2 θ − cot 2 θ = 1
31. cos 2 θ (1 + tan 2 θ ) = cos 2 θ ⋅ sec2 θ
1
= cos 2 θ ⋅
cos 2 θ
=1
21. 1 + tan 2 (−θ ) = 1 + (− tan θ ) 2 = 1 + tan 2 θ = sec2 θ
22. 1 + cot 2 (−θ ) = 1 + (− cot θ ) 2 = 1 + cot 2 θ = csc2 θ
32. (1 − cos 2 θ )(1 + cot 2 θ ) = sin 2 θ ⋅ csc2 θ
1
= sin 2 θ ⋅ 2
sin θ
=1
⎛ sin θ cos θ ⎞
23. cos θ (tan θ + cot θ ) = cos θ ⎜
+
⎟
⎝ cos θ sin θ ⎠
⎛ sin 2 θ + cos 2 θ ⎞
= cos θ ⎜
⎟
⎝ cos θ sin θ ⎠
33. (sin θ + cos θ ) 2 + (sin θ − cos θ ) 2
= sin 2 θ + 2sin θ cos θ + cos 2 θ
1
⎛
⎞
= cos θ ⎜
⎟
⎝ cos θ sin θ ⎠
1
=
sin θ
= csc θ
+ sin 2 θ − 2sin θ cos θ + cos 2 θ
= 2sin 2 θ + 2 cos 2 θ
= 2(sin 2 θ + cos 2 θ )
= 2 ⋅1
=2
⎛ cos θ sin θ ⎞
24. sin θ (cot θ + tan θ ) = sin θ ⎜
+
⎟
⎝ sin θ cos θ ⎠
⎛ cos 2 θ + sin 2 θ ⎞
= sin θ ⎜
⎟
⎝ sin θ cos θ ⎠
34. tan 2 θ cos 2 θ + cot 2 θ sin 2 θ
sin 2 θ
cos 2 θ
2
cos
θ
⋅
+
⋅ sin 2 θ
cos 2 θ
sin 2 θ
= sin 2 θ + cos 2 θ
=
1
⎛
⎞
= sin θ ⎜
⎟
⎝ sin θ cos θ ⎠
1
=
cos θ
= sec θ
=1
35. sec 4 θ − sec2 θ = sec 2 θ (sec2 θ − 1)
= (tan 2 θ + 1) tan 2 θ
= tan 4 θ + tan 2 θ
1
− cos 2 u
tan u
= 1 − cos 2 u
25. tan u cot u − cos 2 u = tan u ⋅
36. csc 4 θ − csc 2 θ = csc2 θ (csc2 θ − 1)
= (cot 2 θ + 1) cot 2 θ
= sin 2 u
= cot 4 θ + cot 2 θ
230
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.3: Trigonometric Identities
1
sin u
−
cos u cos u
⎛ 1 − sin u ⎞ ⎛ 1 + sin u ⎞
=⎜
⎟ ⋅⎜
⎟
⎝ cos u ⎠ ⎝ 1 + sin u ⎠
37. sec u − tan u =
=
42. 1 −
1 − sin 2 u
cos u (1 + sin u )
= 1 − 1 − cos θ
= − cos θ
cos 2 u
cos u (1 + sin u )
cos u
=
1 + sin u
=
1
1+
1 + tan v
cot v
=
43.
1 − tan v 1 − 1
cot v
1 ⎞
⎛
⎜1 +
⎟ cot v
= ⎝ cot v ⎠
1 ⎞
⎛
⎜1 −
⎟ cot v
⎝ cot v ⎠
cot v + 1
=
cot v − 1
1
cos u
−
sin u sin u
⎛ 1 − cos u ⎞ ⎛ 1 + cos u ⎞
=⎜
⎟⋅⎜
⎟
⎝ sin u ⎠ ⎝ 1 + cos u ⎠
1 − cos 2 u
=
sin u (1 + cos u )
38. csc u − cot u =
sin 2 u
sin u (1 + cos u )
sin u
=
1 + cos u
1
−1
csc v − 1 sin v
44.
=
1
csc v + 1
+1
sin v
⎛ 1
⎞
− 1⎟ sin v
⎜
= ⎝ sin v ⎠
⎛ 1
⎞
+ 1⎟ sin v
⎜
⎝ sin v ⎠
1 − sin v
=
1 + sin v
=
39. 3sin 2 θ + 4 cos 2 θ = 3sin 2 θ + 3cos 2 θ + cos 2 θ
= 3(sin 2 θ + cos 2 θ ) + cos 2 θ
= 3 ⋅1 + cos 2 θ
= 3 + cos 2 θ
40. 9sec 2 θ − 5 tan 2 θ = 4sec2 θ + 5sec2 θ − 5 tan 2 θ
1
sec θ sin θ cos θ sin θ
+
=
+
45.
1
csc θ cos θ
cos θ
sin θ
sin θ sin θ
=
+
cos θ cos θ
= tan θ + tan θ
= 2 tan θ
= 4sec 2 θ + 5(sec2 θ − tan 2 θ )
= 4sec 2 θ + 5 ⋅1
= 5 + 4sec 2 θ
41. 1 −
sin 2 θ
1 − cos 2 θ
= 1−
1 − cos θ
1 − cos θ
(1 − cos θ )(1 + cos θ )
= 1−
1 − cos θ
= 1 − (1 + cos θ )
cos 2 θ
1 − sin 2 θ
= 1−
1 + sin θ
1 + sin θ
(1 − sin θ )(1 + sin θ )
= 1−
1 + sin θ
= 1 − (1 − sin θ )
46.
= 1 − 1 + sin θ
= sin θ
csc θ − 1 csc θ − 1 csc θ + 1
=
⋅
cot θ
cot θ csc θ + 1
csc 2 θ − 1
=
cot θ (csc θ + 1)
cot 2 θ
cot θ (csc θ + 1)
cot θ
=
csc θ + 1
=
231
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
1
1+
1 + sin θ
csc θ
=
47.
1 − sin θ 1 − 1
csc θ
csc θ + 1
= csc θ
csc θ − 1
csc θ
csc θ + 1 csc θ
=
⋅
csc θ csc θ − 1
csc θ + 1
=
csc θ − 1
50.
cos v
1 + sin v cos 2 v + (1 + sin v) 2
+
=
1 + sin v
cos v
cos v(1 + sin v)
=
cos 2 v + 1 + 2sin v + sin 2 v
cos v(1 + sin v)
2 + 2sin v
cos v(1 + sin v)
2(1 + sin v)
=
cos v(1 + sin v)
2
=
cos v
= 2sec v
=
1
(cos θ + 1) ⋅
cos θ + 1
θ
cos
48.
=
cos θ − 1 (cos θ − 1) ⋅ 1
cos θ
1
1+
cos θ
=
1
1−
cos θ
1 + sec θ
=
1 − sec θ
1
sin θ
sin θ
θ
sin
51.
=
⋅
1
sin θ − cos θ sin θ − cos θ
sin θ
1
=
cos θ
1−
sin θ
1
=
1 − cot θ
1 − sin v
cos v
(1 − sin v) 2 + cos 2 v
49.
+
=
cos v
1 − sin v
cos v(1 − sin v)
52. 1 −
=
1 − 2sin v + sin 2 v + cos 2 v
cos v(1 − sin v)
1 − 2sin v + 1
cos v(1 − sin v)
2 − 2sin v
=
cos v(1 − sin v)
2(1 − sin v)
=
cos v(1 − sin v)
2
=
cos v
= 2sec v
=
sin 2 θ
1 + cos θ − sin 2 θ
=
1 + cos θ
1 + cos θ
1 + cos θ − (1 − cos 2 θ )
=
1 + cos θ
1 + cos θ − 1 + cos 2 θ
=
1 + cos θ
cos θ + cos 2 θ
=
1 + cos θ
cos θ (1 + cos θ )
=
1 + cos θ
= cos θ
232
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Section 3.3: Trigonometric Identities
53.
(sec θ − tan θ ) 2
56.
= sec θ − 2sec θ tan θ + tan θ
2
2
1
1 sin θ sin 2 θ
−
2
⋅
⋅
+
cos θ cos θ cos 2 θ
cos 2 θ
1 − 2sin θ + sin 2 θ
=
cos 2 θ
(1 − sin θ )(1 − sin θ )
=
1 − sin 2 θ
(1 − sin θ )(1 − sin θ )
=
(1 − sin θ )(1 + sin θ )
1 − sin θ
=
1 + sin θ
=
54. (csc θ − cot θ ) 2
= csc 2 θ − 2 csc θ cot θ + cot 2 θ
1
1 cos θ cos 2 θ
−
⋅
⋅
+
2
sin θ sin θ sin 2 θ
sin 2 θ
1 − 2 cos θ + cos 2 θ
=
sin 2 θ
(1 − cos θ )(1 − cos θ )
=
1 − cos 2 θ
(1 − cos θ )(1 − cos θ )
=
(1 − cos θ )(1 + cos θ )
1 − cos θ
=
1 + cos θ
=
cot θ
tan θ
+
1 − tan θ 1 − cot θ
cos θ
sin θ
= sin θ + cos θ
sin θ
cos θ
1−
1−
cos θ
sin θ
cos θ
sin θ
sin θ
cos θ
=
+
cos θ − sin θ sin θ − cos θ
cos θ
sin θ
cos 2 θ
sin 2 θ
=
+
sin θ (cos θ − sin θ ) cos θ (sin θ − cos θ )
=
− cos 2 θ ⋅ cos θ + sin 2 θ ⋅ sin θ
sin θ cos θ (sin θ − cos θ )
=
sin 3 θ − cos3 θ
sin θ cos θ (sin θ − cos θ )
=
(sin θ − cos θ )(sin 2 θ + sin θ cos θ + cos 2 θ )
sin θ cos θ (sin θ − cos θ )
sin 2 θ + sin θ cos θ + cos 2 θ
sin θ cos θ
2
sin θ cos θ
cos 2 θ
sin θ
+
+
=
sin θ cos θ sin θ cos θ sin θ cos θ
sin θ
cos θ
=
+1+
cos θ
sin θ
= 1 + tan θ + cot θ
=
57. tan θ +
cos θ
sin θ
55.
+
1 − tan θ 1 − cot θ
cos θ
sin θ
=
+
sin θ
cos θ
1−
1−
cos θ
sin θ
cos θ
sin θ
=
+
cos θ − sin θ sin θ − cos θ
cos θ
sin θ
cos 2 θ
sin 2 θ
=
+
cos θ − sin θ sin θ − cos θ
cos 2 θ − sin 2 θ
=
cos θ − sin θ
(cos θ − sin θ )(cos θ + sin θ )
=
cos θ − sin θ
= sin θ + cos θ
cos θ
sin θ
cos θ
=
+
1 + sin θ cos θ 1 + sin θ
sin θ (1 + sin θ ) + cos 2 θ
=
cos θ (1 + sin θ )
sin θ + sin 2 θ + cos 2 θ
cos θ (1 + sin θ )
sin θ + 1
=
cos θ (1 + sin θ )
1
=
cos θ
= sec θ
=
233
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Chapter 3: Analytic Trigonometry
sin θ cos θ
−
tan θ − cot θ cos θ sin θ
=
61.
tan θ + cot θ sin θ + cos θ
cos θ sin θ
sin 2 θ − cos 2 θ
θ sin θ
= cos
sin 2 θ + cos 2 θ
cos θ sin θ
sin 2 θ − cos 2 θ
=
1
= sin 2 θ − cos 2 θ
1
(sin θ cos θ ) ⋅
sin θ cos θ
cos 2 θ
=
58.
2
2
cos θ − sin θ (cos 2 θ − sin 2 θ ) ⋅ 1
cos 2 θ
sin θ
= cos θ2
sin θ
1−
cos 2 θ
tan θ
=
1 − tan 2 θ
59.
tan θ + sec θ − 1
tan θ − sec θ + 1
tan θ + (sec θ − 1) tan θ + (sec θ − 1)
=
⋅
tan θ − (sec θ − 1) tan θ + (sec θ − 1)
=
1
cos 2 θ
−
sec θ − cos θ cos θ cos θ
=
62.
sec θ + cos θ
1
cos 2 θ
+
cos θ cos θ
1 − cos 2 θ
= cos θ2
1 + cos θ
cos θ
1 − cos 2 θ
=
1 + cos 2 θ
sin 2 θ
=
1 + cos 2 θ
tan 2 θ + 2 tan θ (sec θ − 1) + sec2 θ − 2sec θ + 1
tan 2 θ − (sec2 θ − 2sec θ + 1)
sec 2 θ − 1 + 2 tan θ (sec θ − 1) + sec2 θ − 2secθ + 1
sec2 θ − 1 − sec 2 θ + 2sec θ − 1
2
2sec θ − 2sec θ + 2 tan θ (sec θ − 1)
=
2sec θ − 2
2sec θ (sec θ − 1) + 2 tan θ (sec θ − 1)
=
2sec θ − 2
2(sec θ − 1)(sec θ + tan θ )
=
2(sec θ − 1)
= tan θ + sec θ
=
60.
sin u cos u
−
tan u − cot u
63.
+ 1 = cos u sin u + 1
sin u cos u
tan u + cot u
+
cos u sin u
sin 2 u − cos 2 u
u sin u + 1
= cos
sin 2 u + cos 2 u
cos u sin u
sin 2 u − cos 2 u
=
+1
1
= sin 2 u − cos 2 u + 1
sin θ − cos θ + 1
sin θ + cos θ − 1
(sin θ − cos θ ) + 1 (sin θ + cos θ ) + 1
=
⋅
(sin θ + cos θ ) − 1 (sin θ + cos θ ) + 1
=
sin 2 θ − cos 2 θ + sin θ + cos θ + sin θ − cos θ + 1
(sin θ + cos θ ) 2 − 1
sin 2 θ − cos 2 θ + 2sin θ + 1
sin 2 θ + 2sin θ cos θ + cos 2 θ − 1
sin 2 θ − (1 − sin 2 θ ) + 2sin θ + 1
=
2sin θ cos θ + 1 − 1
2sin 2 θ + 2sin θ
=
2sin θ cos θ
2sin θ (sin θ + 1)
=
2sin θ cos θ
sin θ + 1
=
cos θ
=
= sin 2 u + (1 − cos 2 u )
= sin 2 u + sin 2 u
= 2sin 2 u
234
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Section 3.3: Trigonometric Identities
sin u cos u
−
tan u − cot u
64.
+ 2 cos 2 u = cos u sin u + 2 cos 2 u
sin u cos u
tan u + cot u
+
cos u sin u
sin 2 u − cos 2 u
u sin u + 2 cos 2 u
= cos
sin 2 u + cos 2 u
cos u sin u
sin 2 u − cos 2 u
=
+ 2 cos 2 u
1
= sin 2 u + cos 2 u
67.
68.
=1
1
sin θ
+
sec θ + tan θ cos θ cos θ
=
65.
cot θ + cos θ cos θ + cos θ
sin θ
1 + sin θ
cos θ
=
cos θ + cos θ sin θ
sin θ
1 + sin θ
sin θ
=
⋅
cos θ cos θ (1 + sin θ )
sin θ
1
=
⋅
cos θ cos θ
= tan θ sec θ
1
sec θ
cos
θ
66.
=
1 + sec θ 1 + 1
cos θ
1
= cos θ
cos θ + 1
cos θ
1
⎛
=⎜
⎝ 1 + cos θ
1 − cos θ
=
1 − cos 2 θ
1 − cos θ
=
sin 2 θ
1 − tan 2 θ
1 − tan 2 θ + 1 + tan 2 θ
+1 =
2
1 + tan θ
1 + tan 2 θ
2
=
sec 2 θ
1
= 2⋅ 2
sec θ
= 2 cos 2 θ
1 − cot 2 θ
1 − cot 2 θ
+ 2 cos 2 θ =
+ 2 cos 2 θ
2
2
1 + cot θ
csc θ
1
cot 2 θ
=
−
+ 2 cos 2 θ
csc 2 θ csc 2 θ
cos 2 θ
2
= sin 2 θ − sin θ + 2 cos 2 θ
1
sin 2 θ
2
= sin θ − cos 2 θ + 2 cos 2 θ
= sin 2 θ + cos 2 θ
=1
1
1
−
sec θ − csc θ cos θ sin θ
69.
=
1
1
sec θ csc θ
⋅
cos θ sin θ
sin θ − cos θ
= cos θ sin θ
1
cos θ sin θ
sin θ − cos θ cos θ sin θ
=
⋅
cos θ sin θ
1
= sin θ − cos θ
⎞ ⎛ 1 − cos θ ⎞
⎟ ⋅⎜
⎟
⎠ ⎝ 1 − cos θ ⎠
235
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Chapter 3: Analytic Trigonometry
70.
sin 2 θ − tan θ
cos 2 θ − cot θ
sin θ
sin 2 θ −
θ
cos
=
cos θ
2
cos θ −
sin θ
2
sin θ cos θ − sin θ
cos θ
=
2
cos θ sin θ − cos θ
sin θ
2
sin θ cos θ − sin θ
sin θ
=
⋅
cos θ
cos 2 θ sin θ − cos θ
sin θ (sin θ cos θ − 1)
sin θ
=
⋅
cos θ
cos θ (cos θ sin θ − 1)
74.
1 + 2sin θ + sin 2 θ − (1 − 2sin θ + sin 2 θ )
1 − sin 2 θ
4sin θ
=
cos 2 θ
sin θ
1
= 4⋅
⋅
cos θ cos θ
= 4 tan θ sec θ
=
75.
sin 2 θ
cos 2 θ
= tan 2 θ
=
1
− cos θ
cos θ
1 − cos 2 θ
=
cos θ
sin 2 θ
=
cos θ
sin θ
= sin θ ⋅
cos θ
= sin θ tan θ
71. sec θ − cos θ =
76.
sec θ
⎛ sec θ ⎞ ⎛ 1 + sin θ ⎞
=⎜
⎟ ⋅⎜
⎟
1 − sin θ ⎝ 1 − sin θ ⎠ ⎝ 1 + sin θ ⎠
sec θ (1 + sin θ )
=
1 − sin 2 θ
sec θ (1 + sin θ )
=
cos 2 θ
1 1 + sin θ
=
⋅
cos θ cos 2 θ
1 + sin θ
=
cos3 θ
1 + sin θ (1 + sin θ )(1 + sin θ )
=
1 − sin θ (1 − sin θ )(1 + sin θ )
(1 + sin θ ) 2
1 − sin 2 θ
(1 + sin θ ) 2
=
cos 2 θ
=
sin θ cos θ
+
cos θ sin θ
sin 2 θ + cos 2 θ
=
sin θ cos θ
1
=
sin θ cos θ
1
1
=
⋅
cos θ sin θ
= sec θ csc θ
72. tan θ + cot θ =
73.
1 + sin θ 1 − sin θ
−
1 − sin θ 1 + sin θ
(1 + sin θ ) 2 − (1 − sin θ ) 2
=
(1 − sin θ )(1 + sin θ )
⎛ 1 + sin θ ⎞
=⎜
⎟
⎝ cos θ ⎠
2
sin θ ⎞
⎛ 1
=⎜
+
θ
θ ⎟⎠
cos
cos
⎝
= (sec θ + tan θ ) 2
2
1
1
1 + sin θ + 1 − sin θ
+
=
1 − sin θ 1 + sin θ (1 − sin θ )(1 + sin θ )
2
=
1 − sin 2 θ
2
=
cos 2 θ
= 2sec 2 θ
236
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Section 3.3: Trigonometric Identities
77.
(sec v − tan v) 2 + 1
csc v(sec v − tan v)
=
sec2 v − 2sec v tan v + tan 2 v + 1
csc v(sec v − tan v)
=
sec 2 v − 2sec v tan v + sec2 v
csc v(sec v − tan v)
80.
2sec 2 v − 2sec v tan v
csc v(sec v − tan v)
2sec v(sec v − tan v)
=
csc v(sec v − tan v)
2sec v
=
csc v
1
2⋅
= cos v
1
sin v
1 sin v
= 2⋅
⋅
cos v 1
sin v
= 2⋅
cos v
= 2 tan v
=
78.
79.
81.
82.
sec2 v − tan 2 v + tan v 1 + tan v
=
sec v
sec v
sin v
1+
cos v
=
1
cos v
⎛ sin v ⎞
= ⎜1 +
⎟ cos v
⎝ cos v ⎠
= cos v + sin v
sin θ + cos θ sin θ − cos θ
−
cos θ
sin θ
sin θ cos θ sin θ cos θ
=
+
−
+
cos θ cos θ sin θ sin θ
sin θ
cos θ
=
+1−1+
cos θ
sin θ
2
2
sin θ + cos θ
=
cos θ sin θ
1
=
cos θ sin θ
= sec θ csc θ
83.
sin θ + cos θ cos θ − sin θ
−
sin θ
cos θ
sin θ cos θ cos θ sin θ
=
+
−
+
sin θ sin θ cos θ cos θ
cos θ
sin θ
= 1+
−1+
sin θ
cos θ
cos 2 θ + sin 2 θ
=
cos θ sin θ
1
=
cos θ sin θ
= sec θ csc θ
sin 3 θ + cos3 θ
sin θ + cos θ
(sin θ + cos θ )(sin 2 θ − sin θ cos θ + cos 2 θ )
=
sin θ + cos θ
= sin 2 θ + cos 2 θ − sin θ cos θ
= 1 − sin θ cos θ
sin 3 θ + cos3 θ
1 − 2 cos 2 θ
(sin θ + cos θ )(sin 2 θ − sin θ cos θ + cos 2 θ )
=
1 − cos 2 θ − cos 2 θ
(sin θ + cos θ )(sin 2 θ + cos 2 θ − sin θ cos θ )
=
sin 2 θ − cos 2 θ
(sin θ + cos θ )(1 − sin θ cos θ )
=
(sin θ + cos θ )(sin θ − cos θ )
1
1 − sin θ cos θ cos θ
=
⋅
1
sin θ − cos θ
cos θ
1
− sin θ
= cos θ
sin θ
−1
cos θ
sec θ − sin θ
=
tan θ − 1
cos 2 θ − sin 2 θ cos 2 θ − sin 2 θ
=
1 − tan 2 θ
sin 2 θ
1−
cos 2 θ
2
cos θ − sin 2 θ
=
cos 2 θ − sin 2 θ
cos 2 θ
= ( cos 2 θ − sin 2 θ ) ⋅
cos 2 θ
cos 2 θ − sin 2 θ
= cos 2 θ
237
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Chapter 3: Analytic Trigonometry
84.
cos θ + sin θ − sin 3 θ cos θ sin θ sin 3 θ
=
+
−
sin θ
sin θ sin θ sin θ
= cot θ + 1 − sin 2 θ
88.
= cot θ + cos 2 θ
1 + 2cosθ + cos 2 θ + 2sin θ (1 + cosθ ) + sin 2 θ
1 + 2cosθ + cos 2 θ − sin 2 θ
1 + 2cosθ + cos 2 θ + 2sin θ (1 + cosθ ) + 1 − cos 2 θ
=
1 + 2cosθ + cos 2 θ − (1 − cos 2 θ )
=
2
85.
⎡⎣ 2 cos 2 θ − (sin 2 θ + cos 2 θ ) ⎤⎦
(2 cos 2 θ − 1)2
=
cos 4 θ − sin 4 θ (cos 2 θ − sin 2 θ )(cos 2 θ + sin 2 θ )
=
(cos 2 θ − sin 2 θ ) 2
(cos 2 θ − sin 2 θ )(cos 2 θ + sin 2 θ )
2 + 2 cos θ + 2sin θ (1 + cos θ )
2 cos θ + 2 cos 2 θ
2(1 + cos θ ) + 2sin θ (1 + cos θ )
=
2 cos θ (1 + cos θ )
2(1 + cos θ )(1 + sin θ )
=
2 cos θ (1 + cos θ )
1 + sin θ
=
cos θ
1
sin θ
=
+
cos θ cos θ
= sec θ + tan θ
=
cos 2 θ − sin 2 θ
cos 2 θ + sin 2 θ
= cos 2 θ − sin 2 θ
=
= 1 − sin 2 θ − sin 2 θ
= 1 − 2sin 2 θ
86.
87.
1 + cos θ + sin θ
1 + cos θ − sin θ
(1 + cos θ ) + sin θ (1 + cos θ ) + sin θ
=
⋅
(1 + cos θ ) − sin θ (1 + cos θ ) + sin θ
1 − 2 cos 2 θ 1 − cos 2 θ − cos 2 θ
=
sin θ cos θ
sin θ cos θ
2
sin θ − cos 2 θ
=
sin θ cos θ
sin 2 θ
cos 2 θ
=
−
sin θ cos θ sin θ cos θ
sin θ cos θ
=
−
cos θ sin θ
= tan θ − cot θ
89. (a sin θ + b cos θ ) 2 + (a cos θ − b sin θ ) 2
= a 2 sin 2 θ + 2ab sin θ cos θ + b 2 cos 2 θ
+ a 2 cos 2 θ − 2ab sin θ cos θ + b 2 sin 2 θ
= a 2 (sin 2 θ + cos 2 θ ) + b 2 (sin 2 θ + cos 2 θ )
= a 2 + b2
1 + sin θ + cos θ
1 + sin θ − cos θ
(1 + sin θ ) + cos θ (1 + sin θ ) + cos θ
=
⋅
(1 + sin θ ) − cos θ (1 + sin θ ) + cos θ
90. (2a sin θ cosθ ) 2 + a 2 (cos 2 θ − sin 2 θ ) 2
= 4a 2 sin 2 θ cos 2 θ
(
+ a 2 cos 4 θ − 2cos 2 θ sin 2 θ + sin 4 θ
1 + 2sin θ + sin 2 θ + 2cosθ (1 + sin θ ) + cos 2 θ
=
1 + 2sin θ + sin 2 θ − cos 2 θ
1 + 2sin θ + sin 2 θ + 2cosθ (1 + sin θ ) + (1 − sin 2 θ )
=
1 + 2sin θ + sin 2 θ − (1 − sin 2 θ )
2 + 2sin θ + 2cosθ (1 + sin θ )
=
2sin θ + 2sin 2 θ
2(1 + sin θ ) + 2cosθ (1 + sin θ )
=
2sin θ (1 + sin θ )
2(1 + sin θ )(1 + cosθ )
=
2sin θ (1 + sin θ )
1 + cosθ
=
sin θ
(
( cos θ + 2cos θ sin θ + sin θ )
( cos θ + sin θ )
)
= a 4sin θ cos θ + cos θ − 2cos θ sin θ + sin θ
2
= a2
= a2
= a 2 (1)
2
2
4
2
4
2
2
2
2
2
4
4
2
2
= a2
238
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)
Section 3.3: Trigonometric Identities
91.
tan α + tan β
tan α + tan β
=
1
1
cot α + cot β
+
tan α tan β
tan α + tan β
=
tan β + tan α
tan α tan β
98. ln sec θ + tan θ + ln sec θ − tan θ
= ln ( sec θ + tan θ ⋅ sec θ − tan θ
)
= ln sec2 θ − tan 2 θ
= ln 1
=0
⎛ tan α tan β ⎞
= (tan α + tan β ) ⋅ ⎜
⎟
⎝ tan α + tan β ⎠
= tan α tan β
99.
f ( x ) = sin x ⋅ tan x
= sin x ⋅
92. (tan α + tan β )(1 − cot α cot β )
+ (cot α + cot β )(1 − tan α tan β )
= tan α + tan β − tan α cot α cot β
− tan β cot α cot β + cot α + cot β
− cot α tan α tan β − cot β tan α tan β
= tan α + tan β − cot β − cot α + cot α
+ cot β − tan β − tan α
=0
sin x
cos x
sin 2 x
cos x
1 − cos 2 x
=
cos x
1
cos 2 x
=
−
cos x cos x
= sec x − cos x
=
= g ( x)
100.
93. (sin α + cos β ) 2 + (cos β + sin α )(cos β − sin α )
= sin 2 α + 2sin α cos β + cos 2 β + cos 2 β − sin 2 α
f ( x ) = cos x ⋅ cot x
= cos x ⋅
= 2sin α cos β + 2 cos 2 β
cos x
sin x
cos 2 x
sin x
1 − sin 2 x
=
sin x
1
sin 2 x
=
−
sin x sin x
= csc x − sin x
=
= 2 cos β (sin α + cos β )
94. (sin α − cos β ) 2 + (cos β + sin α )(cos β − sin α )
= sin 2 α − 2sin α cos β + cos 2 β + cos 2 β − sin 2 α
= − 2sin α cos β + 2cos 2 β = − 2cos β (sin α − cos β )
95. ln sec θ = ln
1
= ln cos θ
cos θ
96. ln tan θ = ln
sin θ
= ln sin θ − ln cos θ
cos θ
−1
= − ln cos θ
= g ( x)
97. ln 1 + cos θ + ln 1 − cos θ
= ln ( 1 + cos θ ⋅ 1 − cos θ
)
= ln 1 − cos 2 θ
= ln sin 2 θ
= 2 ln sin θ
239
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Chapter 3: Analytic Trigonometry
101.
f (θ ) =
=
=
=
=
=
1 − sin θ
cos θ
−
cos θ
1 + sin θ
(1 − sin θ )(1 + sin θ )
cos θ (1 + sin θ )
(
−
cos θ ⋅ cos θ
1
+
( sin θ ) ⋅ cos θ
1 − sin 2 θ − cos 2 θ
cos θ (1 + sin θ )
cos θ (1 + sin θ )
1
cos θ
⎛ 2
cos 2 θ
−
⎜⎜
2
2
⎝ cos θ cos θ
= 1200
1
cos θ
⎛ 2 − cos 2 θ
⎜⎜
2
⎝ cos θ
=
1−1
cos θ (1 + sin θ )
=
0
cos θ (1 + sin θ )
(
1200 1 + 1 − cos 2 θ
cos θ
)
1 ⎛ 2
⎞
− 1⎟
⎜
cos θ ⎝ cos 2 θ
⎠
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
3
(
1200 1 + sin 2 θ
)
cos θ
3
104. I t = 4 A2
( cscθ − 1)( secθ + tan θ )
csc θ sec θ
csc
−
1
sec θ + tan θ
θ
= 4 A2
⋅
csc θ
sec θ
1
tan θ ⎞
⎛
⎞
⎛
= 4 A2 ⎜ 1 −
⎟ ⎜1 +
⎟
⎝ csc θ ⎠ ⎝ sec θ ⎠
= g (θ )
f (θ ) = tan θ + sec θ
sin θ
1
+
cos θ cos θ
1 + sin θ
=
cos θ
1 + sin θ 1 − sin θ
=
⋅
cos θ 1 − sin θ
1 − sin 2 θ
=
cos θ (1 − sin θ )
=
=
= 1200
1 − ( sin 2 θ + cos 2 θ )
=0
102.
)
103. 1200sec θ 2sec2 θ − 1 = 1200
= 4 A2 (1 − sin θ )(1 + sin θ )
(
= 4 A2 1 − sin 2 θ
)
= 4 A2 cos 2 θ = ( 2 A cos θ )
2
105. Answers will vary.
cos 2 θ
cos θ (1 − sin θ )
106. sin 2 θ + cos 2 θ = 1
tan 2 θ + 1 = sec 2 θ
1 + cot 2 θ = csc 2 θ
cos θ
1 − sin θ
= g (θ )
=
107 – 108. Answers will vary.
240
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Section 3.4: Sum and Difference Formulas
Section 3.4
12. tan
(5 − 2)
1.
2
+ (1 − ( −3) )
2
= 32 + 42 = 9 + 16 = 25 = 5
2. −
3
5
2 1
2
⋅ =
2 2
4
3. a.
1−
b.
⎛ 1+ 3 ⎞ ⎛ 1+ 3 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 1− 3 ⎠ ⎝ 1+ 3 ⎠
1 1
=
2 2
4. −
5. −
=
4+2 3
−2
= cos120º ⋅ cos 45º − sin120º ⋅ sin 45º
8. False
1 2
3 2
=− ⋅
−
⋅
2 2
2 2
1
=−
2+ 6
4
5π
⎛ 3π 2π ⎞
= sin ⎜ +
⎟
12
⎝ 12 12 ⎠
π
π
π
π
= sin ⋅ cos + cos ⋅ sin
4
6
4
6
2 3
2 1
=
⋅
+
⋅
2 2
2 2
1
=
6+ 2
4
(
(
(
= sin 60º ⋅ cos 45º + cos 60º ⋅ sin 45º
)
3 2 1 2
⋅
+ ⋅
2 2 2 2
1
6+ 2
=
4
=
(
)
15. tan15º = tan(45º − 30º )
tan 45º − tan 30º
=
1 + tan 45º ⋅ tan 30º
3
1−
3 ⋅3
=
3 3
1 + 1⋅
3
3− 3 3− 3
=
⋅
3+ 3 3− 3
)
7π
⎛ 4π 3π ⎞
= cos ⎜
+ ⎟
12
⎝ 12 12 ⎠
π
π
π
π
= cos ⋅ cos − sin ⋅ sin
3
4
3
4
1 2
3 2
= ⋅
−
⋅
2 2
2 2
1
2− 6
=
4
(
)
14. sin105º = sin ( 60º + 45º )
π
⎛ 3π 2π ⎞
= sin ⎜ − ⎟
12
⎝ 12 12 ⎠
π
π
π
π
= sin ⋅ cos − cos ⋅ sin
4
6
4
6
2 3
2 1
=
⋅
−
⋅
2 2
2 2
1
=
6− 2
4
11. cos
1+ 2 3 + 3
1− 3
13. cos165º = cos (120º + 45º )
7. False
10. sin
=
= −2− 3
6. False
9. sin
7π
⎛ 3π 4π ⎞
= tan ⎜ +
⎟
12
⎝ 12 12 ⎠
π
π
tan + tan
4
3
=
π
π
1 − tan ⋅ tan
4
3
1+ 3
=
1 − 1⋅ 3
)
=
9−6 3 +3
9−3
=
12 − 6 3
6
= 2− 3
241
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Chapter 3: Analytic Trigonometry
16. tan195º = tan(135º +60º )
tan135º + tan 60º
=
1 − tan135º ⋅ tan 60º
−1 + 3
=
1 − (−1) ⋅ 3
=
⎛
π⎞
−1 + 3 1 − 3
⋅
1+ 3 1− 3
−1 + 2 3 − 3
1− 3
−4+ 2 3
=
−2
=
= 2− 3
17. sin
18. tan
4 2 −4 6
2−6
4 2 −4 6
=
−4
= 6− 2
17π
⎛ 15π 2π ⎞
= sin ⎜
+
⎟
12
⎝ 12 12 ⎠
π
π
5π
5π
= sin ⋅ cos + cos ⋅ sin
4
6
4
6
2 3 ⎛
2⎞ 1
=−
⋅
+ ⎜⎜ −
⎟⋅
2 2 ⎝ 2 ⎟⎠ 2
1
=−
6+ 2
4
(
=
⎛ 5π ⎞
=
1+ 2 3 + 3
1− 3
=
4+2 3
−2
−1
12
−1
=
⎛ 3π 2π ⎞
tan ⎜ +
⎟
⎝ 12 12 ⎠
−1
=
π
π
tan + tan
4
6
π
π
1 − tan ⋅ tan
4
6
π
π⎞
⎛
⎜ 1 − tan 4 ⋅ tan 6 ⎟
= −⎜
⎟
⎜⎜ tan π + tan π ⎟⎟
4
6 ⎠
⎝
1
1 − 1⋅
3⋅ 3
=−
1
3
1+
3
19π
⎛ 15π 4π ⎞
= tan ⎜
+
⎟
12
⎝ 12 12 ⎠
5π
π
tan
+ tan
4
3
=
5π
π
1 − tan ⋅ tan
4
3
1+ 3
=
1 − 1⋅ 3
1+ 3 1+ 3
⋅
1− 3 1+ 3
5π
20. cot ⎜ − ⎟ = − cot =
12 tan 5π
⎝ 12 ⎠
)
=
1
1
=
⎛ π⎞
⎛ 3π 4π ⎞
cos ⎜ − ⎟ cos ⎜ −
⎟
⎝ 12 ⎠
⎝ 12 12 ⎠
1
=
π
π
π
π
cos ⋅ cos + sin ⋅ sin
4
3
4
3
1
=
2 1
2 3
⋅ +
⋅
2 2 2 2
1
=
2+ 6
4
4
2− 6
=
⋅
2+ 6 2− 6
19. sec ⎜ − ⎟ =
⎝ 12 ⎠
=−
= −2− 3
3 −1 3 −1
⋅
3 +1 3 −1
3 − 3 − 3 +1
3 −1
4−2 3
=−
2
=−
= −2 + 3
242
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Section 3.4: Sum and Difference Formulas
21. sin 20º ⋅ cos10º + cos 20º ⋅ sin10º = sin(20º + 10º )
= sin 30º
1
=
2
29. cos
π
5π
5π
π
⎛ π 5π ⎞
⋅ cos + sin ⋅ sin = cos ⎜ − ⎟
12
12
12
12
⎝ 12 12 ⎠
⎛ 4π ⎞
= cos ⎜ − ⎟
⎝ 12 ⎠
⎛ π⎞
= cos ⎜ − ⎟
⎝ 3⎠
π
= cos
3
1
=
2
22. sin 20º ⋅ cos80º − cos 20º ⋅ sin 80º = sin(20º − 80º )
= sin(− 60º )
= − sin 60º
=−
3
2
23. cos 70º ⋅ cos 20º − sin 70º ⋅ sin 20º = cos(70º + 20º )
= cos 90º
=0
30. sin
24. cos 40º ⋅ cos10º + sin 40º ⋅ sin10º = cos(40º − 10º )
= cos 30º
=
25.
26.
3
2
=
tan 20º + tan 25º
= tan ( 20º +25º )
1 − tan 20º tan 25º
= tan 45º
=1
3
2
3
π
31. sin α = , 0 < α <
5
2
cos β =
2 5
π
, − <β <0
5
2
y
tan 40º − tan10º
= tan ( 40º −10º )
1 + tan 40º tan10º
= tan 30º
y
(x, 3)
5
3
=
3
27. sin
π
5π
π
5π
⎛ π 5π ⎞
⋅ cos + cos ⋅ sin
= sin ⎜ + ⎟
18
18
18
18
⎝ 18 18 ⎠
6π
= sin
18
π
= sin
3
α
π
π
7π
7π
⎛ π 7π ⎞
⋅ cos
− cos ⋅ sin
= sin ⎜ −
⎟
12
12
12
12
⎝ 12 12 ⎠
2 5
3
x
x
β
y
5
(2
x
5, y
x 2 + 32 = 52 , x > 0
x 2 = 25 − 9 = 16, x > 0
x=4
⎛ 6π ⎞
= sin ⎜ − ⎟
⎝ 12 ⎠
⎛ π⎞
= sin ⎜ − ⎟
⎝ 2⎠
= −1
cos α =
(2 5 )
2
4
,
5
tan α =
3
4
+ y 2 = 52 , y < 0
y 2 = 25 − 20 = 5, y < 0
5π
7π
5π
7π
⎛ 5π 7 π ⎞
− sin ⋅ sin
= cos ⎜ +
28. cos ⋅ cos
⎟
12
12
12
12
⎝ 12 12 ⎠
12π
= cos
12
= cos π
y=− 5
sin β = −
5
− 5
1
, tan β =
=−
5
2
2 5
= −1
243
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)
Chapter 3: Analytic Trigonometry
a.
sin(α + β ) = sin α cos β + cos α sin β
5
π
, 0<α <
5
2
4
π
sin β = − , − < β < 0
5
2
32. cos α =
3 2 5 4 ⎛
5⎞
= ⋅
+ ⋅ ⎜⎜ −
⎟
5 5
5 ⎝ 5 ⎟⎠
=
6 5−4 5
25
y
y
(
2 5
=
25
b.
=
8 5 +3 5
25
=
11 5
25
( 5)
−4
5
(x, −4)
+ y 2 = 52 , y > 0
sin α =
2 5
,
5
tan α =
2 5
5
=2
x 2 + (− 4) 2 = 52 , x > 0
x 2 = 25 − 16 = 9, x > 0
x=3
6 5+4 5
25
10 5
=
25
=
tan(α − β ) =
x
y = 20 = 2 5
3 2 5 4 ⎛
5⎞
= ⋅
− ⋅ ⎜⎜ −
⎟
5 5
5 ⎝ 5 ⎟⎠
d.
x
y 2 = 25 − 5 = 20, y > 0
sin(α − β ) = sin α cos β − cos α sin β
=
2
β
x
5
4 2 5 3 ⎛
5⎞
⋅
− ⋅ ⎜⎜ −
⎟
5 5
5 ⎝ 5 ⎟⎠
)
y
α
cos(α + β ) = cos α cos β − sin α sin β
=
c.
5
5, y
3
cos β = ,
5
a.
tan β =
−4
4
=−
3
3
sin(α + β ) = sin α cos β + cos α sin β
2 5
5
⎛ 2 5 ⎞ ⎛3⎞ ⎛ 5 ⎞ ⎛ 4⎞
= ⎜⎜
⎟⎟ ⋅ ⎜ ⎟ + ⎜⎜
⎟⎟ ⋅ ⎜ − ⎟
⎝ 5 ⎠ ⎝5⎠ ⎝ 5 ⎠ ⎝ 5⎠
tan α − tan β
1 + tan α ⋅ tan β
=
6 5−4 5
25
2 5
=
25
3 ⎛ 1⎞
−⎜− ⎟
4 ⎝ 2⎠
=
⎛ 3 ⎞⎛ 1 ⎞
1 + ⎜ ⎟⎜ − ⎟
⎝ 4 ⎠⎝ 2 ⎠
5
= 4
5
8
=2
b.
cos(α + β ) = cos α cos β − sin α sin β
⎛ 5 ⎞ ⎛3⎞ ⎛ 2 5 ⎞ ⎛ 4⎞
= ⎜⎜
⎟⎟ ⋅ ⎜ ⎟ − ⎜⎜
⎟⎟ ⋅ ⎜ − ⎟
⎝ 5 ⎠ ⎝5⎠ ⎝ 5 ⎠ ⎝ 5⎠
3 5 +8 5
25
11 5
=
25
=
244
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Section 3.4: Sum and Difference Formulas
sin(α − β ) = sin α cos β − cos α sin β
c.
a.
⎛ 2 5 ⎞ ⎛3⎞ ⎛ 5 ⎞ ⎛ 4⎞
= ⎜⎜
⎟⎟ ⋅ ⎜ ⎟ − ⎜⎜
⎟⎟ ⋅ ⎜ − ⎟
⎝ 5 ⎠ ⎝5⎠ ⎝ 5 ⎠ ⎝ 5⎠
⎛ 4⎞ ⎛ 1⎞ ⎛ 3⎞ ⎛ 3 ⎞
= ⎜ ⎟ ⋅ ⎜ ⎟ + ⎜ − ⎟ ⋅ ⎜⎜
⎟
⎝ 5 ⎠ ⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 2 ⎟⎠
6 5+4 5
25
10 5
=
25
2 5
=
5
=
=
b.
c.
d.
y
1
x
4
,
5
cos α =
tan α − tan β
1 + tan α tan β
4
− − 3
3
=
⎛ 4⎞
1+ ⎜ − ⎟ ⋅ 3
⎝ 3⎠
tan(α − β ) =
− 48 − 25 3
−39
25 3 + 48
=
39
=
r 2 = (−3) 2 + 42 = 25
r =5
sin α =
4+3 3
10
−4−3 3
3
=
3− 4 3
3
⎛ −4−3 3 ⎞ ⎛ 3+ 4 3 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 3− 4 3 ⎠ ⎝ 3+ 4 3 ⎠
(1, y)
β
x
−3
sin(α − β ) = sin α cos β − cos α sin β
=
2
α
−3 − 4 3
10
⎛ 4⎞ ⎛ 1⎞ ⎛ 3⎞ ⎛ 3 ⎞
= ⎜ ⎟ ⋅ ⎜ ⎟ − ⎜ − ⎟ ⋅ ⎜⎜
⎟
⎝ 5 ⎠ ⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 2 ⎟⎠
y
r
4
cos(α + β ) = cos α cos β − sin α sin β
=
⎛ 4⎞
2−⎜− ⎟
⎝ 3⎠
=
⎛ 4⎞
1+ 2 ⋅⎜ − ⎟
⎝ 3⎠
10
= 3
5
−
3
= −2
4 π
33. tan α = − , < α < π
3 2
1
π
cos β = , 0 < β <
2
2
y
(−3, 4)
4−3 3
10
⎛ 3⎞ ⎛1⎞ ⎛ 4⎞ ⎛ 3 ⎞
= ⎜ − ⎟ ⋅ ⎜ ⎟ − ⎜ ⎟ ⋅ ⎜⎜
⎟
⎝ 5 ⎠ ⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 2 ⎟⎠
tan α − tan β
tan(α − β ) =
1 + tan α tan β
d.
sin(α + β ) = sin α cos β + cos α sin β
−3
3
=−
5
5
12 + y 2 = 22 , y > 0
y 2 = 4 − 1 = 3, y > 0
y= 3
sin β =
3
,
2
tan β =
3
= 3
1
245
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Chapter 3: Analytic Trigonometry
5
3π
, π<α <
12
2
1
3π
sin β = − , π < β <
2
2
34. tan α =
−12 α
−5
y
x
r
tan(α − β ) =
d.
β
x
1
5
3
−
12
3
=
5
3
1+ ⋅
12 3
5−4 3
12
=
36 + 5 3
36
⎛ 15 − 12 3 ⎞ ⎛ 36 − 5 3 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 36 + 5 3 ⎠ ⎝ 36 − 5 3 ⎠
y
x
2
(x, −1)
(−12,−5)
r 2 = (−12) 2 + (−5) 2 = 169
r = 13
540 − 507 3 + 180
1296 − 75
720 − 507 3
=
1221
240 − 169 3
=
407
−5
5
−12
12
= − , cos α =
=−
13
13
13
13
2
2
2
x + (−1) = 2 , x < 0
=
sin α =
x 2 = 4 − 1 = 3, x < 0
x=− 3
cos β = −
a.
3
,
2
tan β =
−1
− 3
=
3
3
5
3π
, −
< α < −π
13
2
π
tan β = − 3, < β < π
2
35. sin α =
sin(α + β ) = sin α cos β + cos α sin β
3 ⎞ ⎛ 12 ⎞ ⎛ 1 ⎞
⎛ 5⎞ ⎛
= ⎜ − ⎟ ⋅ ⎜⎜ −
⎟⎟ + ⎜ − ⎟ ⋅ ⎜ − ⎟
13
2
⎝
⎠ ⎝
⎠ ⎝ 13 ⎠ ⎝ 2 ⎠
=
b.
5 3 + 12 12 + 5 3
=
26
26
(−1, 3 )
y
(x, 5)
13
5
3
cos(α + β ) = cos α cos β − sin α sin β
x
3⎞ ⎛ 5 ⎞ ⎛ 1⎞
⎛ 12 ⎞ ⎛
= ⎜ − ⎟ ⋅ ⎜⎜ −
⎟ − ⎜− ⎟⋅⎜− ⎟
⎝ 13 ⎠ ⎝ 2 ⎟⎠ ⎝ 13 ⎠ ⎝ 2 ⎠
y
r
β
x
α
−1
x
x 2 + 52 = 132 , x < 0
12 3 − 5
=
26
c.
tan α − tan β
1 + tan α ⋅ tan β
x 2 = 169 − 25 = 144, x < 0
x = −12
sin(α − β ) = sin α cos β − cos α sin β
3 ⎞ ⎛ 12 ⎞ ⎛ 1 ⎞
⎛ 5⎞ ⎛
= ⎜ − ⎟ ⋅ ⎜⎜ −
⎟ − ⎜− ⎟⋅⎜− ⎟
⎝ 13 ⎠ ⎝ 2 ⎟⎠ ⎝ 13 ⎠ ⎝ 2 ⎠
cos α =
−12
12
=− ,
13
13
tan α = −
5
12
2
r 2 = (−1) 2 + 3 = 4
r=2
5 3 − 12
=
26
sin β =
3
−1
1
, cos β =
=−
2
2
2
246
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Section 3.4: Sum and Difference Formulas
sin(α + β ) = sin α cos β + cos α sin β
a.
12 + y 2 = 22 , y < 0
⎛ 5 ⎞ ⎛ 1 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞
= ⎜ ⎟ ⋅ ⎜ − ⎟ + ⎜ − ⎟ ⋅ ⎜⎜
⎟
⎝ 13 ⎠ ⎝ 2 ⎠ ⎝ 13 ⎠ ⎝ 2 ⎟⎠
=
y 2 = 4 − 1 = 3, y < 0
y=−
−5 − 12 3
5 + 12 3
or −
26
26
sin α =
cos(α + β ) = cos α cos β − sin α sin β
b.
12 − 5 3
26
cos β =
a.
−5 + 12 3
26
(
b.
3
tan β =
1
2 2
=
2
4
sin(α + β ) = sin α cos β + cos α sin β
1− 2 6
6
cos(α + β ) = cos α cos β − sin α sin β
3 ⎞ ⎛1⎞
⎛1⎞ ⎛2 2 ⎞ ⎛
= ⎜ ⎟ ⋅ ⎜⎜
⎟⎟ − ⎜⎜ −
⎟⋅⎜ ⎟
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎟⎠ ⎝ 3 ⎠
)
(
)
=
−5 + 12 3
12
=
12 + 5 3
12
⎛ −5 + 12 3 ⎞ ⎛ 12 − 5 3 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 12 + 5 3 ⎠ ⎝ 12 − 5 3 ⎠
=
2 2
,
3
=
tan α − tan β
1 + tan α tan β
5
− − − 3
12
=
⎛ 5⎞
1+ ⎜ − ⎟ ⋅ − 3
⎝ 12 ⎠
3
− 3
, tan α =
=−
2
1
⎛
3 ⎞ ⎛ 2 2 ⎞ ⎛1⎞ ⎛1⎞
= ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟ + ⎜ ⎟ ⋅ ⎜ ⎟
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝ 3⎠
tan(α − β ) =
d.
=−
x= 8=2 2
⎛ 5 ⎞ ⎛ 1 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞
= ⎜ ⎟ ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ ⎜⎜
⎟
⎝ 13 ⎠ ⎝ 2 ⎠ ⎝ 13 ⎠ ⎝ 2 ⎟⎠
=
3
2
x 2 = 9 − 1 = 8. x > 0
sin(α − β ) = sin α cos β − cos α sin β
c.
−
x 2 + 12 = 32 , x > 0
⎛ 12 ⎞ ⎛ 1 ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞
= ⎜ − ⎟ ⋅ ⎜ − ⎟ − ⎜ ⎟ ⋅ ⎜⎜
⎟
⎝ 13 ⎠ ⎝ 2 ⎠ ⎝ 13 ⎠ ⎝ 2 ⎟⎠
=
3
c.
3+2 2
6
sin(α − β ) = sin α cos β − cos α sin β
⎛
3 ⎞ ⎛ 2 2 ⎞ ⎛1⎞ ⎛1⎞
= ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟ − ⎜ ⎟ ⋅ ⎜ ⎟
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2⎠ ⎝3⎠
=
−1 − 2 6
6
− 240 + 169 3
69
1
π
, − <α < 0
2
2
1
π
sin β = , 0 < β <
3
2
36. cos α =
y
α
2
y
1
y
x
(1, y)
(x, 1)
3
β
1
x
x
247
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Chapter 3: Analytic Trigonometry
d.
tan(α − β ) =
π
tan α − tan β
1 + tan α tan β
d.
2
− 3−
4
=
(
)
1+ − 3 ⋅
2
4
−4 3 − 2
4
=
4− 6
4
⎛ −4 3 − 2 ⎞ ⎛ 4+ 6 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 4− 6 ⎠ ⎝ 4+ 6 ⎠
−1 + 2 2
= 2 2
2 2 +1
2 2
⎛ 2 2 −1 ⎞ ⎛ 2 2 −1 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 2 2 + 1 ⎠ ⎝ 2 2 −1 ⎠
=
−16 3 − 4 2 − 4 18 − 12
16 − 6
=
−18 3 − 16 2
10
=
8 − 4 2 +1
8 −1
=
−9 3 − 8 2
5
=
9−4 2
7
38. cos θ =
1
37. sin θ = , θ in quadrant II
3
a.
⎛1⎞
cos θ = − 1 − sin θ = − 1 − ⎜ ⎟
⎝3⎠
a.
2
2
= − 1−
b.
1
9
=−
8
9
=−
=−
2 2
3
15
16
=−
15
4
b.
2
1
16
π⎞
π
π
⎛
sin ⎜ θ − ⎟ = sin θ ⋅ cos − cos θ ⋅ sin
6⎠
6
6
⎝
⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
= ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟ − ⎜ ⎟ ⋅ ⎜ ⎟
⎝ 4 ⎠ ⎝ 2 ⎠ ⎝4⎠ ⎝2⎠
3 − 2 2 −2 2 + 3
=
6
6
=
π⎞
π
π
⎛
cos ⎜ θ − ⎟ = cos θ ⋅ cos + sin θ ⋅ sin
3⎠
3
3
⎝
c.
⎛ 2 2 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 3 ⎞
= ⎜⎜ −
⎟⎜ ⎟ + ⎜ ⎟⎜
⎟
3 ⎟⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎜⎝ 2 ⎟⎠
⎝
=
sin θ = − 1 − cos 2 θ
= − 1−
π⎞
π
π
⎛
sin ⎜ θ + ⎟ = sin θ ⋅ cos + cos θ ⋅ sin
6⎠
6
6
⎝
=
1
, θ in quadrant IV
4
⎛1⎞
= − 1− ⎜ ⎟
⎝4⎠
⎛ 1 ⎞⎛ 3 ⎞ ⎛ 2 2 ⎞⎛ 1 ⎞
= ⎜ ⎟ ⎜⎜
⎟+⎜−
⎟⎜ ⎟
3 ⎟⎠ ⎝ 2 ⎠
⎝ 3 ⎠ ⎝ 2 ⎟⎠ ⎜⎝
c.
π ⎞ tan θ + tan 4
⎛
tan ⎜ θ + ⎟ =
4 ⎠ 1 − tan θ ⋅ tan π
⎝
4
1
−
+1
2 2
=
1 ⎞
⎛
1− ⎜ −
⎟ ⋅1
⎝ 2 2⎠
−1 − 3 5
8
π⎞
π
π
⎛
cos ⎜ θ + ⎟ = cos θ ⋅ cos − sin θ ⋅ sin
3⎠
3
3
⎝
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 15 ⎞ ⎛ 3 ⎞
= ⎜ ⎟ ⋅ ⎜ ⎟ − ⎜⎜ −
⎟⋅⎜
⎟
⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 4 ⎟⎠ ⎜⎝ 2 ⎟⎠
−2 2 + 3
6
=
1+ 3 5
8
248
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Section 3.4: Sum and Difference Formulas
d.
⎛
tan ⎜ θ
⎝
π
tan θ − tan
π⎞
4
− ⎟=
4 ⎠ 1 + tan θ ⋅ tan π
4
− 15 − 1
=
1 + − 15 ⋅1
(
40. From the solution to Problem 39, we have
−2 2
1
3
sin α = , cos α =
, sin β =
, and
3
2
2
1
cos β = . Thus,
3
g (α + β ) = cos (α + β )
)
= cos α ⋅ cos β − sin α ⋅ sin β
⎛ −1 − 15 ⎞ ⎛ 1 + 15 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 1 − 15 ⎠ ⎝ 1 + 15 ⎠
=
⎛ 3 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 2 2 ⎞
= ⎜⎜
⎟⎟ ⎜ ⎟ − ⎜ ⎟ ⎜⎜ −
⎟
3 ⎟⎠
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝
−1 − 2 15 − 15
1 − 15
=
−16 − 2 15
−14
8 + 15
=
7
=
41. From the solution to Problem 39, we have
−2 2
1
3
, sin β =
, and
sin α = , cos α =
3
2
2
1
cos β = . Thus,
3
g (α − β ) = cos (α − β )
39. α lies in quadrant I . Since x 2 + y 2 = 4 ,
r = 4 = 2 . Now, ( x, 1) is on the circle, so
x 2 + 12 = 4
= cos α ⋅ cos β + sin α ⋅ sin β
x 2 = 4 − 12
⎛ 3 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 2 2 ⎞
= ⎜⎜
⎟⎟ ⎜ ⎟ + ⎜ ⎟ ⎜⎜ −
⎟
3 ⎟⎠
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝
x = 4 − 12 = 3
y 1
x
3
= and cos α = =
.
r 2
2
r
β lies in quadrant IV . Since x 2 + y 2 = 1 ,
Thus, sin α =
⎛1
r = 1 = 1 . Now, ⎜ ,
⎝3
=
⎞
y ⎟ is on the circle, so
⎠
3 2 2
3−2 2
−
=
6
6
6
42. From the solution to Problem 39, we have
−2 2
1
3
, sin β =
, and
sin α = , cos α =
3
2
2
1
cos β = . Thus,
3
f (α − β ) = sin (α − β )
2
⎛1⎞
2
⎜ ⎟ + y =1
⎝3⎠
⎛1⎞
y2 = 1− ⎜ ⎟
⎝3⎠
3 2 2
3+2 2
+
=
6
6
6
2
= sin α ⋅ cos β − cos α ⋅ sin β
2
8
2 2
⎛1⎞
y = − 1− ⎜ ⎟ = −
=−
9
3
⎝3⎠
⎛ 1 ⎞⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 2 2 ⎞
= ⎜ ⎟⎜ ⎟ − ⎜⎜
⎟⎜ −
⎟
3 ⎟⎠
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎟⎠ ⎜⎝
y −23 2 −2 2
and
=
=
r
1
3
x 1 1
cos β = = 3 = . Thus,
r 1 3
f (α + β ) = sin (α + β )
Thus, sin β =
=
1 2 6 1+ 2 6
+
=
6
6
6
= sin α ⋅ cos β + cos α ⋅ sin β
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞⎛ 2 2 ⎞
= ⎜ ⎟ ⎜ ⎟ + ⎜⎜
⎟⎜ −
⎟
3 ⎟⎠
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎟⎜
⎠⎝
=
1 2 6 1− 2 6
−
=
6
6
6
249
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Chapter 3: Analytic Trigonometry
43. From the solution to Problem 39, we have
−2 2
1
3
, sin β =
, and
sin α = , cos α =
3
2
2
1
cos β = . Thus,
3
1
sin α
1
3
and
tan α =
= 2 =
=
cos α
3
3
3
2
2 2
−
sin β
3 = −2 2 . Finally,
=
tan β =
1
cos β
3
tan α + tan β
h (α + β ) = tan (α + β ) =
1 − tan α tan β
(
)
(
)
44. From the solution to Problem 43, we have
3
and tan β = −2 2 . Thus,
tan α =
3
tan α − tan β
h (α − β ) = tan (α − β ) =
1 + tan α tan β
3
+ −2 2
= 3
3
−2 2
1−
3
3
−2 2
3
= 3
⋅
2 6 3
1+
3
3 −6 2 3− 2 6
=
⋅
3+ 2 6 3− 2 6
=
=
(
)
(
)
3
− −2 2
3
=
3
1+
−2 2
3
3
+2 2
3
3
=
⋅
2 6 3
1−
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
π
π
⎛π
⎞
45. sin ⎜ + θ ⎟ = sin ⋅ cos θ + cos ⋅ sin θ
2
2
⎝2
⎠
= 1 ⋅ cos θ + 0 ⋅ sin θ
= cos θ
3 3 − 6 2 − 18 2 + 24 3
π
π
⎛π
⎞
46. cos ⎜ + θ ⎟ = cos ⋅ cos θ − sin ⋅ sin θ
2
2
⎝2
⎠
= 0 ⋅ cos θ − 1 ⋅ sin θ
= − sin θ
9 − 6 6 + 6 6 − 24
27 3 − 24 2 8 2 − 9 3
=
−15
5
47. sin ( π − θ ) = sin π ⋅ cos θ − cos π ⋅ sin θ
= 0 ⋅ cos θ − ( −1) sin θ
= sin θ
48. cos ( π − θ ) = cos π ⋅ cos θ + sin π ⋅ sin θ
= −1 ⋅ cos θ + 0 ⋅ sin θ
= − cos θ
49. sin ( π + θ ) = sin π ⋅ cos θ + cos π ⋅ sin θ
= 0 ⋅ cos θ + ( −1) sin θ
= − sin θ
50. cos ( π + θ ) = cos π ⋅ cos θ − sin π ⋅ sin θ
= −1 ⋅ cos θ − 0 ⋅ sin θ
= − cos θ
250
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Section 3.4: Sum and Difference Formulas
tan π − tan θ
1 + tan π ⋅ tan θ
0 − tan θ
=
1 + 0 ⋅ tan θ
− tan θ
=
1
= − tan θ
51. tan ( π − θ ) =
58.
59.
tan 2π − tan θ
1 + tan 2π ⋅ tan θ
0 − tan θ
=
1 + 0 ⋅ tan θ
− tan θ
=
1
= − tan θ
52. tan ( 2π − θ ) =
60.
3π
3π
⎛ 3π
⎞
53. sin ⎜ + θ ⎟ = sin ⋅ cos θ + cos ⋅ sin θ
2
2
⎝ 2
⎠
= −1 ⋅ cos θ + 0 ⋅ sin θ
= − cos θ
61.
3π
3π
⎛ 3π
⎞
54. cos ⎜ + θ ⎟ = cos ⋅ cos θ − sin ⋅ sin θ
2
2
⎝ 2
⎠
= 0 ⋅ cos θ − (−1) ⋅ sin θ
= sin θ
cos(α + β ) cos α cos β − sin α sin β
=
cos α cos β
cos α cos β
cos α cos β sin α sin β
=
−
cos α cos β cos α cos β
= 1 − tan α tan β
cos(α − β ) cos α cos β + sin α sin β
=
sin α cos β
sin α cos β
cos α cos β sin α sin β
=
+
sin α cos β sin α cos β
= cot α + tan β
sin(α + β ) sin α cos β + cos α sin β
=
sin(α − β ) sin α cos β − cos α sin β
sin α cos β + cos α sin β
cos α cos β
=
sin α cos β − cos α sin β
cos α cos β
sin α cos β cos α sin β
+
cos α cos β cos α cos β
=
sin α cos β cos α sin β
−
cos α cos β cos α cos β
tan α + tan β
=
tan α − tan β
55. sin(α + β ) + sin(α − β )
= sin α cos β + cos α sin β
+ sin α cos β − cos α sin β
= 2sin α cos β
56. cos(α + β ) + cos(α − β )
= cos α cos β − sin α sin β
+ cos α cos β + sin α sin β
= 2 cos α cos β
57.
sin(α + β ) sin α cos β + cos α sin β
=
cos α cos β
cos α cos β
sin α cos β cos α sin β
=
+
cos α cos β cos α cos β
= tan α + tan β
62.
cos(α + β ) cos α cos β − sin α sin β
=
cos(α − β ) cos α cos β + sin α sin β
cos α cos β − sin α sin β
cos α cos β
=
cos α cos β + sin α sin β
cos α cos β
cos α cos β sin α sin β
−
cos α cos β cos α cos β
=
cos α cos β sin α sin β
+
cos α cos β cos α cos β
1 − tan α tan β
=
1 + tan α tan β
sin(α + β ) sin α cos β + cos α sin β
=
sin α cos β
sin α cos β
sin α cos β cos α sin β
=
+
sin α cos β sin α cos β
= 1 + cot α tan β
251
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Chapter 3: Analytic Trigonometry
cos(α + β )
sin(α + β )
cos α cos β − sin α sin β
=
sin α cos β + cos α sin β
cos α cos β − sin α sin β
sin α sin β
=
sin α cos β + cos α sin β
sin α sin β
cos α cos β sin α sin β
−
sin α sin β sin α sin β
=
sin α cos β cos α sin β
+
sin α sin β sin α sin β
cot α cot β − 1
=
cot β + cot α
63. cot(α + β ) =
66. sec(α − β ) =
1
cos α cos β + sin α sin β
1
cos α cos β
=
cos α cos β + sin α sin β
cos α cos β
1
1
⋅
cos α cos β
=
cos α cos β sin α sin β
+
cos α cos β cos α cos β
sec α sec β
=
1 + tan α tan β
=
67. sin(α − β )sin(α + β )
= ( sin α cos β − cos α sin β )( sin α cos β + cos α sin β )
cos(α − β )
sin(α − β )
cos α cos β + sin α sin β
=
sin α cos β − cos α sin β
cos α cos β + sin α sin β
sin α sin β
=
sin α cos β − cos α sin β
sin α sin β
cos α cos β sin α sin β
+
sin α sin β sin α sin β
=
sin α cos β cos α sin β
−
sin α sin β sin α sin β
cot α cot β + 1
=
cot β − cot α
64. cot(α − β ) =
65. sec(α + β ) =
1
cos(α − β )
= sin 2 α cos 2 β − cos 2 α sin 2 β
= sin 2 α (1 − sin 2 β ) − (1 − sin 2 α )sin 2 β
= sin 2 α − sin 2 α sin 2 β − sin 2 β + sin 2 α sin 2 β
= sin 2 α − sin 2 β
68. cos(α − β )cos(α + β )
= ( cos α cos β + sin α sin β )( cos α cos β − sin α sin β )
= cos 2 α cos 2 β − sin 2 α sin 2 β
= cos 2 α (1 − sin 2 β ) − (1 − cos 2 α )sin 2 β
= cos 2 α − cos 2 α sin 2 β − sin 2 β + cos 2 α sin 2 β
= cos 2 α − sin 2 β
69. sin(θ + k π) = sin θ ⋅ cos k π + cos θ ⋅ sin k π
1
= (sin θ )(−1) k + (cos θ )(0)
cos(α + β )
= (−1) k sin θ , k any integer
1
cos α cos β − sin α sin β
1
sin α sin β
=
cos α cos β − sin α sin β
sin α sin β
1
1
⋅
sin α sin β
=
cos α cos β sin α sin β
−
sin α sin β sin α sin β
csc α csc β
=
cot α cot β − 1
=
70. cos(θ + k π) = cos θ ⋅ cos k π − sin θ ⋅ sin k π
= (cos θ )(−1) k − (sin θ )(0)
= (−1) k cos θ , k any integer
1
⎛
⎞
⎛π π ⎞
71. sin ⎜ sin −1 + cos −1 0 ⎟ = sin ⎜ + ⎟
2
⎝
⎠
⎝6 2⎠
⎛ 2π ⎞
= sin ⎜
⎟
⎝ 3 ⎠
=
3
2
252
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.4: Sum and Difference Formulas
⎛
⎞
3
⎛π
⎞
+ cos −1 1⎟⎟ = sin ⎜ + 0 ⎟
72. sin ⎜⎜ sin −1
2
⎝3
⎠
⎝
⎠
π
= sin
3
3
=
2
cos α = 1 − sin 2 α
2
16
⎛ 4⎞
= 1− ⎜ − ⎟ = 1−
=
25
⎝ 5⎠
sec β = 1 + tan 2 β
2
9
25 5
⎛3⎞
= 1+ ⎜ ⎟ = 1−
=
=
16
16 4
⎝4⎠
⎡
3
⎛ 4 ⎞⎤
73. sin ⎢sin −1 − cos −1 ⎜ − ⎟ ⎥
5
⎝ 5 ⎠⎦
⎣
3
⎛ 4⎞
Let α = sin −1 and β = cos −1 ⎜ − ⎟ . α is in
5
⎝ 5⎠
3
quadrant I; β is in quadrant II. Then sin α = ,
5
4 π
π
0 ≤ α ≤ , and cos β = − , ≤ β ≤ π .
5 2
2
cos β =
2
16
⎛4⎞
= 1− ⎜ ⎟ = 1−
=
25
⎝5⎠
9
3
=
25 5
⎡
3⎤
⎛ 4⎞
sin ⎢sin −1 ⎜ − ⎟ − tan −1 ⎥
4⎦
⎝ 5⎠
⎣
= sin (α − β )
cos α = 1 − sin α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
5
25
25 5
⎝ ⎠
= sin α cos β − cos α sin β
⎛ 4⎞ ⎛ 4⎞ ⎛ 3⎞ ⎛ 3⎞
= ⎜ − ⎟⋅⎜ ⎟ − ⎜ ⎟⋅⎜ ⎟
⎝ 5⎠ ⎝ 5⎠ ⎝5⎠ ⎝5⎠
16 9
25
=− −
=−
25 25
25
= −1
sin β = 1 − cos 2 β
2
4
5
sin β = 1 − cos 2 β
2
16
⎛ 4⎞
= 1− ⎜ − ⎟ = 1−
=
25
⎝ 5⎠
9
3
=
25 5
9
3
=
25 5
⎡
3
⎛ 4 ⎞⎤
sin ⎢sin −1 − cos −1 ⎜ − ⎟ ⎥ = sin (α − β )
5
⎝ 5 ⎠⎦
⎣
= sin α cos β − cos α sin β
4
5⎞
⎛
75. cos ⎜ tan −1 + cos −1 ⎟
3
13
⎝
⎠
5
−1 4
and β = cos −1 . α is in
Let α = tan
3
13
⎛ 3⎞ ⎛ 4⎞ ⎛ 4⎞ ⎛ 3⎞
= ⎜ ⎟ ⋅⎜ − ⎟ − ⎜ ⎟ ⋅⎜ ⎟
⎝5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝5⎠
12 12
=− −
25 25
24
=−
25
quadrant I; β is in quadrant I. Then tan α =
0<α <
π
2
, and cos β =
4
,
3
5
π
, 0≤β ≤ .
13
2
sec α = 1 + tan 2 α
⎡
3⎤
⎛ 4⎞
74. sin ⎢sin −1 ⎜ − ⎟ − tan −1 ⎥
5
4⎦
⎝
⎠
⎣
3
⎛ 4⎞
Let α = sin −1 ⎜ − ⎟ and β = tan −1 . α is in
4
⎝ 5⎠
quadrant IV; β is in quadrant I. Then
2
16
⎛4⎞
= 1+ ⎜ ⎟ = 1+
=
9
⎝3⎠
cos α =
4
π
3
, − ≤ α ≤ 0 , and tan β = ,
5
4
2
π
0<β < .
2
sin α = −
25 5
=
9
3
3
5
sin α = 1 − cos 2 α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
25
25 5
⎝5⎠
253
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
5
3⎞
⎛
77. cos ⎜ sin −1 − tan −1 ⎟
13
4⎠
⎝
5
3
and β = tan −1 . α is in
Let α = sin −1
13
4
sin β = 1 − cos 2 β
2
25
144 12
⎛5⎞
= 1− ⎜ ⎟ = 1−
=
=
169
169 13
⎝ 13 ⎠
4
5⎞
⎛
cos ⎜ tan −1 + cos −1 ⎟
3
13 ⎠
⎝
= cos (α + β )
quadrant I; β is in quadrant I. Then sin α =
0≤α ≤
= cos α cos β − sin α sin β
2
25
144 12
⎛5⎞
= 1− ⎜ ⎟ = 1−
=
=
13
169
169
13
⎝ ⎠
sec β = 1 + tan 2 β
⎡
5
⎛ 3 ⎞⎤
76. cos ⎢ tan −1 − sin −1 ⎜ − ⎟ ⎥
12
⎝ 5 ⎠⎦
⎣
5
⎛ 3⎞
Let α = tan −1
and β = sin −1 ⎜ − ⎟ . α is in
12
⎝ 5⎠
quadrant I; β is in quadrant IV. Then
−
2
9
25 5
⎛3⎞
= 1+ ⎜ ⎟ = 1+
=
=
16
16 4
⎝4⎠
cos β =
3
5
π
, 0 < α < , and sin β = − ,
5
12
2
4
5
sin β = 1 − cos 2 β
π
<α < 0.
2
2
16
⎛4⎞
= 1− ⎜ ⎟ = 1−
=
25
⎝5⎠
sec α = 1 + tan 2 α
9
3
=
25 5
5
3⎤
⎡
cos ⎢sin −1 − tan −1 ⎥
13
4⎦
⎣
= cos (α − β )
2
25
169 13
⎛ 5⎞
= 1+ ⎜ ⎟ = 1+
=
=
144
144 12
⎝ 12 ⎠
cos α =
π
3
π
, and tan β = , 0 < β < .
4
2
2
cos α = 1 − sin 2 α
⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 12 ⎞
= ⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
15 48
33
=
−
=−
65 65
65
tan α =
5
,
13
= cos α cos β + sin α sin β
12 4 5 3
= ⋅ + ⋅
13 5 13 5
48 15
=
+
65 65
63
=
65
12
13
sin α = 1 − cos 2 α
2
144
25
5
⎛ 12 ⎞
= 1− ⎜ ⎟ = 1−
=
=
169
169 13
⎝ 13 ⎠
cos β = 1 − sin 2 β
4
12 ⎞
⎛
78. cos ⎜ tan −1 + cos −1 ⎟
3
13 ⎠
⎝
4
12
Let α = tan −1 and β = cos −1 . α is in
3
13
2
9
16 4
⎛ 3⎞
= 1− ⎜ − ⎟ = 1−
=
=
25
25 5
⎝ 5⎠
⎡
5
⎛ 3 ⎞⎤
cos ⎢ tan −1 − sin −1 ⎜ − ⎟ ⎥
12
⎝ 5 ⎠⎦
⎣
= cos (α − β )
quadrant I; β is in quadrant I. Then tan α =
= cos α cos β + sin α sin β
0<α <
⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞ 48 15 33
= ⎜ ⎟⋅⎜ ⎟ + ⎜ ⎟⋅⎜ − ⎟ =
−
=
⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ 65 65 65
4
,
3
12
π
π
, and cos β = , 0 ≤ β ≤ .
13
2
2
254
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Section 3.4: Sum and Difference Formulas
3⎞
π
⎛
tan ⎜ sin −1 ⎟ + tan
5⎠
6
⎛ −1 3 π ⎞
⎝
tan ⎜ sin
+ ⎟=
π
5 6⎠
⎛ −1 3 ⎞
⎝
1 − tan ⎜ sin
⎟ ⋅ tan
5
6
⎝
⎠
sec α = 1 + tan 2 α
2
16
⎛4⎞
= 1+ ⎜ ⎟ = 1+
=
9
⎝3⎠
cos α =
25 5
=
9
3
3
3
+
4
3
=
3 3
1− ⋅
4 3
9+ 3
12
=
12 − 3 3
12
9 + 3 12 + 3 3
=
⋅
12 − 3 3 12 + 3 3
3
5
sin α = 1 − cos 2 α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
25
25 5
⎝5⎠
sin β = 1 − cos 2 β
2
144
25
5
⎛ 12 ⎞
= 1− ⎜ ⎟ = 1−
=
=
169
169 13
⎝ 13 ⎠
108 + 75 3 + 36
144 − 27
144 + 75 3
=
117
48 + 25 3
=
39
=
4
12 ⎞
⎛
cos ⎜ tan −1 + cos −1 ⎟
3
13 ⎠
⎝
= cos (α + β )
= cos α cos β − sin α sin β
⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞
= ⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
36 20
=
−
65 65
16
=
65
3⎞
⎛π
80. tan ⎜ − cos −1 ⎟
4
5
⎝
⎠
3
Let α = cos −1 . α is in quadrant I. Then
5
3
π
cos α = , 0 ≤ α ≤ .
2
5
3 π⎞
⎛
79. tan ⎜ sin −1 + ⎟
5
6⎠
⎝
3
Let α = sin −1 . α is in quadrant I. Then
5
3
π
sin α = , 0 ≤ α ≤ .
5
2
sin α = 1 − cos 2 α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
5
25
25 5
⎝ ⎠
4
sin α 5 4 5 4
tan α =
= = ⋅ =
cos α 3 5 3 3
5
cos α = 1 − sin 2 α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
25
25 5
⎝5⎠
π
3⎞
⎛
tan − tan ⎜ cos −1 ⎟
3
4
5
⎛π
⎞
⎝
⎠
tan ⎜ − cos −1 ⎟ =
π
5⎠
⎛ −1 3 ⎞
⎝4
1 + tan ⋅ tan ⎜ cos
⎟
4
5⎠
⎝
4
1
−
1−
1 3
1
3
=
= 3 =− ⋅ =−
4
7
3
7
7
1 + 1⋅
3
3
3
sin α 5 3 5 3
tan α =
= = ⋅ =
cos α 4 5 4 4
5
255
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Chapter 3: Analytic Trigonometry
4
⎛
⎞
81. tan ⎜ sin −1 + cos −1 1⎟
5
⎝
⎠
−1 4
and β = cos −11 ; α is in
Let α = sin
5
4
π
quadrant I. Then sin α = , 0 ≤ α ≤ , and
5
2
4
⎛
⎞
sin ⎜ cos −1 + sin −1 1⎟
5
⎛ −1 4
⎞
⎝
⎠
tan ⎜ cos
+ sin −1 1⎟ =
5
⎝
⎠ tan ⎛ cos −1 4 + sin −1 1⎞
⎜
⎟
5
⎝
⎠
sin (α + β )
=
cos (α + β )
cos β = 1 , 0 ≤ β ≤ π . So, β = cos −1 1 = 0 .
=
cos α = 1 − sin 2 α
2
16
⎛4⎞
= 1− ⎜ ⎟ = 1−
=
25
⎝5⎠
⎛3⎞
⎛4⎞
⎜ ⎟ (0) + ⎜ ⎟ (1)
5
⎝5⎠
=⎝ ⎠
⎛4⎞
⎛3⎞
⎜ ⎟ (0) − ⎜ ⎟ (1)
⎝5⎠
⎝5⎠
4
4
= 5 =−
3
3
−
5
9
3
=
25 5
4
sin α 5 4 5 4
tan α =
= = ⋅ =
cos α 3 5 3 3
5
⎛ −1 4
⎞
tan ⎜ sin
− cos −1 1⎟
5
⎝
⎠
4
⎛
⎞
tan ⎜ sin −1 ⎟ + tan cos −1 1
5⎠
⎝
=
4⎞
⎛
1 − tan ⎜ sin −1 ⎟ ⋅ tan cos −1 1
5⎠
⎝
4
4
+0
4
= 3
= 3=
4
1− ⋅ 0 1 3
3
(
(
83. cos cos −1 u + sin −1 v
)
(
sin β = 1 , −
2
≤β ≤
π
2
)
sin α = 1 − cos 2 α = 1 − u 2
cos β = 1 − sin 2 β = 1 − v 2
. So, β = sin 1 =
(
−1
π
2
2
)
cos cos −1 u + sin −1 v = cos(α + β )
= cos α cos β − sin α sin β
= u 1 − v2 − v 1 − u 2
(
84. sin sin −1 u − cos −1 v
)
Let α = sin u and β = cos −1 v . Then
−1
.
π
π
≤ α ≤ , and
2
2
cos β = v, 0 ≤ β ≤ π .
−1 ≤ u ≤ 1 , −1 ≤ v ≤ 1
sin α = u , −
sin α = 1 − cos 2 α
16
⎛4⎞
= 1− ⎜ ⎟ = 1−
=
5
25
⎝ ⎠
)
Let α = cos −1 u and β = sin −1 v .
Then cos α = u, 0 ≤ α ≤ π , and
π
π
sin β = v, − ≤ β ≤
2
2
−1 ≤ u ≤ 1 , −1 ≤ v ≤ 1
4
⎛
⎞
82. tan ⎜ cos −1 + sin −1 1⎟
5
⎝
⎠
−1 4
and β = sin -1 1 ; α is in
Let α = cos
5
4
π
quadrant I. Then cos α = , 0 ≤ α ≤ , and
2
5
π
sin α cos β + cos α sin β
cos α cos β − sin α sin β
9
3
=
25 5
cos α = 1 − sin 2 α = 1 − u 2
sin β = 1 − cos 2 β = 1 − v 2
3
sin α 5 3 5 3
π
tan α =
= = ⋅ = , but tan is
cos α 4 5 4 4
2
5
undefined. Therefore, we cannot use the sum
formula for tangent. Rewriting using sine and
cosine, we obtain:
(
)
sin sin −1 u − cos −1 v = sin(α − β )
= sin α cos β − cos α sin β
= uv − 1 − u 2 1 − v 2
256
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Section 3.4: Sum and Difference Formulas
(
85. sin tan −1 u − sin −1 v
)
sin α = 1 − cos 2 α
Let α = tan −1 u and β = sin −1 v . Then
= 1−
π
π
< α < , and
2
2
π
π
sin β = v, − ≤ β ≤ .
2
2
−∞ < u < ∞ , −1 ≤ v ≤ 1
tan α = u, −
=
=
sec α = tan 2 α + 1 = u 2 + 1
=
1
cos α =
u2 + 1
cos β =
sin α = 1 − cos 2 α
=
u +1
u
=
2
=
u2 +1
(
sin tan −1 u − sin −1 v
=
)
v2 + 1 − 1
v2 + 1
v2
v +1
v
2
v2 + 1
)
= cos α cos β − sin α sin β
=
u +1
2
86. cos tan −1 u + tan −1 v
1
v +1
2
= cos(α + β )
u 1− v − v
=
)
1
1
⋅
u +1 v +1
1 − uv
2
2
−
u
u +1
2
⋅
v
v +1
2
u 2 + 1 ⋅ v2 + 1
(
87. tan sin −1 u − cos −1 v
Let α = tan u and β = tan −1 v . Then
−1
)
Let α = sin u and β = cos −1 v . Then
−1
π
π
tan α = u, − < α < , and
2
2
π
π
tan β = v, − < β < .
2
2
−∞ < u < ∞ , −∞ < v < ∞
π
π
≤ α ≤ , and
2
2
cos β = v, 0 ≤ β ≤ π .
−1 ≤ u ≤ 1 , −1 ≤ v ≤ 1
sin α = u , −
cos α = 1 − sin 2 α = 1 − u 2
sec α = tan 2 α + 1 = u 2 + 1
cos α =
v +1
cos tan −1 u + tan −1 v
2
(
1
2
(
= sin(α − β )
= sin α cos β − cos α sin β
1
u
=
⋅ 1 − v2 −
⋅v
2
u +1
u2 +1
=
u2 +1
= 1−
2
u
u2
u +1
u
2
sin β = 1 − cos 2 β
1
2
u +1
u2 +1−1
=
u2 +1
=
u2 +1−1
u2 +1
sec β = tan 2 β + 1 = v 2 + 1
cos β = 1 − sin 2 β = 1 − v 2
= 1−
1
u2 +1
1
tan α =
u +1
2
sin α
u
=
cos α
1− u2
257
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Chapter 3: Analytic Trigonometry
(
sec tan −1 u + cos −1 v
sin β = 1 − cos 2 β = 1 − v 2
tan β =
= sec(α + β )
sin β
1 − v2
=
cos β
v
(
=
)
tan sin −1 u − cos −1 v = tan(α − β )
=
1− u2
=
1+
u
1− u2
1 − v2
v
⋅
uv − 1 − u 2 1 − v 2
v 1− u2
v 1 − u 2 + u 1 − v2
v 1− u
=
(
−1
−1
88. sec tan u + cos v
uv − 1 − u
u2 +1
2
1− v
2
v 1− u + u 1− v
2
2
2
=
−1
⎛π
⎞
⎛π
⎞
sin α = cos ⎜ − α ⎟ , cos ⎜ − α ⎟ = cos β . If
2
2
⎝
⎠
⎝
⎠
π
⎛π
⎞
v ≥ 0 , then 0 ≤ α ≤ , so that ⎜ − α ⎟ and β
2
2
⎝
⎠
⎡ π⎤
both lie in the interval ⎢ 0, ⎥ . If v < 0 , then
⎣ 2⎦
π
π
⎛
⎞
− ≤ α < 0 , so that ⎜ − α ⎟ and β both lie in
2
⎝2
⎠
⎛π ⎤
the interval ⎜ , π ⎥ . Either way,
⎝2 ⎦
π
⎛π
⎞
cos ⎜ − α ⎟ = cos β implies − α = β , or
2
2
⎝
⎠
π
π
α + β = . Thus, sin −1 v + cos −1 v = .
2
2
π
π
< α < , and
2
2
cos β = v, 0 ≤ β ≤ π .
−∞ < u < ∞ , −1 ≤ v ≤ 1
tan α = u, −
sec α = tan 2 α + 1 = u 2 + 1
1
u2 + 1
sin α = 1 − cos 2 α
= 1−
=
=
=
v − u 1 − v2
sin α = v = cos β , and since
Let α = tan u and β = cos v . Then
cos α =
u2 +1
89. Let α = sin −1 v and β = cos −1 v . Then
)
−1
cos(α + β )
1
cos α cos β − sin α sin β
1
=
u
1
⋅v −
⋅ 1 − v2
2
2
u +1
u +1
1
=
v
u 1 − v2
−
u2 +1
u2 +1
1
=
v − u 1 − v2
1 − v2
v
−
1
=
tan α − tan β
1 + tan α tan β
u
=
)
1
u +1
2
u2 +1−1
u2 +1
u2
u +1
u
2
u2 +1
sin β = 1 − cos 2 β = 1 − v 2
258
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Section 3.4: Sum and Difference Formulas
(
90. Let α = tan −1 v and β = cot −1 v . Then
tan α = v = cot β , and since
93. sin sin −1 v + cos −1 v
) (
)
+ cos ( sin v ) sin ( cos v )
= sin sin −1 v cos cos −1 v
⎛π
⎞
⎛π
⎞
tan α = cot ⎜ − α ⎟ , cot ⎜ − α ⎟ = cot β . If
⎝2
⎠
⎝2
⎠
π
⎛π
⎞
v ≥ 0 , then 0 ≤ α < , so that ⎜ − α ⎟ and β
2
2
⎝
⎠
⎛ π⎤
both lie in the interval ⎜ 0, ⎥ . If v < 0 , then
⎝ 2⎦
π
⎛π
⎞
− < α < 0 , so that ⎜ − α ⎟ and β both lie in
2
⎝2
⎠
π
⎛
⎞
the interval ⎜ , π ⎟ . Either way,
⎝2 ⎠
π
⎛π
⎞
cot ⎜ − α ⎟ = cot β implies − α = β , or
2
2
⎝
⎠
π
π
α + β = . Thus, tan −1 v + cot −1 v = . Note
2
2
that v ≠ 0 since cot −1 0 is undefined.
−1
−1
= v ⋅ v + 1 − v2 1 − v2
= v2 + 1 − v2
=1
(
94. cos sin −1 v + cos −1 v
(
) (
)
)
− sin ( sin v ) sin ( cos v )
= cos sin −1 v cos cos −1 v
−1
−1
= 1 − v2 ⋅ v − v ⋅ 1 − v2
=0
95.
1
⎛1⎞
91. Let α = tan −1 ⎜ ⎟ and β = tan −1 v . Because
v
⎝v⎠
must be defined, v ≠ 0 and so α , β ≠ 0 . Then
tan α =
(
)
1
1
=
= cot β , and since
v tan β
⎛π
⎞
⎛π
⎞
tan α = cot ⎜ − α ⎟ , cot ⎜ − α ⎟ = cot β .
2
2
⎝
⎠
⎝
⎠
π
⎛π
⎞
Because v > 0 , 0 < α < and so ⎜ − α ⎟ and
2
2
⎝
⎠
⎛ π⎞
β both lie in the interval ⎜ 0, ⎟ . Then
⎝ 2⎠
96.
π
⎛π
⎞
cot ⎜ − α ⎟ = cot β implies − α = β or
2
⎝2
⎠
π
α = − β . Thus,
2
⎛1⎞ π
tan −1 ⎜ ⎟ = − tan −1 v, if v > 0 .
⎝v⎠ 2
92. Let θ = tan −1 e− v . Then tan θ = e− v , so
1
π
cot θ = − v = ev . Because 0 < θ < , we know
2
e
−v
that e > 0 , which means
cot −1 ev = cot −1 ( cot θ ) = θ = tan −1 e− v .
f ( x + h) − f ( x )
h
sin( x + h) − sin x
=
h
sin x cos h + cos x sin h − sin x
=
h
cos x sin h − sin x + sin x cos h
=
h
cos x sin h − sin x (1 − cos h )
=
h
sin h
1 − cos h
= cos x ⋅
− sin x ⋅
h
h
f ( x + h) − f ( x )
h
cos( x + h) − cos x
h
cos x cos h − sin x sin h − cos x
=
h
− sin x sin h + cos x cos h − cos x
=
h
− sin x sin h − cos x (1 − cos h )
=
h
sin h
1 − cos h
= − sin x ⋅
− cos x ⋅
h
h
259
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
97. a.
(
−1
−1
−1
)
((
−1
−1
)
−1
)
tan tan 1 + tan 2 + tan 3 = tan tan 1 + tan 2 + tan 3 =
(
(
)
( )
2 ) tan ( tan 3)
tan tan −1 1 + tan −1 2 + tan tan −1 3
(
−1
1 − tan tan 1 + tan
)
−1
−1
(
) +3
3
1+ 2
+3
+3
1 − tan ( tan 1) tan ( tan 2 )
−3 + 3 0
−
1
−
1
⋅
2
= 1
=
=
=0
=
=
+
2
3
1
+
1
9
10
tan ( tan 1) + tan ( tan 2 )
⋅3 1− ⋅3
1−
1−
⋅3
−1
1 − 1⋅ 2
1 − tan ( tan 1) tan ( tan 2 )
tan tan −1 1 + tan tan −1 2
−1
−1
−1
−1
−1
−1
b. From the definition of the inverse tangent function we know 0 < tan −1 1 <
π
, 0 < tan −1 2 <
π
, and
2
2
π
3π
⎛ 3π ⎞
0 < tan −1 3 < . Thus, 0 < tan −1 1 + tan −1 2 + tan −1 3 <
. On the interval ⎜ 0, ⎟ , tan θ = 0 if and only if
2
2
⎝ 2 ⎠
θ = π . Therefore, from part (a), tan −1 1 + tan −1 2 + tan −1 3 = π .
(
)
= sin (ωt ) ( sin (ωt ) cos φ − cos (ω t ) sin φ )
98. cos φ sin 2 (ωt ) − sin φ sin (ω t ) cos (ωt ) = sin (ω t ) cos φ sin (ω t ) − sin φ cos (ω t )
= sin (ωt ) sin (ωt − φ )
99. Note that θ = θ 2 − θ1 .
Then tan θ = tan (θ 2 − θ1 ) =
tan θ 2 − tan θ1
m − m1
= 2
1 + tan θ 2 tan θ1 1 + m2 m1
100. sin(α − θ )sin( β − θ )sin(γ − θ )
= ( sin α cosθ − cos α sin θ )( sin β cosθ − cos β sin θ )( sin γ cosθ − cos γ sin θ )
⎛
⎞
⎛
⎞
⎛
⎞
⎛ cosθ ⎞
⎛ cosθ ⎞
⎛ cosθ ⎞
= sin θ ⎜ sin α ⎜
⎟ − cos α ⎟ sin θ ⎜ sin β ⎜
⎟ − cos β ⎟ sin θ ⎜ sin γ ⎜
⎟ − cos γ ⎟
θ
θ
θ
sin
sin
sin
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎛ cosθ cos β
⎛
⎛ cosθ cos α ⎞ ⎞ ⎛
−
= sin 3 θ ⎜ sin α ⎜
−
⎟ ⎟ ⎜ sin β ⎜
θ
α
sin
sin
⎝
⎠
⎝
⎠⎝
⎝ sin θ sin β
⎞ ⎞⎛
⎛ cosθ cos γ ⎞ ⎞
−
⎟ ⎟⎜ sin γ ⎜
⎟⎟
⎠ ⎠⎝
⎝ sin θ sin γ ⎠ ⎠
= sin 3 θ ( sin α ( cot θ − cot α ) ) ( sin β ( cot θ − cot β ) ) ( sin γ ( cot θ − cot γ ) )
= sin 3 θ sin α sin β sin γ ( cot β + cot γ )( cot α + cot γ )( cot α + cot β )
⎛ cos β cos γ ⎞⎛ cos α cos γ ⎞⎛ cos α cos β ⎞
= sin 3 θ sin α sin β sin γ ⎜
+
+
+
⎟⎜
⎟⎜
⎟
⎝ sin β sin γ ⎠⎝ sin α sin γ ⎠⎝ sin α sin β ⎠
⎛ sin(γ + β ) ⎞⎛ sin(γ + α ) ⎞⎛ sin( β + α ) ⎞
= sin 3 θ sin α sin β sin γ ⎜
⎟⎜
⎟⎜
⎟
⎝ sin β sin γ ⎠⎝ sin α sin γ ⎠⎝ sin α sin β ⎠
⎛ sin(180º − α ) ⎞⎛ sin(180º − β ) ⎞⎛ sin(180º − γ ) ⎞
= sin 3 θ sin α sin β sin γ ⎜
⎟⎜
⎟⎜
⎟
⎝ sin β sin γ ⎠⎝ sin α sin γ ⎠⎝ sin α sin β ⎠
⎛ sin α ⎞⎛ sin β ⎞⎛ sin γ ⎞
= sin 3 θ sin α sin β sin γ ⎜
⎟⎜
⎟⎜
⎟
⎝ sin β sin γ ⎠⎝ sin α sin γ ⎠⎝ sin α sin β ⎠
= sin 3 θ
260
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.5: Double-angle and Half-angle Formulas
101. If tan α = x + 1 and tan β = x − 1 , then
2cot (α − β ) = 2 ⋅
=
Section 3.5
1
1. sin 2 θ , 2 cos 2 θ , 2sin 2 θ
tan (α − β )
2
tan α − tan β
1 + tan α tan β
2. 1 − cos θ
3. sin θ
2 (1 + tan α tan β )
=
tan α − tan β
=
=
4. True
2 (1 + ( x + 1)( x − 1) )
5. False
x + 1 − ( x − 1)
( (
6. False
))
2 1 + x2 − 1
3
π
θ π
7. sin θ = , 0 < θ < . Thus, 0 < < , which
5
2
2 4
x +1− x +1
2x2
=
2
= x2
means
lies in quadrant I.
2
y = 3, r = 5
102. The first step in the derivation,
x 2 + 32 = 52 , x > 0
π
π ⎞ tan θ + tan 2
⎛
tan ⎜ θ + ⎟ =
, is impossible
2 ⎠ 1 − tan θ ⋅ tan π
⎝
2
because tan
θ
x 2 = 25 − 9 = 16, x > 0
x=4
4
So, cos θ = .
5
π
is undefined.
2
103. If formula (7) is used, we obtain
π
tan − tan θ
⎛π
⎞
2
tan ⎜ − θ ⎟ =
. However, this is
2
⎝
⎠ 1 + tan π ⋅ tan θ
2
π
impossible because tan is undefined. Using
2
formulas (3a) and (3b), we obtain
⎛π
⎞
sin ⎜ − θ ⎟
2
⎛π
⎞
⎝
⎠.
tan ⎜ − θ ⎟ =
π
2
⎛
⎝
⎠ cos − θ ⎞
⎜
⎟
⎝2
⎠
cos θ
=
sin θ
= cot θ
a.
3 4 24
sin(2θ ) = 2sin θ cos θ = 2 ⋅ ⋅ =
5 5 25
b.
cos(2θ ) = cos 2 θ − sin 2 θ
2
2
16 9
7
⎛ 4⎞ ⎛3⎞
= ⎜ ⎟ −⎜ ⎟ =
−
=
5
5
25
25
25
⎝ ⎠ ⎝ ⎠
c.
sin
θ
2
=
=
d.
cos
θ
2
=
=
1 − cosθ
2
1−
2
4
5 =
1
5 = 1 = 1 10 = 10
2
10
10 10 10
1 + cosθ
2
1+
2
4
5 =
9
5 = 9 = 3 10 = 3 10
2
10
10
10 10
261
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Chapter 3: Analytic Trigonometry
3
π
θ π
8. cos θ = , 0 < θ < . Thus, 0 < < , which
5
2
2 4
means
c.
sin
θ
2
θ
⎛ 3⎞
1− ⎜ − ⎟
⎝ 5⎠
=
2
lies in quadrant I.
2
x = 3, r = 5
32 + y 2 = 52 , y > 0
y 2 = 25 − 9 = 16, y > 0
y=4
So, sin θ =
d.
a.
4 3 24
sin(2θ ) = 2sin θ cos θ = 2 ⋅ ⋅ =
5 5 25
b.
cos(2θ ) = cos 2 θ − sin 2 θ
cos
θ
2
sin
θ
2
1−
=
d.
cos
θ
2
2
3
5 =
3
1+
5 =
=
2
9. tan θ =
10. tan θ =
2
5 = 1= 1 5= 5
2
5
5
5 5
θ
2
x = −3, y = − 4
8
5 =
2
r= 5
4
2 5 2 5
=
=
5
5
5 5
5
2
2 5
=−
, cos θ = −
5
5
5
5
sin(2θ ) = 2sin θ cos θ
sin θ = −
a.
1
=−
⎛
5⎞ ⎛ 2 5⎞ 4
= 2 ⋅ ⎜⎜ −
⎟⎟ ⋅ ⎜⎜ −
⎟=
5 ⎟⎠ 5
⎝ 5 ⎠ ⎝
lies in quadrant II.
b.
cos(2θ ) = cos 2 θ − sin 2 θ
2
⎛ 2 5⎞ ⎛
= ⎜⎜ −
⎟ −⎜−
5 ⎟⎠ ⎜⎝
⎝
20 5 15
=
−
=
=
25 25 25
sin(2θ ) = 2sin θ cos θ
c.
⎛ 4 ⎞ ⎛ 3 ⎞ 24
= 2⋅⎜ − ⎟⋅⎜ − ⎟ =
⎝ 5 ⎠ ⎝ 5 ⎠ 25
b.
lie in quadrant II.
r 2 = (− 2) 2 + (−1) 2 = 4 + 1 = 5
r 2 = (−3) 2 + (− 4) 2 = 9 + 16 = 25
r =5
4
3
sin θ = − , cos θ = −
5
5
a.
θ
2
x = − 2, y = −1
4
3π
π θ 3π
, π <θ <
. Thus, < <
,
3
2
2 2 4
which means
1
3π
π θ 3π
. Thus, < <
,
, π <θ <
2
2
2 2 4
which means
1 + cos θ
2
=
1 + cos θ
2
2
1
1 5
5
5
=−
=−
=−
=−
2
5
5
5 5
2
1 − cos θ
2
=
=−
⎛ 3⎞
1+ ⎜ − ⎟
⎝ 5⎠
=−
2
9 16
7
⎛3⎞ ⎛ 4⎞
= ⎜ ⎟ −⎜ ⎟ =
−
=−
5
5
25
25
25
⎝ ⎠ ⎝ ⎠
c.
8
5 = 4= 2 5=2 5
2
5
5
5 5
=
4
.
5
2
1 − cos θ
2
=
2
2
9 16
7
⎛ 3⎞ ⎛ 4⎞
= ⎜− ⎟ −⎜− ⎟ =
−
=−
25 25
25
⎝ 5⎠ ⎝ 5 ⎠
2
3
5
⎛ −2 5 ⎞
1 − ⎜⎜
⎟⎟
θ
1 − cos θ
⎝ 5 ⎠
sin =
=
2
2
2
=
cos(2θ ) = cos 2 θ − sin 2 θ
5⎞
⎟
5 ⎟⎠
=
5+ 2 5
5
2
5+ 2 5
10
262
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Section 3.5: Double-angle and Half-angle Formulas
d.
⎛ −2 5 ⎞
1 + ⎜⎜
⎟⎟
θ
1 + cos θ
⎝ 5 ⎠
cos = −
=−
2
2
2
d.
⎛
6⎞
1 + ⎜⎜ −
⎟⎟
θ
1 + cos θ
⎝ 3 ⎠
cos =
=
2
2
2
5−2 5
5
=−
2
=−
which means
θ
2
x = − 6, r = 3
2
5−2 5
10
=
6 π
π θ π
,
< θ < π . Thus, < < ,
3
2
4 2 2
11. cos θ = −
(− 6 )
=
which means
(
x2 + − 3
y2 = 9 − 6 = 3
)
a.
cos θ =
a.
sin(2θ ) = 2sin θ cos θ
6
3
sin(2θ ) = 2sin θ cos θ
⎛
3⎞ ⎛ 6⎞
= 2 ⋅ ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 3 ⎠ ⎝ 3 ⎠
⎛ 3⎞ ⎛
6⎞
= 2 ⋅ ⎜⎜
⎟⎟ ⋅ ⎜⎜ −
⎟⎟
⎝ 3 ⎠ ⎝ 3 ⎠
b.
2 18
6 2
2 2
=−
=−
9
9
3
=−
b.
cos(2θ ) = cos 2 θ − sin 2 θ
2
⎛
6⎞ ⎛ 3⎞
= ⎜⎜ −
⎟⎟ − ⎜⎜
⎟⎟
⎝ 3 ⎠ ⎝ 3 ⎠
6 3 3 1
= − = =
9 9 9 3
c.
=3
x= 6
3
3
=−
2
lies in quadrant II.
x2 = 9 − 3 = 6
y= 3
sin θ =
θ
2
y = − 3, r = 3
+ y 2 = 32
3− 6
6
3 3π
3π θ
< θ < 2π . Thus,
< <π,
,
3
2
4 2
12. sin θ = −
lies in quadrant I.
3− 6
3
2
=
cos(2θ ) = cos 2 θ − sin 2 θ
2
2
⎛ 6⎞ ⎛
3⎞
= ⎜⎜
⎟⎟ − ⎜⎜ −
⎟⎟
⎝ 3 ⎠ ⎝ 3 ⎠
6 3 3 1
= − = =
9 9 9 3
⎛
6⎞
1 − ⎜⎜ −
⎟⎟
3
θ
1 − cos θ
⎝
⎠
sin =
=
2
2
2
=
2 18
6 2
2 2
=−
=−
9
9
3
c.
3+ 6
3
2
θ
1 − cos θ
=
sin =
2
2
=
3+ 6
6
=
2
6
3
1−
2
3− 6
3
2
3− 6
6
263
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Chapter 3: Analytic Trigonometry
d.
cos
θ
2
=−
6
3
1+
1 + cos θ
=−
2
14. csc θ = − 5, cos θ < 0 , so π < θ <
π θ 3π
θ
< <
, which means
lies in quadrant II.
2 2 4
2
−1
5
sin θ =
=−
, r = 5, y = −1
5
5
2
3+ 6
3
=−
2
x 2 + (−1) 2 =
3+ 6
=−
6
13. sec θ = 3, sin θ > 0 , so 0 < θ <
θ
2
cos θ =
a.
y2 = 9 −1 = 8
b.
sin(2θ ) = 2sin θ cos θ = 2 ⋅
b.
cos(2θ ) = cos θ − sin θ
2 2 1 4 2
⋅ =
3 3
9
2
c.
2
2
1 8
7
⎛1⎞ ⎛ 2 2 ⎞
= ⎜ ⎟ − ⎜⎜
⎟⎟ = − = −
9 9
9
⎝3⎠ ⎝ 3 ⎠
θ
2
=
d.
cos
θ
2
=
=
2
1
3 =
cos(2θ ) = cos 2 θ − sin 2 θ
d.
4
3 =
2
2
=
3
2
3
3
3
=
5⎞
⎟
5 ⎟⎠
2
3
5
⎛ 2 5⎞
1 − ⎜⎜ −
⎟
5 ⎟⎠
θ
1 − cos θ
⎝
=
sin =
2
2
2
=
2
3 = 1= 1 3= 3
2
3
3
3 3
1 + cos θ
2
1+
sin(2θ ) = 2sin θ cos θ
=
1 − cos θ
2
1
1−
3 =
=
2
2 5
5
⎛ 2 5⎞ ⎛
= ⎜⎜ −
⎟ −⎜−
5 ⎟⎠ ⎜⎝
⎝
20 5 15
=
−
=
=
25 25 25
2 2
3
a.
sin
5
=−
2
y= 8=2 2
c.
−2
⎛
5⎞ ⎛ 2 5⎞ 4
= 2 ⋅ ⎜⎜ −
⎟⎟ ⋅ ⎜⎜ −
⎟=
5 ⎟⎠ 5
⎝ 5 ⎠ ⎝
12 + y 2 = 32
2
2
x = −2
π
. Thus,
2
<
sin θ =
( 5)
x2 = 5 − 1 = 4
π
θ
, which means
lies in quadrant I.
2
4
1
cos θ = , x = 1 , r = 3 .
3
0<
3π
. Thus,
2
5+2 5
5
2
5+2 5
10
⎛ 2 5⎞
1 + ⎜⎜ −
⎟
5 ⎟⎠
θ
1 + cos θ
⎝
=−
cos = −
2
2
2
5−2 5
5
=−
2
6
3
=−
5−2 5
10
264
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Section 3.5: Double-angle and Half-angle Formulas
15. cot θ = −2, sec θ < 0 , so
π
< θ < π . Thus,
2
16. sec θ = 2, csc θ < 0 , so
π θ π
θ
< < , which means
lies in quadrant I.
4 2 2
2
x = − 2, y = 1
3π θ
θ
< < π , which means
lies in quadrant II.
4
2
2
1
cos θ = , x = 1, r = 2
2
12 + y 2 = 22
r 2 = (− 2) 2 + 12 = 4 + 1 = 5
r= 5
y2 = 4 −1 = 3
5
−2
2 5
, cos θ =
=−
sin θ =
=
5
5
5
5
1
a.
y= 3
sin(2θ ) = 2sin θ cos θ
2
⎛
3⎞ ⎛1⎞
3
= 2 ⋅ ⎜⎜ −
⎟⎟ ⋅ ⎜ ⎟ = −
2
⎝ 2 ⎠ ⎝2⎠
2
b.
=
2
c.
=
sin
θ
2
=
=
5+2 5
5
2
d.
cos
θ
2
5+2 5
10
=
=
⎛ 2 5⎞
1 + ⎜⎜ −
⎟
5 ⎟⎠
θ
1 + cos θ
⎝
cos =
=
2
2
2
=
cos(2θ ) = cos 2 θ − sin 2 θ
2
3⎞
1 3
1
⎛1⎞ ⎛
= ⎜ ⎟ − ⎜⎜ −
⎟⎟ = − = −
4 4
2
⎝2⎠ ⎝ 2 ⎠
⎛ 2 5⎞
1 − ⎜⎜ −
⎟
5 ⎟⎠
θ
1 − cos θ
⎝
sin =
=
2
2
2
=
d.
a.
cos(2θ ) = cos 2 θ − sin 2 θ
⎛ 2 5⎞ ⎛ 5⎞
= ⎜⎜ −
⎟ −⎜
⎟
5 ⎟⎠ ⎜⎝ 5 ⎟⎠
⎝
20 5 15 3
=
−
=
=
25 25 25 5
c.
3
2
sin(2θ ) = 2sin θ cos θ
sin θ = −
⎛ 5⎞ ⎛ 2 5⎞
20
4
= 2 ⋅ ⎜⎜
=−
⎟⎟ ⋅ ⎜⎜ −
⎟⎟ = −
5 ⎠
25
5
⎝ 5 ⎠ ⎝
b.
3π
< θ < 2π . Thus,
2
1 − cos θ
2
1−
2
1
2 =
1
2 = 1 =1
2
4 2
1 + cos θ
2
1+
2
1
2 =
3
2 =
2
17. tan θ = − 3, sin θ < 0 , so
3
3
=
4
2
3π
< θ < 2π . Thus,
2
3π θ
θ
< < π , which means
lies in quadrant II.
4 2
2
x = 1, y = −3
5−2 5
5
2
r 2 = 12 + (−3) 2 = 1 + 9 = 10
5−2 5
10
r = 10
sin θ =
a.
−3
10
=−
3 10
1
10
, cos θ =
=
10
10
10
sin ( 2θ ) = 2sin θ cos θ
⎛ 3 10 ⎞ ⎛ 10 ⎞
= 2 ⋅ ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 10 ⎠ ⎝ 10 ⎠
6
3
=− =−
10
5
265
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Chapter 3: Analytic Trigonometry
b.
cos(2θ ) = cos 2 θ − sin 2 θ
b.
2
cos(2θ ) = cos 2 θ − sin 2 θ
2
2
⎛ 3 10 ⎞ ⎛
10 ⎞
= ⎜⎜ −
⎟⎟ − ⎜⎜ −
⎟⎟
⎝ 10 ⎠ ⎝ 10 ⎠
90 10
80 4
=
−
=
=
100 100 100 5
⎛ 10 ⎞ ⎛ 3 10 ⎞
= ⎜⎜
⎟⎟ − ⎜⎜ −
⎟⎟
⎝ 10 ⎠ ⎝ 10 ⎠
10 90
80
4
=
−
=−
=−
100 100
100
5
c.
1−
θ
1 − cos θ
sin =
=
2
2
10
10
2
c.
⎛ 3 10 ⎞
1 − ⎜⎜ −
⎟
10 ⎟⎠
θ
1 − cos θ
⎝
=
sin =
2
2
2
10 − 10
10
=
2
d.
10 + 3 10
10
=
2
=
10 − 10
20
=
10 + 3 10
20
=
1 10 − 10
2
5
=
1 10 + 3 10
2
5
1+
θ
1 + cos θ
cos = −
=−
2
2
10
10
2
d.
10 + 10
10
=−
2
=−
10 + 10
20
=−
1 10 + 10
2
5
18. cot θ = 3, cos θ < 0 , so π < θ <
⎛ 3 10 ⎞
1 + ⎜⎜ −
⎟
10 ⎟⎠
θ
1 + cos θ
⎝
=−
cos = −
2
2
2
10 − 3 10
10
=−
2
3π
. Thus,
2
=
r 2 = (−3) 2 + (−1) 2 = 9 + 1 = 10
r = 10
=
1
10
=−
10
,
10
10 − 3 10
20
=−
1 10 − 3 10
2
5
1 − cos 45°
2
1−
2
2
2 =
2− 2
=
4
2− 2
2
2+ 2
=
4
2+ 2
2
⎛ 45° ⎞
20. cos 22.5° = cos ⎜
⎟
⎝ 2 ⎠
3
3 10
=−
cos θ = −
10
10
a.
=−
⎛ 45° ⎞
19. sin 22.5° = sin ⎜
⎟
⎝ 2 ⎠
π θ 3π
θ
which means
is in quadrant II.
< <
2
2 2 4
x = −3, y = −1
sin θ = −
2
=
sin ( 2θ ) = 2sin θ cos θ
⎛ 10 ⎞ ⎛ 3 10 ⎞ 6 3
= 2 ⋅ ⎜⎜ −
=
⎟⎟ ⋅ ⎜⎜ −
⎟⎟ =
⎝ 10 ⎠ ⎝ 10 ⎠ 10 5
=
1 + cos 45°
2
1+
2
2
2 =
266
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Section 3.5: Double-angle and Half-angle Formulas
7π
⎛ 7π ⎞
1 − cos
⎜ 4 ⎟
7π
4
21. tan
= tan ⎜
⎟ =−
7π
8
⎝ 2 ⎠
1 + cos
4
1 − cos 390°
⎛ 390° ⎞
24. sin195° = sin ⎜
⎟ =−
2
⎝ 2 ⎠
=−
2
2 ⋅2
=−
2 2
1+
2
1−
⎛ 2− 2 ⎞ ⎛ 2− 2 ⎞
= − ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 2+ 2 ⎠ ⎝ 2− 2 ⎠
(2 − 2 )
=−
2
25. sec
2
⎛ 2− 2 ⎞
= − ⎜⎜
⎟
2 ⎟⎠
⎝
=−
(
)
2 −1
= 1− 2
9π
⎛ 9π ⎞
1 − cos
⎜ 4 ⎟
9π
4
= tan ⎜
22. tan
⎟=
9π
⎝ 2 ⎠
8
1 + cos
4
2
2
2
=
⋅
2 2
1+
2
1−
⎛ 2− 2 ⎞ ⎛ 2− 2 ⎞
= ⎜⎜
⎟⎟
⎟⎟ ⋅ ⎜⎜
⎝ 2+ 2 ⎠ ⎝ 2− 2 ⎠
(2 − 2 )
=
=
3
2
1−
2
=−
2− 3
4
=−
2− 3
2
15π
1
1
=
=
15π
8
⎛ 15π ⎞
cos
⎜
⎟
8
cos ⎜ 4 ⎟
⎝ 2 ⎠
1
=
15π
1 + cos
4
2
1
=
2
1+
2
2
1
=
2+ 2
4
2
=
2+ 2
⎞ ⎛ 2+
⎟⋅⎜
⎟ ⎜ 2+
⎠ ⎝
⎛ 2 2+ 2 ⎞ ⎛ 2−
⎟⋅⎜
=⎜
⎜ 2 + 2 ⎟ ⎜⎝ 2 −
⎝
⎠
⎛
2
=⎜
⎜ 2+ 2
⎝
2
2
2− 2
2
=
= 2 −1
(
2 2− 2
(
= 2− 2
= −1 + 2
)
)
2⎞
⎟
2 ⎟⎠
2⎞
⎟
2 ⎟⎠
2+ 2
2
2+ 2
⎛ 330° ⎞
23. cos165° = cos ⎜
⎟
⎝ 2 ⎠
=−
=−
1 + cos 330°
2
1+
3
2 = − 2+ 3 = − 2+ 3
2
4
2
267
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Chapter 3: Analytic Trigonometry
26. csc
1
7π
1
=
=
⎛ 7π ⎞
8 sin 7 π
⎜
⎟
8
sin ⎜ 4 ⎟
⎝ 2 ⎠
1
=
7π
1 − cos
4
2
1
=
2
1−
2
2
1
=
2− 2
4
2
=
2− 2
⎛
⎞ ⎛ 2−
2
=⎜
⎟⋅⎜
⎜ 2− 2 ⎟ ⎜ 2−
⎝
⎠ ⎝
⎛2 2− 2 ⎞ ⎛2+
⎟⋅⎜
=⎜
⎜ 2 − 2 ⎟ ⎜⎝ 2 +
⎝
⎠
=
(
2 2+ 2
(
= 2+ 2
)
)
29. θ lies in quadrant II. Since x 2 + y 2 = 5 , r = 5 .
Now, the point (a, 2) is on the circle, so
a 2 + 22 = 5
a 2 = 5 − 22
a = − 5 − 2 2 = − 1 = −1
(a is negative because θ lies in quadrant II.)
Thus, sin θ =
a −1
5
=
=−
. Thus,
r
5
5
f ( 2θ ) = sin ( 2θ ) = 2sin θ cos θ
cos θ =
⎛2 5⎞ ⎛
5⎞
20
4
= 2 ⋅ ⎜⎜
=−
⎟⎟ ⋅ ⎜⎜ −
⎟⎟ = −
25
5
⎝ 5 ⎠ ⎝ 5 ⎠
30. From the solution to Problem 29, we have
2 5
5
and cos θ = −
.
sin θ =
5
5
2⎞
⎟
2 ⎟⎠
Thus, g ( 2θ ) = cos ( 2θ ) = cos 2 θ − sin 2 θ
2⎞
⎟
2 ⎟⎠
2
2− 2
2
2− 2
31. Note: Since θ lies in quadrant II,
quadrant I. Therefore, cos
⎛ π⎞
1 − cos ⎜ − ⎟
⎝ 4⎠
=−
2
=−
θ
2
θ
2
must lie in
is positive. From the
solution to Problem 29, we have cos θ = −
⎛− 5⎞
1 + ⎜⎜
⎟⎟
⎝ 5 ⎠
=
2
⎛ 3π ⎞
⎜− ⎟
⎛ 3π ⎞
28. cos ⎜ − ⎟ = cos ⎜ 4 ⎟
⎝ 2 ⎠
⎝ 8 ⎠
⎛ 3π ⎞
1 + cos ⎜ − ⎟
⎝ 4 ⎠
=
2
=
=
2− 2
=
4
5
.
5
θ
1 + cos θ
⎛θ ⎞
Thus, g ⎜ ⎟ = cos =
2
2
⎝2⎠
2
2 = − 2− 2 = − 2− 2
2
4
2
⎛
2⎞
1 + ⎜⎜ −
⎟
2 ⎟⎠
⎝
=
=
2
2
⎛
5⎞ ⎛2 5⎞
= ⎜⎜ −
⎟⎟ − ⎜⎜
⎟⎟
⎝ 5 ⎠ ⎝ 5 ⎠
5 20
15
3
=
−
=−
=−
25 25
25
5
⎛ π⎞
⎜− ⎟
⎛ π⎞
27. sin ⎜ − ⎟ = sin ⎜ 4 ⎟
⎝ 2 ⎠
⎝ 8⎠
1−
b
2
2 5
=
=
and
r
5
5
2− 2
2
=
5− 5
5
2
5− 5
10
(
10 5 − 5
)
10
268
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Section 3.5: Double-angle and Half-angle Formulas
32. Note: Since θ lies in quadrant II,
quadrant I. Therefore, sin
θ
2
θ
2
34. From the solution to Problem 29, we have
2 5
5
and cos θ = −
. Thus,
sin θ =
5
5
⎛− 5⎞
1 − ⎜⎜
⎟⎟
θ 1 − cos θ
⎛θ ⎞
⎝ 5 ⎠
h ⎜ ⎟ = tan =
=
2
sin θ
2 5
⎝2⎠
5
5+ 5
= 5
2 5
5
5+ 5
=
2 5
must lie in
is positive. From the
solution to Problem 29, we have cos θ = −
5
.
5
θ
1 − cos θ
⎛θ ⎞
Thus, f ⎜ ⎟ = sin =
2
2
⎝2⎠
⎛− 5⎞
1 − ⎜⎜
⎟
5 ⎟⎠
⎝
=
2
=
=
=
5+ 5
5
2
=
5+ 5
10
(
10 5 + 5
)
10
⎛ 1 ⎞
r = 1 = 1 . Now, the point ⎜ − , b ⎟ is on the
⎝ 4 ⎠
circle, so
2
⎛ 1⎞
2
⎜− ⎟ +b =1
⎝ 4⎠
a = − 5 − 2 2 = − 1 = −1
(a is negative because θ lies in quadrant II.)
b 2
= −2 .
Thus, tan θ = =
a −1
h ( 2θ ) = tan ( 2θ )
1 − ( −2 )
2
=
5
35. α lies in quadrant III. Since x 2 + y 2 = 1 ,
a 2 = 5 − 22
=
5
5 5 +5
10
5 +1 1+ 5
=
=
2
2
a 2 + 22 = 5
2 tan θ
1 − tan 2 θ
2 ( −2 )
2 5
⋅
=
33. θ lies in quadrant II. Since x 2 + y 2 = 5 , r = 5 .
Now, the point (a, 2) is on the circle, so
=
5+ 5
⎛ 1⎞
b2 = 1 − ⎜ − ⎟
⎝ 4⎠
2
2
15
15
⎛ 1⎞
b = − 1− ⎜ − ⎟ = −
=−
16
4
⎝ 4⎠
(b is negative because α lies in quadrant III.)
1
−
1
a
Thus, cos α = = 4 = − and
r
1
4
15
−
b
4 = − 15 . Thus,
sin α = =
r
1
4
2
g ( 2α ) = cos ( 2α ) = cos α − sin 2 α
−4
−4 4
=
=
1 − 4 −3 3
2
2
⎛ 1 ⎞ ⎛ 15 ⎞
= ⎜ − ⎟ − ⎜⎜
⎟
⎝ 4 ⎠ ⎝ 4 ⎟⎠
1 15
14
7
= − =− =−
16 16
16
8
269
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Chapter 3: Analytic Trigonometry
36. From the solution to Problem 35, we have
15
1
and cos α = − . Thus,
sin α = −
4
4
f ( 2α ) = sin ( 2α )
39. From the solution to Problem 35, we have
15
1
and cos α = − . Thus,
sin α = −
4
4
α 1 − cos α
⎛α ⎞
h ⎜ ⎟ = tan =
2
sin α
⎝2⎠
= 2sin α cos α
⎛ 1⎞
1− ⎜ − ⎟
⎝ 4⎠
=
15
−
4
5
= 4
15
−
4
5
=−
15
⎛ 15 ⎞ ⎛ 1 ⎞
15
= 2 ⋅ ⎜⎜ −
⎟⎟ ⋅ ⎜ − ⎟ =
8
⎝ 4 ⎠ ⎝ 4⎠
37. Note: Since α lies in quadrant III,
quadrant II. Therefore, sin
α
2
α
2
must lie in
is positive. From
the solution to Problem 35, we have cos α = −
α
⎛α ⎞
Thus, f ⎜ ⎟ = sin
2
⎝2⎠
=
1
.
4
1 − cos α
2
⎛ 1⎞
1− ⎜ − ⎟
⎝ 4⎠
=
2
=
5
4 = 5 = 5 ⋅ 2 = 10 = 10
2
8
8 2
16
4
38. Note: Since α lies in quadrant III,
quadrant II. Therefore, cos
α
2
α
2
15
⋅
=−
5 15
15
=−
15
3
15
15
40. α lies in quadrant III. Since x 2 + y 2 = 1 ,
⎛ 1 ⎞
r = 1 = 1 . Now, the point ⎜ − , b ⎟ is on the
⎝ 4 ⎠
circle, so
2
⎛ 1⎞
2
⎜− ⎟ +b =1
⎝ 4⎠
must lie in
is negative. From
⎛ 1⎞
b2 = 1 − ⎜ − ⎟
⎝ 4⎠
1
the solution to Problem 35, we have cos α = − .
4
Thus,
α
⎛α ⎞
g ⎜ ⎟ = cos
2
2
⎝ ⎠
=−
5
=−
2
2
15
15
⎛ 1⎞
b = − 1− ⎜ − ⎟ = −
=−
16
4
⎝ 4⎠
(b is negative because α lies in quadrant III.)
1 + cos α
2
15
−
b
4 = 15 .
Thus, tan θ = =
1
a
−
4
h ( 2α ) = tan ( 2α )
⎛ 1⎞
1+ ⎜ − ⎟
⎝ 4⎠
=−
2
=
3
3
3 2
6
6
=− 4 =−
=− ⋅ =−
=−
2
8
8 2
16
4
=
2 tan α
1 − tan 2 α
( 15 )
1 − ( 15 )
2
2
=
2 15 2 15
15
=
=−
1 − 15
−14
7
270
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Section 3.5: Double-angle and Half-angle Formulas
(
41. sin 4 θ = sin 2 θ
)
45. We use the result of problem 42 to help solve
this problem:
sin ( 5θ ) = sin(4θ + θ )
2
⎛ 1 − cos ( 2θ ) ⎞
=⎜
⎟
2
⎝
⎠
1
= ⎡⎣1 − 2 cos ( 2θ ) + cos 2 ( 2θ ) ⎤⎦
4
1 1
1
= − cos ( 2θ ) + cos 2 ( 2θ )
4 2
4
1 1
1 ⎛ 1 + cos ( 4θ ) ⎞
= − cos ( 2θ ) + ⎜
⎟
4 2
4⎝
2
⎠
1 1
1 1
= − cos ( 2θ ) + + cos ( 4θ )
4 2
8 8
3 1
1
= − cos ( 2θ ) + cos ( 4θ )
8 2
8
2
= sin ( 4θ ) cosθ + cos ( 4θ ) sin θ
(
(
) (
)
= cos θ 4sin θ − 8sin θ + 1 − 2sin 2 ( 2θ ) sin θ
2
(
3
)(
= 1 − sin θ 4sin θ − 8sin θ
2
3
(
)
2
)
+ sin θ 1 − 8sin 2 θ cos 2 θ
)
+ sin θ 1 − 2 ( 2sin θ cosθ )
= 4sin θ − 12sin θ + 8sin θ
3
5
(
= 4sin θ − 12sin θ + 8sin θ
3
5
(
+ sin θ − 8sin 3 θ 1 − sin 2 θ
)
= 5sin θ − 12sin θ + 8sin θ − 8sin θ + 8sin θ
3
42. sin ( 4θ ) = sin ( 2 ⋅ 2θ )
5
3
5
= 16sin 5 θ − 20sin 3 θ + 5sin θ
= 2sin ( 2θ ) cos ( 2θ )
(
= 2(2sin θ cos θ ) 1 − 2sin 2 θ
(
= 4sin θ cos θ 1 − 2sin 2 θ
)
46. We use the results from problems 42 and 44 to
help solve this problem:
cos(5θ ) = cos(4θ + θ )
)
(
= cos ( 4θ ) cos θ − sin ( 4θ ) sin θ
)
= ( cos θ ) ⎡ 4sin θ 1 − 2sin 2 θ ⎤
⎣
⎦
(
= ( cos θ ) 4sin θ − 8sin 3 θ
(
)
(
))
(
− cos θ 4sin θ − 8sin 3 θ sin θ
= 8cos θ − 8cos θ + cos θ
5
= cos ( 2θ ) cos θ − sin ( 2θ ) sin θ
3
− 4 cos θ sin 2 θ + 8cos θ sin 4 θ
)
= 2 cos 2 θ − 1 cos θ − 2sin θ cos θ sin θ
= 8cos5 θ − 8cos3 θ + cos θ
− 4 cos θ (1 − cos 2 θ ) + 8cos θ (1 − cos 2 θ ) 2
= 2 cos3 θ − cos θ − 2sin 2 θ cos θ
(
)
= 8cos 4 θ − 8cos 2 θ + 1 cos θ
43. cos(3θ ) = cos(2θ + θ )
(
)
= cosθ 4sin θ − 8sin 3 θ cosθ + cos ( 2(2θ ) ) sin θ
)
= 8cos5 θ − 8cos3 θ + cos θ − 4 cos θ
= 2 cos3 θ − cos θ − 2 1 − cos 2 θ cos θ
+ 4 cos3 θ + 8cos θ (1 − 2 cos 2 θ + cos 4 θ )
= 2 cos3 θ − cos θ − 2 cos θ + 2 cos3 θ
= 8cos5 θ − 4 cos3 θ − 3cos θ
= 4 cos3 θ − 3cos θ
+ 8cos θ − 16 cos3 θ + 8cos5 θ
44. cos ( 4θ ) = cos ( 2 ⋅ 2θ )
= 16 cos5 θ − 20 cos3 θ + 5cos θ
= 2 cos (2θ ) − 1
2
(
= 2 ( 4 cos
)
(
47. cos 4 θ − sin 4 θ = cos 2 θ + sin 2 θ
2
= 2 2 cos 2 θ − 1 − 1
4
)( cos
2
θ − sin 2 θ )
= 1 ⋅ cos ( 2θ )
θ − 4 cos 2 θ + 1) − 1
= cos ( 2θ )
= 8cos 4 θ − 8cos 2 θ + 2 − 1
= 8cos 4 θ − 8cos 2 θ + 1
271
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Chapter 3: Analytic Trigonometry
cos θ sin θ
−
cot θ − tan θ sin θ cos θ
=
48.
cot θ + tan θ cos θ + sin θ
sin θ cos θ
cos 2 θ − sin 2 θ
θ cos θ
= sin
cos 2 θ + sin 2 θ
sin θ cos θ
cos 2 θ − sin 2 θ
sin θ cos θ
=
⋅
sin θ cos θ
cos 2 θ + sin 2 θ
2
2
cos θ − sin θ
=
1
= cos ( 2θ )
49. cot(2θ ) =
50. cot(2θ ) =
1
1
=
2 tan θ
tan(2θ )
1 − tan 2 θ
1 − tan 2 θ
=
2 tan θ
1
1−
2
cot
θ
=
2
cot θ
cot 2 θ − 1
2
= cot θ
2
cot θ
cot 2 θ − 1 cot θ
=
⋅
2
cot 2 θ
2
cot θ − 1
=
2 cot θ
51. sec(2θ ) =
1
1
=
cos(2θ ) 2 cos 2 θ − 1
1
=
2
−1
sec 2 θ
1
=
2 − sec 2 θ
sec 2 θ
sec 2 θ
=
2 − sec 2 θ
52. csc ( 2θ ) =
1
1
=
sin ( 2θ ) 2sin θ cos θ
1 1
1
= ⋅
⋅
2 cos θ sin θ
1
= sec θ csc θ
2
53. cos 2 (2u ) − sin 2 (2u ) = cos [ 2(2u ) ] = cos(4u )
54. (4sin u cos u )(1 − 2sin 2 u )
= 2(2sin u cos u )(1 − 2sin 2 u )
= 2sin 2u cos 2u
= sin ( 2 ⋅ 2u )
= sin ( 4u )
55.
1
1
=
2 tan θ
tan(2θ )
1 − tan 2 θ
1 − tan 2 θ
=
2 tan θ
1⎛ 1
tan 2 θ ⎞
= ⎜
−
⎟
2 ⎝ tan θ tan θ ⎠
1
= ( cot θ − tan θ )
2
cos(2θ )
cos 2 θ − sin 2 θ
=
1 + sin(2θ ) 1 + 2sin θ cos θ
(cos θ − sin θ )(cos θ + sin θ )
=
cos 2 θ + sin 2 θ + 2sin θ cos θ
(cos θ − sin θ )(cos θ + sin θ )
=
(cos θ + sin θ )(cos θ + sin θ )
cos θ − sin θ
sin θ
=
cos θ + sin θ
sin θ
cos θ − sin θ
=
cos θ + sin θ
cos θ sin θ
−
= sin θ sin θ
cos θ sin θ
+
sin θ sin θ
cot θ − 1
=
cot θ + 1
272
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Section 3.5: Double-angle and Half-angle Formulas
1
4sin 2 θ cos 2 θ
4
1
2
= ( 2sin θ cos θ )
4
2
1
= ⎡⎣sin ( 2θ ) ⎤⎦
4
1 ⎡1 − cos ( 4θ ) ⎤
= ⋅⎢
⎥
4 ⎣
2
⎦
1
= ⎡⎣1 − cos ( 4θ ) ⎤⎦
8
56. sin 2 θ cos 2 θ =
(
)
60. tan
v 1 − cos v
1
cos v
=
=
−
= csc v − cot v
2
sin v
sin v sin v
1 − cos θ
1
+ cos θ
61.
1
− cos θ
θ
1 + tan 2
1+
2
1 + cos θ
1 + cos θ − (1 − cos θ )
1 + cos θ
=
1 + cos θ + 1 − cos θ
1 + cos θ
2 cos θ
1
= + cos θ
2
1 + cos θ
2 cos θ 1 + cos θ
=
⋅
1 + cos θ
2
= cos θ
1 − tan 2
1
1
2
⎛θ ⎞
=
=
57. sec 2 ⎜ ⎟ =
⎝ 2 ⎠ cos 2 ⎛ θ ⎞ 1 + cos θ 1 + cos θ
⎜ ⎟
2
⎝2⎠
1
1
2
⎛θ ⎞
=
=
58. csc 2 ⎜ ⎟ =
⎝ 2 ⎠ sin 2 ⎛ θ ⎞ 1 − cos θ 1 − cos θ
⎜ ⎟
2
⎝2⎠
62.
1
1
⎛v⎞
59. cot 2 ⎜ ⎟ =
=
1
cos
−
v
v
2
⎝ ⎠ tan 2 ⎛ ⎞
⎜ ⎟ 1 + cos v
⎝2⎠
1 + cos v
=
1 − cos v
1
1+
v
sec
=
1
1−
sec v
sec v + 1
= sec v
sec v − 1
sec v
sec v + 1 sec v
=
⋅
sec v sec v − 1
sec v + 1
=
sec v − 1
θ
2 =
1−
sin 3 θ + cos3 θ
sin θ + cos θ
=
( sin θ + cos θ ) ( sin 2 θ − sin θ cos θ + cos 2 θ )
sin θ + cos θ
= sin θ − sin θ cos θ + cos 2 θ
1
= sin 2 θ + cos 2 θ − ( 2sin θ cos θ )
2
1
= 1 − sin ( 2θ )
2
2
(
63.
)
sin(3θ ) cos(3θ ) sin ( 3θ ) cos θ − cos ( 3θ ) sin θ
−
=
sin θ
cos θ
sin θ cos θ
sin(3θ − θ )
=
sin θ cos θ
sin 2θ
=
sin θ cos θ
2sin θ cos θ
=
sin θ cos θ
=2
273
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Chapter 3: Analytic Trigonometry
cos θ + sin θ cos θ − sin θ ( cos θ + sin θ ) − ( cos θ − sin θ )
−
=
cos θ − sin θ cos θ + sin θ
( cos θ − sin θ )( cos θ + sin θ )
2
64.
=
2
(
cos 2 θ + 2 cos θ sin θ + sin 2 θ − cos 2 θ − 2 cos θ sin θ + sin 2 θ
)
cos θ − sin θ
cos 2 θ + 2 cos θ sin θ + sin 2 θ − cos 2 θ + 2 cos θ sin θ − sin 2 θ
=
cos 2 θ − sin 2 θ
4 cos θ sin θ
=
cos ( 2θ )
=
=
2
2
2(2sin θ cos θ )
cos ( 2θ )
2sin ( 2θ )
cos ( 2θ )
= 2 tan ( 2θ )
65. tan ( 3θ ) = tan(2θ + θ )
2 tan θ
2 tan θ + tan θ − tan 3 θ
+
tan
θ
2
tan ( 2θ ) + tan θ
3 tan θ − tan 3 θ 1 − tan 2 θ
3 tan θ − tan 3 θ
1 − tan 2 θ
=
= 1 − tan θ
=
=
⋅
=
2
2
2
2
1 − tan ( 2θ ) tan θ 1 − 2 tan θ ⋅ tan θ
1 − tan θ − 2 tan θ
1 − tan θ
1 − 3 tan θ
1 − 3 tan 2 θ
2
1 − tan θ
1 − tan 2 θ
66. tan θ + tan(θ + 120º ) + tan(θ + 240º )
tan θ + tan120º
tan θ + tan 240º
= tan θ +
+
1 − tan θ tan120º 1 − tan θ tan 240º
tan θ − 3
tan θ + 3
= tan θ +
+
1 − tan θ − 3 1 − tan θ 3
(
= tan θ +
=
(
)
tan θ − 3
1 + 3 tan θ
+
) (
( )
tan θ + 3
1 − 3 tan θ
)(
) (
)(
tan θ 1 − 3 tan θ + tan θ − 3 1 − 3 tan θ + tan θ + 3 1 + 3 tan θ
2
)
1 − 3 tan θ
tan θ − 3 tan 3 θ + tan θ − 3 tan 2 θ − 3 + 3 tan θ + tan θ + 3 tan 2 θ + 3 + 3 tan θ
=
1 − 3 tan 2 θ
−3 tan 3 θ + 9 tan θ
=
1 − 3 tan 2 θ
=
2
(
3 3 tan θ − tan 3 θ
)
1 − 3 tan θ
= 3 tan ( 3θ ) (from Problem 65)
2
274
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Section 3.5: Double-angle and Half-angle Formulas
67.
1
⋅ ln 1 − cos ( 2θ ) − ln 2
2
1
1 − cos 2θ
= ⋅ ln
2
2
(
⎛ 1 − cos ( 2θ )
= ln ⎜
⎜
2
⎝
(
= ln sin 2 θ
1/ 2
)
1/ 2
⎡
⎛ 3 ⎞⎤
73. tan ⎢ 2 cos −1 ⎜ − ⎟ ⎥
⎝ 5 ⎠⎦
⎣
⎛ 3⎞
Let α = cos −1 ⎜ − ⎟ . α lies in quadrant II.
⎝ 5⎠
3 π
≤α ≤ π.
Then cos α = − ,
5 2
5
sec α = −
3
)
⎞
⎟
⎟
⎠
tan α = − sec2 α − 1
= ln sin θ
2
68.
1
⋅ ln 1 + cos ( 2θ ) − ln 2
2
1
1 + cos 2θ
= ⋅ ln
2
2
(
⎛ 1 + cos ( 2θ )
= ln ⎜
⎜
2
⎝
(
= ln cos 2 θ
1/ 2
)
1/ 2
25
16
4
⎛ 5⎞
= − ⎜ − ⎟ −1 = −
−1 = −
=−
9
9
3
⎝ 3⎠
)
⎡
2 tan α
⎛ 3 ⎞⎤
tan ⎢ 2 cos −1 ⎜ − ⎟ ⎥ = tan 2α =
1 − tan 2 α
⎝ 5 ⎠⎦
⎣
⎛ 4⎞
2⎜ − ⎟
3⎠
= ⎝
2
⎛ 4⎞
1− ⎜ − ⎟
⎝ 3⎠
8
−
3 ⋅9
=
16 9
1−
9
−24
=
9 − 16
−24
=
−7
24
=
7
⎞
⎟
⎟
⎠
= ln cos θ
1⎞
π
3
⎛
⎛ π⎞
69. sin ⎜ 2sin −1 ⎟ = sin ⎜ 2 ⋅ ⎟ = sin =
2⎠
3
2
⎝
⎝ 6⎠
⎡
3⎤
2π
3
⎛ π⎞
=
70. sin ⎢ 2sin −1
⎥ = sin ⎜ 2 ⋅ ⎟ = sin
2 ⎦
3
2
⎝ 3⎠
⎣
3⎞
3⎞
⎛
⎛
71. cos ⎜ 2sin −1 ⎟ = 1 − 2sin 2 ⎜ sin −1 ⎟
5
5
⎝
⎠
⎝
⎠
⎛3⎞
= 1− 2⎜ ⎟
⎝5⎠
18
= 1−
25
7
=
25
2
3⎞
⎛
74. tan ⎜ 2 tan −1 ⎟ =
4⎠
⎝
3⎞
⎛
2 tan ⎜ tan −1 ⎟
4⎠
⎝
2⎛
−1 3 ⎞
1 − tan ⎜ tan
⎟
4⎠
⎝
⎛3⎞
2⋅⎜ ⎟
⎝4⎠
=
2
⎛3⎞
1− ⎜ ⎟
⎝4⎠
3
16
= 2 ⋅
9 16
1−
16
24
=
16 − 9
24
=
7
4⎞
4⎞
⎛
⎛
72. cos ⎜ 2 cos −1 ⎟ = 2 cos 2 ⎜ cos −1 ⎟ − 1
5
5
⎝
⎠
⎝
⎠
2
⎛4⎞
= 2 ⎜ ⎟ −1
⎝5⎠
32
=
−1
25
7
=
25
275
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Chapter 3: Analytic Trigonometry
4⎞
⎛
75. sin ⎜ 2 cos −1 ⎟
5
⎝
⎠
−1 4
Let α = cos
. α is in quadrant I.
5
4
π
Then cos α = , 0 ≤ α ≤ .
5
2
3⎞
⎛1
78. cos 2 ⎜ sin −1 ⎟
2
5
⎝
⎠
−1 3
Let α = sin
. α is in quadrant I. Then
5
3
π
sin α = , 0 < α < .
5
2
sin α = 1 − cos 2 α
2
16
⎛4⎞
= 1− ⎜ ⎟ = 1−
=
5
25
⎝ ⎠
cos α = 1 − sin 2 α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
5
25
25 5
⎝ ⎠
9
3
=
25 5
3⎞
⎛1
⎛1 ⎞
cos 2 ⎜ sin −1 ⎟ = cos 2 ⎜ ⋅ α ⎟
5⎠
⎝2
⎝2 ⎠
4⎞
⎛
sin ⎜ 2 cos −1 ⎟ = sin 2α
5⎠
⎝
3 4 24
= 2sin α cos α = 2 ⋅ ⋅ =
5 5 25
1 + cos α
=
=
2
⎡
⎛ 4 ⎞⎤
76. cos ⎢ 2 tan −1 ⎜ − ⎟ ⎥
⎝ 3 ⎠⎦
⎣
⎛ 4⎞
Let α = tan −1 ⎜ − ⎟ . α is in quadrant IV.
⎝ 3⎠
4
π
Then tan α = − , − < α < 0 .
3
2
3⎞
⎛
79. sec ⎜ 2 tan −1 ⎟
4⎠
⎝
⎛3⎞
Let α = tan −1 ⎜ ⎟ . α is in quadrant I.
⎝4⎠
3
π
Then tan α = , 0 < α < .
4
2
sec α = tan 2 α + 1
2
16
⎛ 4⎞
= ⎜ − ⎟ +1 =
+1 =
3
9
⎝
⎠
cos α =
4 9
5=5= 9
2
2 10
1+
sec α = tan 2 α + 1
25 5
=
9
3
2
9
25 5
⎛3⎞
= ⎜ ⎟ +1 =
+1 =
=
16
16 4
⎝4⎠
3
5
cos α =
⎡
⎛ 4 ⎞⎤
cos ⎢ 2 tan −1 ⎜ − ⎟ ⎥ = cos 2α = 2 cos 2 α − 1
⎝ 3 ⎠⎦
⎣
2
⎛3⎞
= 2 ⎜ ⎟ −1
⎝5⎠
18
=
−1
25
7
=−
25
4
5
3⎞
1
⎛
sec ⎜ 2 tan −1 ⎟ = sec ( 2α ) =
4⎠
cos ( 2α )
⎝
1
=
2 cos 2 α − 1
1
=
2
⎛4⎞
2 ⎜ ⎟ −1
⎝5⎠
1
=
32
−1
25
1
=
7
25
25
=
7
3⎞
⎛
1 − cos ⎜ cos −1 ⎟ 1 − 3
3⎞
5⎠
⎛1
⎝
5
77. sin 2 ⎜ cos −1 ⎟ =
=
2
5
2
2
⎝
⎠
2
=5
2
1
=
5
276
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Section 3.5: Double-angle and Half-angle Formulas
b. Here we have D = 15 and W = 6.5 .
⎡
⎛ 3 ⎞⎤
80. csc ⎢ 2sin −1 ⎜ − ⎟ ⎥
⎝ 5 ⎠⎦
⎣
⎛ 3⎞
Let α = sin −1 ⎜ − ⎟ . α is in quadrant IV.
⎝ 5⎠
3
π
Then sin α = − , − ≤ α ≤ 0 .
5
2
⎛ 3⎞
cos α = 1 − sin 2 α = 1 − ⎜ − ⎟
⎝ 5⎠
= 1−
6.5 = 2 (15 ) tan
2
13
tan =
2 60
θ
13
= tan −1
2
60
−1 13
θ = 2 tan
≈ 24.45°
60
θ
2
9
25
82.
=
1
⎡
⎛ 3 ⎞⎤
csc ⎢ 2sin −1 ⎜ − ⎟ ⎥ = csc ( 2α ) =
sin ( 2α )
⎝ 5 ⎠⎦
⎣
1
=
2sin α cos α
1
=
⎛ 3 ⎞⎛ 4 ⎞
2 ⎜ − ⎟⎜ ⎟
⎝ 5 ⎠⎝ 5 ⎠
1
=
24
−
25
25
=−
24
D=
(
) ( sin θ cosθ ) + I xy ( cos2 θ − sin 2 θ )
(
) 12 sin 2θ + I xy cos 2θ
= Ix − I y
4
=
5
81. a.
(
I x sin θ cos θ − I y sin θ cos θ + I xy cos 2 θ − sin 2 θ
= Ix − I y
16
25
=
83. a.
Ix − I y
2
sin 2θ + I xy cos 2θ
v02 2
cos θ (sin θ − cos θ )
16
v2 2
(cos θ sin θ − cos 2 θ )
= 0
16
v2 2 1
= 0
⋅ (2 cos θ sin θ − 2 cos 2 θ )
16 2
v2 2 ⎡
⎛ 1 + cos 2θ ⎞ ⎤
= 0
sin 2θ − 2 ⎜
⎟⎥
⎢
32 ⎣
2
⎝
⎠⎦
R(θ ) =
v02 2
⎡sin ( 2θ ) − 1 − cos ( 2θ ) ⎤⎦
32 ⎣
v2 2
= 0
⎡sin ( 2θ ) − cos ( 2θ ) − 1⎦⎤
32 ⎣
=
1W
2
csc θ − cot θ
W = 2 D ( csc θ − cot θ )
b. Let Y1 =
1
cos θ 1 − cos θ
csc θ − cot θ =
−
=
sin θ sin θ
sin θ
= tan
20
θ
2
Therefore, W = 2 D tan
θ
2
θ
322 2
[sin(2 x) − cos(2 x)]
32
.
45°
0
90°
277
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)
Chapter 3: Analytic Trigonometry
c.
Using the MAXIMUM feature on the
calculator:
c.
20
The largest value of the sine function is 1.
Solve:
sin 2θ = 1
2θ =
45°
θ=
90°
0
R has the largest value when θ = 67.5º .
d.
1
1
84. y = sin(2π x) + sin(4π x)
2
4
1
1
= sin(2π x) + sin(2 ⋅ 2π x)
2
4
1
1
= sin(2π x) + [ 2sin(2π x) cos(2π x) ]
2
4
1
1
= sin(2π x) + [sin(2π x) cos(2π x) ]
2
2
1
1
= sin(2π x) + ⎡⎣sin(2π x) ⋅ 2 cos 2 (π x) − 1 ⎤⎦
2
2
1
1
= sin(2π x) + sin(2π x) cos 2 (π x) − sin(2π x)
2
2
= sin(2π x) cos 2 (π x)
(
85. Let b represent the base of the triangle.
θ h
θ b/2
cos =
sin =
2 s
s
2
h = s cos
θ
b = 2 s sin
2
)
θ
2
b.
2 cos θ sin θ = 2sin θ cos θ = sin(2θ )
π
2
π
2
y = sin =
=
4
2
4
2
The dimensions of the largest rectangle are
2
.
2 by
2
x = cos
cos 2 θ − sin 2 θ
cos 2 θ + sin 2 θ
cos 2 θ − sin 2 θ
cos 2 θ
=
2
cos θ + sin 2 θ
cos 2 θ
1 − tan 2 θ 4
=
⋅
1 + tan 2 θ 4
4 − 4 tan 2 θ
=
4 + 4 tan 2 θ
y
x
= y; cos θ = = x
1
1
A = 2 xy = 2 cos θ sin θ
= 45°
88. cos ( 2θ ) = cos 2 θ − sin 2 θ =
θ
a.
4
2sin θ cos 2 θ
⋅
cos θ
1
sin θ
2⋅
= cos θ
1
cos 2 θ
2 tan θ
=
sec2 θ
2 tan θ 4
=
⋅
1 + tan 2 θ 4
4(2 tan θ )
=
4 + (2 tan θ ) 2
4x
=
4 + x2
= s 2 sin cos
2
2
1 2
= s sin θ
2
86. sin θ =
2
π
87. sin ( 2θ ) = 2sin θ cos θ =
1
A = b⋅h
2
1 ⎛
θ ⎞⎛
θ⎞
= ⋅ ⎜ 2s sin ⎟⎜ s cos ⎟
2 ⎝
2 ⎠⎝
2⎠
θ
π
=
=
4 − ( 2 tan θ )
2
4 + ( 2 tan θ )
2
4 − x2
4 + x2
278
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Section 3.5: Double-angle and Half-angle Formulas
89.
1
1
⋅ sin 2 x + C = − ⋅ cos ( 2 x )
2
4
1
1
C = − ⋅ cos ( 2 x ) − ⋅ sin 2 x
4
2
1
= − ⋅ cos ( 2 x ) + 2sin 2 x
4
1
= − ⋅ 1 − 2sin 2 x + 2sin 2 x
4
1
= − ⋅ (1)
4
1
=−
4
(
)
(
90.
⎛α ⎞
92. If z = tan ⎜ ⎟ , then
⎝2⎠
⎛α ⎞
1 − tan 2 ⎜ ⎟
1− z2
⎝2⎠
=
1+ z2
2 ⎛α ⎞
1 + tan ⎜ ⎟
⎝2⎠
1 − cos α
1−
+ cos α
1
=
1 − cos α
1+
1 + cos α
1 + cos α − (1 − cos α )
1 + cos α
=
1 + cos α + 1 − cos α
1 + cos α
1 + cos α − (1 − cos α )
=
1 + cos α + 1 − cos α
2 cos α
=
2
= cos α
)
1
1
⋅ cos 2 x + C = ⋅ cos ( 2 x )
2
4
1
1
C = ⋅ cos ( 2 x ) − ⋅ cos 2 x
4
2
1
1
2
= ⋅ 2 cos x − 1 − cos 2 x
4
2
1
1
1
= cos 2 x − − cos 2 x
2
4 2
1
=−
4
(
)
93.
⎛α ⎞
91. If z = tan ⎜ ⎟ , then
⎝2⎠
⎛α ⎞
2 tan ⎜ ⎟
2z
⎝2⎠
=
2
1+ z
⎛α ⎞
1 + tan 2 ⎜ ⎟
⎝2⎠
⎛α ⎞
2 tan ⎜ ⎟
⎝2⎠
=
⎛α ⎞
sec2 ⎜ ⎟
⎝2⎠
⎛α ⎞
⎛α ⎞
= 2 tan ⎜ ⎟ cos 2 ⎜ ⎟
⎝2⎠
⎝2⎠
f ( x) = sin 2 x =
1 − cos ( 2 x )
2
Starting with the graph of y = cos x , compress
horizontally by a factor of 2, reflect across the xaxis, shift 1 unit up, and shrink vertically by a
factor of 2.
94. g ( x) = cos 2 x =
1 + cos ( 2 x )
2
Starting with the graph of y = cos x , compress
horizontally by a factor of 2, reflect across the xaxis, shift 1 unit up, and shrink vertically by a
factor of 2.
⎛α ⎞
2sin ⎜ ⎟
⎝ 2 ⎠ ⋅ cos 2 ⎛ α ⎞
=
⎜ ⎟
⎛α ⎞
⎝2⎠
cos ⎜ ⎟
⎝2⎠
⎛α ⎞
⎛α ⎞
= 2sin ⎜ ⎟ cos ⎜ ⎟
2
⎝ ⎠
⎝2⎠
⎡ ⎛ α ⎞⎤
= sin ⎢ 2 ⎜ ⎟ ⎥
⎣ ⎝ 2 ⎠⎦
= sin α
279
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Chapter 3: Analytic Trigonometry
π
⎛π⎞
1 + cos
⎜4⎟
π
4
96. cos = cos ⎜ ⎟ =
⎝2⎠
8
2
π
⎛ π ⎞
1 − cos
⎜ 12 ⎟
π
12
= sin ⎜ ⎟ =
95. sin
24
⎝ 2 ⎠
2
(
⎛1
1− ⎜
⎝4
=
=
=
8−2
(
)
⎞
6+ 2 ⎟
⎠ = 1−1
2
2 8
6+ 2
16
( (
2 4−
)=
6+ 2
4
))
8−2
(
(
6+ 2
6+ 2
)
=
)
=
=
8+2
(
(
2
4− 6 − 2
4
=
)
⎞
6+ 2 ⎟
⎠ = 1+1
2
2 8
6+ 2
16
(
2 4+ 6 + 2
4
)=
)=
8+2
(
(
6+ 2
2+ 2
4
2+ 2
2
π
⎛π⎞
1 − cos
⎜8⎟
π
8
sin = sin ⎜ ⎟ =
16
⎝2⎠
2
π
⎛ π ⎞
1 + cos
⎜ 12 ⎟
π
12
cos
= cos ⎜ ⎟ =
24
⎝ 2 ⎠
2
⎛1
1+ ⎜
⎝4
=
2
2
2 =
=
4
=
1+
6+ 2
)
=
)
1−
2+ 2
2
=
2
2− 2+ 2
4
2− 2+ 2
2
π
⎛π⎞
1 + cos
⎜8⎟
π
8
cos = cos ⎜ ⎟ =
16
⎝2⎠
2
4
2
4+ 6 + 2
4
=
=
1+
2+ 2
2
=
2
2+ 2+ 2
4
2+ 2+ 2
2
97. sin 3 θ + sin 3 (θ + 120º ) + sin 3 (θ + 240º )
= sin 3 θ + ( sin θ cos (120º ) + cos θ sin (120º ) ) + ( sin θ cos ( 240º ) + cos θ sin ( 240º ) )
3
3
3
3
⎛ 1
⎞ ⎛ 1
⎞
3
3
= sin θ + ⎜⎜ − ⋅ sin θ +
⋅ cos θ ⎟⎟ + ⎜⎜ − ⋅ sin θ −
⋅ cos θ ⎟⎟
2
2
⎝ 2
⎠ ⎝ 2
⎠
1
= sin 3 θ + ⋅ − sin 3 θ + 3 3 sin 2 θ cos θ − 9sin θ cos 2 θ + 3 3 cos3 θ
8
1
− sin 3 θ + 3 3 sin 2 θ cos θ + 9sin θ cos 2 θ + 3 3 cos3 θ
8
3
(
)
(
)
1
3 3
9
3 3
= sin 3 θ − ⋅ sin 3 θ +
⋅ sin 2 θ cos θ − ⋅ sin θ cos 2 θ +
⋅ cos3 θ
8
8
8
8
1
3 3
9
3 3
− ⋅ sin 3 θ −
⋅ sin 2 θ cos θ − ⋅ sin θ cos 2 θ −
⋅ cos3 θ
8
8
8
8
3
9
3
3
= ⋅ sin 3 θ − ⋅ sin θ cos 2 θ = ⋅ ⎡⎣sin 3 θ − 3sin θ 1 − sin 2 θ ⎤⎦ = ⋅ sin 3 θ − 3sin θ + 3sin 3 θ
4
4
4
4
3
3
= ⋅ 4sin 3 θ − 3sin θ = − ⋅ sin ( 3θ )
4
4
(
(
)
(
)
)
280
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Section 3.6: Product-to-Sum and Sum-to-Product Formulas
⎛ θ⎞
98. tan θ = tan ⎜ 3 ⋅ ⎟
⎝ 3⎠
=
3 tan
θ
3
− tan 3
1 − 3 tan
1
[cos(3θ − 5θ ) − cos(3θ + 5θ )]
2
1
= ⎡⎣ cos(− 2θ ) − cos ( 8θ ) ⎤⎦
2
1
= ⎡⎣ cos ( 2θ ) − cos ( 8θ ) ⎤⎦
2
4. sin(3θ ) sin(5θ ) =
θ
2
θ
3
(from problem 65)
3
θ⎛
θ⎞
tan ⎜ 3 − tan 2 ⎟
3⎝
3⎠
a tan =
3
2 θ
1 − 3 tan
3
θ
θ
θ⎛
θ⎞
3 tan − tan 3 = a tan ⎜ 1 − 3 tan 2 ⎟
3
3
3⎝
3⎠
1
[cos(3θ − 5θ ) + cos(3θ + 5θ )]
2
1
= ⎡⎣ cos(− 2θ ) + cos ( 8θ ) ⎤⎦
2
1
= ⎡⎣ cos ( 2θ ) + cos ( 8θ ) ⎤⎦
2
θ
3 − tan 2
3 − tan
3a tan
2
θ
3
− tan
θ
3
2 θ
3
2
( 3a − 1) tan 2
θ
3
θ
3
θ
5. cos(3θ ) cos(5θ ) =
θ⎞
⎛
= a ⎜ 1 − 3 tan 2 ⎟
3⎠
⎝
= a − 3a tan
1
[sin(4θ + 6θ ) + sin(4θ − 6θ )]
2
1
= ⎡⎣sin (10θ ) + sin(− 2θ ) ⎤⎦
2
1
= ⎡⎣sin (10θ ) − sin ( 2θ ) ⎤⎦
2
6. sin(4θ ) cos(6θ ) =
2θ
3
= a−3
= a−3
1
[cos(θ − 2θ ) − cos(θ + 2θ )]
2
1
= ⎡⎣ cos(−θ ) − cos ( 3θ ) ⎤⎦
2
1
= ⎡⎣ cos θ − cos ( 3θ ) ⎤⎦
2
a −3
3 3a − 1
a −3
θ
tan = ±
3
3a − 1
tan 2
7. sin θ sin(2θ ) =
=
99. Answers will vary.
1
[cos(3θ − 4θ ) + cos(3θ + 4θ )]
2
1
= ⎡⎣ cos(− θ ) + cos ( 7θ ) ⎤⎦
2
1
= ⎡⎣ cos θ + cos ( 7θ ) ⎤⎦
2
8. cos(3θ ) cos(4θ ) =
Section 3.6
1
[cos(4θ − 2θ ) − cos(4θ + 2θ )]
2
1
= ⎡⎣ cos ( 2θ ) − cos ( 6θ ) ⎤⎦
2
1. sin(4θ ) sin(2θ ) =
9. sin
1
[cos(4θ − 2θ ) + cos(4θ + 2θ )]
2
1
= ⎡⎣cos(2θ ) + cos ( 6θ ) ⎤⎦
2
2. cos(4θ ) cos(2θ ) =
3θ
θ 1 ⎡ ⎛ 3θ θ ⎞
⎛ 3θ θ ⎞ ⎤
cos = ⎢sin ⎜ + ⎟ + sin ⎜ − ⎟ ⎥
2
2 2⎣ ⎝ 2 2⎠
⎝ 2 2 ⎠⎦
1
= ⎣⎡sin ( 2θ ) + sin θ ⎦⎤
2
θ
5θ 1 ⎡ ⎛ θ 5θ ⎞
⎛ θ 5θ ⎞ ⎤
= sin ⎜ + ⎟ + sin ⎜ − ⎟ ⎥
10. sin cos
2
2 2 ⎢⎣ ⎝ 2 2 ⎠
⎝ 2 2 ⎠⎦
1
= ⎡⎣sin ( 3θ ) + sin(− 2θ ) ⎤⎦
2
1
= ⎡⎣sin ( 3θ ) − sin ( 2θ ) ⎤⎦
2
1
[sin(4θ + 2θ ) + sin(4θ − 2θ )]
2
1
= ⎡⎣sin ( 6θ ) + sin ( 2θ ) ⎤⎦
2
3. sin(4θ ) cos(2θ ) =
281
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Chapter 3: Analytic Trigonometry
⎛ 4θ − 2θ ⎞
⎛ 4θ + 2θ ⎞
11. sin(4θ ) − sin(2θ ) = 2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
= 2sin θ cos ( 3θ )
⎛ θ + 3θ ⎞
⎛ θ − 3θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
2sin(2θ )
2sin(2θ ) cos(−θ )
=
2sin(2θ )
= cos(−θ )
= cos θ
sin θ + sin(3θ )
=
19.
2sin(2θ )
⎛ 4θ + 2θ ⎞
⎛ 4θ − 2θ ⎞
12. sin(4θ ) + sin(2θ ) = 2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
= 2sin ( 3θ ) cos θ
⎛ θ + 3θ ⎞
⎛ θ − 3θ ⎞
2 cos ⎜
cos ⎜
⎟
⎟
cos θ + cos(3θ )
⎝ 2 ⎠
⎝ 2 ⎠
=
20.
2 cos(2θ )
2 cos(2θ )
2 cos(2θ ) cos(−θ )
=
2 cos(2θ )
= cos(−θ )
⎛ 2θ + 4θ ⎞
⎛ 2θ − 4θ ⎞
13. cos(2θ ) + cos(4θ ) = 2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
= 2 cos ( 3θ ) cos(−θ )
= 2 cos ( 3θ ) cos θ
⎛ 5θ + 3θ ⎞ ⎛ 5θ − 3θ ⎞
14. cos(5θ ) − cos(3θ ) = − 2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
= − 2sin ( 4θ ) sin θ
= cos θ
⎛ 4θ + 2θ ⎞
⎛ 4θ − 2θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
cos(4θ ) + cos(2θ )
2sin(3θ ) cos θ
=
2 cos(3θ ) cos θ
sin(3θ )
=
cos(3θ )
= tan(3θ )
sin(4θ ) + sin(2θ )
=
21.
cos(4θ ) + cos(2θ )
⎛ θ + 3θ ⎞
⎛ θ − 3θ ⎞
15. sin θ + sin(3θ ) = 2sin ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
= 2sin ( 2θ ) cos(−θ )
= 2sin ( 2θ ) cos θ
⎛ θ + 3θ ⎞
⎛ θ − 3θ ⎞
16. cos θ + cos(3θ ) = 2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
= 2 cos ( 2θ ) cos(−θ )
⎛ θ + 3θ ⎞ ⎛ θ − 3θ ⎞
−2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ 3θ − θ ⎞
⎛ 3θ + θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
− 2sin(2θ ) sin(−θ )
=
2sin θ cos(2θ )
−(− sin θ ) sin(2θ )
=
sin θ cos(2θ )
= tan(2θ )
= 2 cos ( 2θ ) cos θ
⎛ θ 3θ ⎞ ⎛ θ 3θ
⎜ +
⎟ ⎜ −
3θ
17. cos − cos
= − 2sin ⎜ 2 2 ⎟ sin ⎜ 2 2
2
2
⎝ 2 ⎠ ⎝ 2
⎛ θ⎞
= − 2sin θ sin ⎜ − ⎟
⎝ 2⎠
θ
cos θ − cos(3θ )
=
22.
sin(3θ ) − sin θ
⎞
⎟
⎟
⎠
θ⎞
⎛
= − 2sin θ ⎜ − sin ⎟
2⎠
⎝
= 2sin θ sin
θ
⎛ θ + 3θ ⎞ ⎛ θ − 3θ ⎞
−2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ θ + 3θ ⎞
⎛ θ − 3θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
− 2sin(2θ ) sin(−θ )
=
2sin(2θ ) cos(−θ )
−(− sin θ )
=
cos θ
= tan θ
cos θ − cos(3θ )
23.
=
sin θ + sin(3θ )
2
⎛ θ 3θ
⎜ −
3θ
= 2sin ⎜ 2 2
18. sin − sin
⎝ 2
2
2
θ
⎞
⎛ θ 3θ
⎟
⎜2+ 2
⎟ cos ⎜
⎠
⎝ 2
⎞
⎟
⎟
⎠
⎛ θ⎞
= 2sin ⎜ − ⎟ cos θ
⎝ 2⎠
= − 2sin
θ
2
cos θ
282
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.6: Product-to-Sum and Sum-to-Product Formulas
⎛ θ + 5θ ⎞ ⎛ θ − 5θ ⎞
−2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ θ + 5θ ⎞
⎛ θ − 5θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
− 2sin(3θ ) sin(− 2θ )
=
2sin(3θ ) cos(− 2θ )
28.
cos θ − cos(5θ )
24.
=
sin θ + sin(5θ )
=
⎛ 4θ − 8θ ⎞
⎛ 4θ + 8θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
=
⎛ 4θ + 8θ ⎞ ⎛ 4θ − 8θ ⎞
−2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
2sin(− 2θ ) cos(6θ )
=
− 2sin(6θ ) sin(− 2θ )
cos(6θ )
=
− sin(6θ )
= − cot(6θ )
−(− sin 2θ )
cos ( 2θ )
= tan ( 2θ )
25. sin θ [sin θ + sin(3θ ) ]
⎡
⎛ θ + 3θ ⎞
⎛ θ − 3θ ⎞ ⎤
= sin θ ⎢ 2sin ⎜
⎟ cos ⎜
⎟⎥
2
⎝
⎠
⎝ 2 ⎠⎦
⎣
= sin θ [ 2sin(2θ ) cos(−θ ) ]
29.
sin(4θ ) + sin(8θ )
sin(4θ ) − sin(8θ )
⎛ 4θ + 8θ ⎞
⎛ 4θ − 8θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
=
⎛ 4θ − 8θ ⎞
⎛ 4θ + 8θ ⎞
−2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
2sin(6θ ) cos(− 2θ )
=
2sin(− 2θ ) cos(6θ )
sin(6θ ) cos(2θ )
=
− sin(2θ ) cos(6θ )
= − tan(6θ ) cot(2θ )
tan(6θ )
=−
tan(2θ )
= cos θ [ 2sin(2θ ) sin θ ]
⎡ 1
⎤
= cos θ ⎢ 2 ⋅ [ cos θ − cos(3θ ) ]⎥
⎣ 2
⎦
= cos θ [ cos θ − cos(3θ ) ]
26. sin θ ⎡⎣sin ( 3θ ) + sin(5θ ) ⎤⎦
⎡
⎛ 3θ + 5θ ⎞
⎛ 3θ − 5θ ⎞ ⎤
= sin θ ⎢ 2sin ⎜
⎟ cos ⎜
⎟⎥
⎝ 2 ⎠
⎝ 2 ⎠⎦
⎣
= sin θ [ 2sin(4θ ) cos(−θ ) ]
= cos θ [ 2sin(4θ ) sin θ ]
⎡ 1
⎤
= cos θ ⎢ 2 ⋅ ⎡⎣ cos ( 3θ ) − cos(5θ ) ⎤⎦ ⎥
⎣ 2
⎦
= cos θ ⎡⎣ cos ( 3θ ) − cos(5θ ) ⎤⎦
27.
sin(4θ ) − sin(8θ )
cos(4θ ) − cos(8θ )
30.
cos(4θ ) − cos(8θ )
cos(4θ ) + cos(8θ )
⎛ 4θ + 8θ ⎞ ⎛ 4θ − 8θ ⎞
−2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
=
⎛ 4θ + 8θ ⎞
⎛ 4θ − 8θ ⎞
2 cos ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
−2sin(6θ )sin(−2θ )
=
2 cos(6θ ) cos(−2θ )
sin(6θ ) sin( −2θ )
=−
⋅
cos(6θ ) cos(−2θ )
= − tan(6θ ) tan(−2θ )
= tan(2θ ) tan(6θ )
sin(4θ ) + sin(8θ )
cos(4θ ) + cos(8θ )
⎛ 4θ + 8θ ⎞
⎛ 4θ − 8θ ⎞
2sin ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
=
⎛ 4θ + 8θ ⎞
⎛ 4θ − 8θ ⎞
2 cos ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
2sin(6θ ) cos(− 2θ )
=
2 cos(6θ ) cos(− 2θ )
sin(6θ )
=
cos(6θ )
= tan(6θ )
283
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Chapter 3: Analytic Trigonometry
sin α + sin β
31.
=
sin α − sin β
⎛α + β ⎞
⎛α − β
2sin ⎜
⎟ cos ⎜
2
⎝
⎠
⎝ 2
⎛α − β ⎞
⎛α + β
2sin ⎜
⎟ cos ⎜
⎝ 2 ⎠
⎝ 2
⎞
⎟
⎠
⎞
⎟
⎠
sin α + sin β
33.
=
cos α + cos β
⎛α + β ⎞
⎛α − β ⎞
2sin ⎜
⎟ cos ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
⎛α + β ⎞
⎛α − β ⎞
2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛α + β ⎞
⎛α − β ⎞
sin ⎜
⎟ cos ⎜
⎟
2
⎝
⎠⋅
⎝ 2 ⎠
=
⎛α + β ⎞
⎛α − β ⎞
cos ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛α + β ⎞
sin ⎜
⎟
⎝ 2 ⎠
=
⎛α + β ⎞
cos ⎜
⎟
⎝ 2 ⎠
⎛α + β ⎞ ⎛α − β ⎞
= tan ⎜
⎟ cot ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛α + β ⎞
= tan ⎜
⎟
⎝ 2 ⎠
cos α + cos β
32.
=
cos α − cos β
⎛α + β ⎞
⎛α − β ⎞
2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛α + β ⎞ ⎛α − β ⎞
− 2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
sin α − sin β
34.
=
cos α − cos β
⎛α − β ⎞
⎛α + β ⎞
2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛α + β ⎞ ⎛α − β ⎞
−2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛α + β ⎞
⎛α − β ⎞
cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠⋅
⎝ 2 ⎠
=−
⎛α + β ⎞
⎛α − β ⎞
sin ⎜
⎟ sin ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠
⎛α + β ⎞
cos ⎜
⎟
⎝ 2 ⎠
=−
⎛α + β ⎞
sin ⎜
⎟
⎝ 2 ⎠
⎛α + β
= − cot ⎜
⎝ 2
⎛α + β ⎞
= − cot ⎜
⎟
⎝ 2 ⎠
⎞ ⎛α − β ⎞
⎟ cot ⎜
⎟
⎠ ⎝ 2 ⎠
35. 1 + cos(2θ ) + cos(4θ ) + cos(6θ ) = cos 0 + cos(6θ ) + cos(2θ ) + cos(4θ )
⎛ 0 + 6θ ⎞
⎛ 0 − 6θ ⎞
⎛ 2θ + 4θ ⎞
⎛ 2θ − 4θ ⎞
= 2 cos ⎜
⎟ cos ⎜
⎟ + 2 cos ⎜
⎟ cos ⎜
⎟
2
2
2
⎝
⎠
⎝
⎠
⎝
⎠
⎝ 2 ⎠
= 2 cos(3θ ) cos(−3θ ) + 2 cos(3θ ) cos(−θ )
= 2 cos 2 (3θ ) + 2 cos(3θ ) cos θ
= 2 cos(3θ ) [ cos(3θ ) + cos θ ]
⎡
⎛ 3θ + θ ⎞
⎛ 3θ − θ
= 2 cos(3θ ) ⎢ 2 cos ⎜
⎟ cos ⎜
⎝ 2 ⎠
⎝ 2
⎣
= 2 cos(3θ ) [ 2 cos(2θ ) cos θ ]
⎞⎤
⎟⎥
⎠⎦
= 4 cos θ cos(2θ ) cos(3θ )
36. 1 − cos(2θ ) + cos(4θ ) − cos(6θ ) = [ cos 0 − cos(6θ ) ] + [ cos(4θ ) − cos(2θ ) ]
⎛ 0 + 6θ ⎞ ⎛ 0 − 6θ ⎞
⎛ 2θ + 4θ ⎞ ⎛ 2θ − 4θ ⎞
= −2sin ⎜
⎟ sin ⎜
⎟ − 2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎝ 2 ⎠ ⎝ 2 ⎠
= −2sin(3θ ) sin(−3θ ) − 2sin(3θ ) sin(θ )
= 2sin 2 (3θ ) − 2sin(3θ ) sin θ
= 2sin(3θ ) [sin(3θ ) − sin θ ]
⎡
⎛ 3θ − θ ⎞
⎛ 3θ + θ ⎞ ⎤
= 2sin(3θ ) ⎢ 2sin ⎜
⎟ cos ⎜
⎟⎥
2
⎝
⎠
⎝ 2 ⎠⎦
⎣
= 2sin(3θ ) [ 2sin θ cos(2θ ) ]
= 4sin θ cos(2θ ) sin(3θ )
284
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.6: Product-to-Sum and Sum-to-Product Formulas
37. a.
y = sin [ 2π (852)t ] + sin [ 2π (1209)t ]
2π (852)t + 2π (1209)t ⎞
⎛ 2π (852)t − 2π (1209)t ⎞
= 2sin ⎛⎜
⎟ cos ⎜
⎟
2
2
⎝
⎠
⎝
⎠
= 2sin(2061π t ) cos(−357π t )
= 2sin(2061π t ) cos(357π t )
b. Because sin θ ≤ 1 and cos θ ≤ 1 for all θ , it follows that sin(2061π t ) ≤ 1 and cos(357π t ) ≤ 1 for all
values of t. Thus, y = 2sin(2061π t ) cos(357π t ) ≤ 2 ⋅1 ⋅1 = 2 . That is, the maximum value of y is 2.
c.
Let Y1 = 2sin(2061π x) cos(357π x) .
38. a.
y = sin [ 2π (941)t ] + sin [ 2π (1477)t ]
2π (941)t + 2π (1477)t ⎞
⎛ 2π (941)t − 2π (1477)t ⎞
= 2sin ⎛⎜
⎟ cos ⎜
⎟
2
2
⎝
⎠
⎝
⎠
= 2sin(2418π t ) cos(−536π t )
= 2sin(2418π t ) cos(536π t )
b.
Because sin θ ≤ 1 and cos θ ≤ 1 for all θ , it follows that sin(2418π t ) ≤ 1 and cos(2418π t ) ≤ 1 for all
values of t. Thus, y = 2sin(2418π t ) cos(536π t ) ≤ 2 ⋅1 ⋅1 = 2 . That is, the maximum value of y is 2.
c.
Let Y1 = 2sin(2418π x) cos(536π x) .
2
2
39. I u = I x cos θ + I y sin θ − 2 I xy sin θ cos θ
⎛ cos 2θ + 1 ⎞
⎛ 1 − cos 2θ ⎞
= Ix ⎜
⎟ + Iy ⎜
⎟ − I xy 2sin θ cos θ
2
2
⎝
⎠
⎝
⎠
I
I
cos
2
θ
I cos 2θ I x
y
y
= x
+ + −
− I xy sin 2θ
2
2 2
2
Ix + I y Ix − I y
=
+
cos 2θ − I xy sin 2θ
2
2
285
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
I v = I x sin 2 θ + I y cos 2 θ + 2 I xy sin θ cos θ
⎛ 1 − cos 2θ ⎞
⎛ cos 2θ + 1 ⎞
= Ix ⎜
⎟ + Iy ⎜
⎟ + I xy 2sin θ cos θ
2
2
⎝
⎠
⎝
⎠
I
cos
2
I
θ
I
I cos 2θ
y
y
= x− x
+
+ + I xy sin 2θ
2
2
2
2
Ix + I y Ix − I y
cos 2θ + I xy sin 2θ
=
−
2
2
40. a.
Since φ and v0 are fixed, we need to maximize sin θ cos (θ − φ ) .
(
)
(
)
1⎡
sin θ + (θ − φ ) + sin θ − (θ − φ ) ⎤⎥
⎦
2 ⎣⎢
1
= ⎡⎣sin ( 2θ − φ ) + sin φ ⎤⎦
2
This quantity will be maximized when sin ( 2θ − φ ) = 1 . So,
sin θ cos (θ − φ ) =
Rmax
b.
1
2v02 ⋅ ⋅ (1 + sin φ ) v 2 (1 + sin φ )
v02
0
2
=
=
=
2
2
g (1 − sin φ )
g cos φ
g cos φ
Rmax =
( 50 )2
9.8 (1 − sin 35° )
≈ 598.24
The maximum range is about 598 meters.
41. sin ( 2α ) + sin ( 2 β ) + sin ( 2γ )
⎛ 2α + 2 β ⎞
⎛ 2α − 2β ⎞
= 2sin ⎜
⎟ cos ⎜
⎟ + sin ( 2γ )
2
2
⎝
⎠
⎝
⎠
= 2sin(α + β ) cos(α − β ) + 2sin γ cos γ
= 2sin(π − γ ) cos(α − β ) + 2sin γ cos γ
= 2sin γ cos(α − β ) + 2sin γ cos γ
= 2sin γ [ cos(α − β ) + cos γ ]
⎡
⎛α − β +γ ⎞
⎛ α − β − γ ⎞⎤
= 2sin γ ⎢ 2 cos ⎜
⎟ cos ⎜
⎟⎥
2
2
⎝
⎠
⎝
⎠⎦
⎣
π
β
2
α
−
π
2
−
⎛
⎞
⎛
⎞
= 4sin γ cos ⎜
⎟
⎟ cos ⎜
⎝ 2 ⎠
⎝ 2 ⎠
π⎞
⎛π
⎞
⎛
= 4sin γ cos ⎜ − β ⎟ cos ⎜ α − ⎟
2
2⎠
⎝
⎠
⎝
= 4sin γ sin β sin α
= 4sin α sin β sin γ
286
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 3.6: Product-to-Sum and Sum-to-Product Formulas
sin α sin β sin γ
+
+
cos α cos β cos γ
sin α cos β cos γ + sin β cos α cos γ + sin γ cos α cos β
=
cos α cos β cos γ
cos γ (sin α cos β + cos α sin β ) + sin γ cos α cos β
=
cos α cos β cos γ
cos γ sin(α + β ) + sin γ cos α cos β
=
cos α cos β cos γ
cos γ sin(π − γ ) + sin γ cos α cos β
=
cos α cos β cos γ
cos γ sin γ + sin γ cos α cos β
=
cos α cos β cos γ
sin γ (cos γ + cos α cos β )
=
cos α cos β cos γ
42. tan α + tan β + tan γ =
=
=
=
sin γ ⎡⎣cos ( π − (α + β ) ) + cos α cos β ⎤⎦
cos α cos β cos γ
sin γ [ − cos(α + β ) + cos α cos β ]
cos α cos β cos γ
sin γ ( − cos α cos β + sin α sin β + cos α cos β )
cos α cos β cos γ
sin γ (sin α sin β )
=
cos α cos β cos γ
= tan α tan β tan γ
43. Add the sum formulas for sin(α + β ) and sin(α − β ) and solve for sin α cosβ :
sin(α + β ) = sin α cos β + cos α sin β
sin(α − β ) = sin α cos β − cos α sin β
sin(α + β ) + sin(α − β ) = 2sin α cos β
sin α cos β =
1
[sin(α + β ) + sin(α − β )]
2
⎛α −β ⎞
⎛α + β ⎞
44. 2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
α −β α +β ⎞
1⎡
⎛α −β
= 2 ⋅ ⎢sin ⎛⎜
+
⎟ + sin ⎜
2⎣ ⎝ 2
2 ⎠
⎝ 2
⎛ 2α ⎞
⎛ −2 β ⎞
= sin ⎜
⎟ + sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
−
α + β ⎞⎤
2
⎟⎥
⎠⎦
= sin α + sin ( − β )
= sin α − sin β
⎛α − β
Thus, sin α − sin β = 2sin ⎜
⎝ 2
⎞
⎛α + β
⎟ cos ⎜
⎠
⎝ 2
⎞
⎟.
⎠
287
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
⎛α + β ⎞
⎛α − β ⎞
45. 2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
1⎡
α +β α −β ⎞
⎛α + β
−
= 2 ⋅ ⎢ cos ⎛⎜
⎟ + cos ⎜
2⎣ ⎝ 2
2 ⎠
⎝ 2
⎛ 2β ⎞
⎛ 2α ⎞
= cos ⎜
⎟ + cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
= cos β + cos α
+
7. 2sin θ + 3 = 2
2sin θ = −1
1
sin θ = −
2
7π
11π
θ=
+ 2k π or θ =
+ 2k π , k is any integer
6
6
⎧ 7 π 11π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ,
⎬.
⎩6 6 ⎭
α − β ⎞⎤
⎟⎥
⎠⎦
2
⎛α + β ⎞
⎛α − β ⎞
Thus, cos α + cos β = 2 cos ⎜
⎟ cos ⎜
⎟.
⎝ 2 ⎠
⎝ 2 ⎠
⎛α + β ⎞ ⎛α − β ⎞
46. −2sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
1⎡
α +β α −β ⎞
⎛α + β
= −2 ⋅ ⎢ cos ⎛⎜
−
⎟ − cos ⎜
2⎣ ⎝ 2
2 ⎠
⎝ 2
1
2
1
1 − cos θ =
2
1
= cos θ
2
π
5π
θ = + 2k π or θ = + 2k π , k is any integer
3
3
⎧ π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩3 3 ⎭
8. 1 − cos θ =
+
α − β ⎞⎤
2
⎟⎥
⎠⎦
⎡ ⎛ 2β ⎞
⎛ 2α ⎞ ⎤
= − ⎢ cos ⎜
⎟ − cos ⎜
⎟⎥
2
⎝
⎠
⎝ 2 ⎠⎦
⎣
= cos α − cos β
⎛α + β
Thus, cos α − cos β = − 2sin ⎜
⎝ 2
⎞ ⎛α − β
⎟ sin ⎜
⎠ ⎝ 2
⎞
⎟.
⎠
9. 4 cos 2 θ = 1
1
cos 2 θ =
4
Section 3.7
1. 3x − 5 = − x + 1
4x = 6
6 3
x= =
4 2
cos θ = ±
π
2π
+ k π or θ =
+ k π , k is any integer
3
3
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 2π 4π 5π ⎫
, , ⎬.
⎨ ,
⎩3 3 3 3 ⎭
θ=
⎧3⎫
The solution set is ⎨ ⎬ .
⎩2⎭
2.
3.
2
1
, −
2
2
π
6
,
1
2
10. tan 2 θ =
1
3
1
3
=±
3
3
π
5π
+ k π, k is any integer
θ = + k π or θ =
6
6
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 5π 7π 11π ⎫
, ,
⎨ ,
⎬.
⎩6 6 6 6 ⎭
5π
6
tan θ = ±
⎧
π
5π
⎫
4. ⎨θ θ = + 2π k , θ =
+ 2π k , k is any integer ⎬
6
6
⎩
⎭
5. False
6. False
288
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Section 3.7: Trigonometric Equations (I)
11. 2sin 2 θ − 1 = 0
15. cos ( 2θ ) = −
2sin θ = 1
2
sin 2 θ =
2π
4π
+ 2kπ or 2θ =
+ 2kπ
3
3
π
2π
θ = + kπ or θ =
+ kπ , k is any integer
3
3
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 2π 4π 5π ⎫
,
, ⎬.
⎨ ,
⎩3 3 3 3 ⎭
2θ =
1
2
1
2
=±
2
2
π
3π
+ kπ , k is any integer
θ = + kπ or θ =
4
4
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 3π 5π 7 π ⎫
⎨ , , , ⎬.
⎩4 4 4 4 ⎭
sin θ = ±
16. tan ( 2θ ) = −1
3π
+ k π, k is any integer
4
3π k π
θ = + , k is any integer
8
2
On the interval 0 ≤ θ < 2π , the solution set is
⎧ 3π 7π 11π 15π ⎫
,
,
⎨ ,
⎬.
8 ⎭
⎩8 8 8
2θ =
12. 4 cos 2 θ − 3 = 0
4 cos 2 θ = 3
cos 2 θ =
3
4
cos θ = ±
3
2
3θ
= −2
2
3θ 2π
3θ 4π
=
+ 2kπ or
=
+ 2kπ
2
3
2
3
4π 4kπ
8π 4kπ
θ=
+
+
or θ =
,
9
3
9
3
k is any integer
On the interval 0 ≤ θ < 2π , the solution set is
⎧ 4π 8π 16π ⎫
⎨ , ,
⎬.
⎩9 9 9 ⎭
π
5π
+ k π or θ =
+ k π , k is any integer
6
6
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 5π 7π 11π ⎫
, ,
⎨ ,
⎬.
⎩6 6 6 6 ⎭
17. sec
θ=
13. sin ( 3θ ) = −1
3π
+ 2k π
2
π 2k π
θ= +
, k is any integer
2
3
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 7 π 11π ⎫
,
⎨ ,
⎬.
⎩2 6 6 ⎭
3θ =
2θ
=− 3
3
2θ 5π
=
+ kπ , k is any integer
3
6
5π 3kπ
θ=
, k is any integer
+
4
2
18. cot
⎛θ ⎞
14. tan ⎜ ⎟ = 3
⎝2⎠
θ
⎧ 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ⎬ .
⎩4⎭
π
+ kπ , k is any integer
3
2π
θ=
+ 2π k , k is any integer
3
⎧ 2π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ⎬ .
⎩3 ⎭
2
=
1
2
19. 2sin θ + 1 = 0
2sin θ = −1
1
sin θ = −
2
7π
11π
θ=
+ 2k π or θ =
+ 2k π , k is any integer
6
6
⎧ 7 π 11π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ,
⎬.
⎩6 6 ⎭
289
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Chapter 3: Analytic Trigonometry
20. cos θ + 1 = 0
cos θ = −1
θ = π + 2k π , k is any integer
26. 4sin θ + 3 3 = 3
4sin θ = − 2 3
2 3
3
=−
4
2
4π
5π
θ=
+ 2k π or θ =
+ 2k π , k is any integer
3
3
⎧ 4 π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩3 3⎭
On the interval 0 ≤ θ < 2π , the solution set is {π } .
sin θ = −
21. tan θ + 1 = 0
tan θ = −1
3π
θ = + k π , k is any integer
4
⎧ 3π 7π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩4 4 ⎭
22.
π⎞
⎛
27. cos ⎜ 2θ − ⎟ = −1
2⎠
⎝
π
2θ − = π + 2k π
2
3π
+ 2k π
2θ =
2
3π
θ = + k π , k is any integer
4
⎧ 3π 7π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩4 4⎭
3 cot θ + 1 = 0
3 cot θ = −1
cot θ = −
θ=
1
3
=−
3
3
2π
+ k π , k is any integer
3
⎧ 2π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩3 3⎭
π⎞
⎛
28. sin ⎜ 3θ + ⎟ = 1
18 ⎠
⎝
π π
3θ + = + 2k π
18 2
4π
+ 2k π
3θ =
9
4π 2k π
+
θ=
, k is any integer
27
3
On the interval 0 ≤ θ < 2π , the solution set is
⎧ 4π 22π 40π ⎫
,
⎨ ,
⎬.
⎩ 27 27 27 ⎭
23. 4sec θ + 6 = − 2
4sec θ = − 8
sec θ = − 2
2π
4π
+ 2k π or θ =
+ 2k π , k is any integer
3
3
⎧ 2π 4 π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩3 3⎭
θ=
24. 5csc θ − 3 = 2
5csc θ = 5
csc θ = 1
π
θ = + 2k π , k is any integer
2
⎛θ π ⎞
29. tan ⎜ + ⎟ = 1
⎝2 3⎠
θ π π
+ = + kπ
2 3 4
θ
π
= − + kπ
2
12
π
θ = − + 2k π , k is any integer
6
⎧11π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨
⎬.
⎩ 6 ⎭
⎧π⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ⎬ .
⎩2⎭
25. 3 2 cos θ + 2 = −1
3 2 cos θ = − 3
cos θ = −
1
2
=−
2
2
3π
5π
+ 2k π or θ =
+ 2k π , k is any integer
4
4
⎧ 3π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ⎬ .
⎩4 4⎭
θ=
290
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Section 3.7: Trigonometric Equations (I)
35. cos θ = 0
⎧
π
3π
⎫
⎨θ θ = + 2k π or θ = + 2k π ⎬ , k is any
2
2
⎩
⎭
integer
π 3π 5π 7 π 9π 11π
Six solutions are θ = , , ,
.
, ,
2 2 2 2 2 2
⎛θ π ⎞ 1
30. cos ⎜ − ⎟ =
⎝3 4⎠ 2
θ π π
θ π 5π
− = + 2k π or
− =
+ 2k π
3 4 3
3 4 3
θ 7π
θ 23π
=
+ 2k π
or
=
+ 2k π
3 12
3 12
7π
23π
+ 6k π
or
+ 6k π ,
θ=
θ=
4
4
k is any integer.
⎧ 7π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ⎬ .
⎩4⎭
31. sin θ =
2
2
⎧
π
3π
⎫
+ 2kπ ⎬ , k is any
⎨θ θ = + 2k π or θ =
4
4
⎩
⎭
integer
π 3π 9π 11π 17π 19π
Six solutions are θ = , , ,
,
,
.
4 4 4 4
4
4
36. sin θ =
1
2
⎧
π
5π
⎫
+ 2kπ ⎬ , k is any
⎨θ θ = + 2kπ or θ =
6
6
⎩
⎭
integer. Six solutions are
π 5π 13π 17π 25π 29π
θ= , ,
,
,
,
.
6 6 6
6
6
6
37. cos ( 2θ ) = −
2π
4π
+ 2k π or 2θ =
+ 2k π, k is any integer
3
3
⎧
π
2π
⎫
+ k π ⎬ , k is any integer
⎨θ θ = + k π or θ =
3
3
⎩
⎭
π 2π 4π 5π 7π 8π
.
,
,
, ,
Six solutions are θ = ,
3 3 3 3 3 3
2θ =
32. tan θ = 1
⎧
π
⎫
⎨θ θ = + k π ⎬ , k is any integer
4
⎩
⎭
π 5π 9π 13π 17 π 21π
.
,
,
Six solutions are θ = , , ,
4 4 4 4
4
4
33. tan θ = −
1
2
38. sin ( 2θ ) = −1
3π
+ 2k π, k is any integer
2
⎧
3π
⎫
+ k π ⎬ , k is any integer
⎨θ θ =
4
⎩
⎭
Six solutions are
3π 7π 11π 15π 19π 23π
θ= , ,
.
,
,
,
4 4 4
4
4
4
2θ =
3
3
⎧
5π
⎫
+ k π ⎬ , k is any integer
⎨θ θ =
6
⎩
⎭
Six solutions are
5π 11π 17 π 23π 29π 35π
θ= ,
.
,
,
,
,
6 6
6
6
6
6
39. sin
θ
=−
3
2
2
4π
θ 5π
=
+ 2k π or =
+ 2k π, k is any integer
2 3
2 3
⎧
8π
10π
⎫
+ 4k π or θ =
+ 4k π ⎬ , k is any
⎨θ θ =
3
3
⎩
⎭
integer. Six solutions are
8π 10π 20π 22π 32π 34π
θ= ,
,
,
,
,
.
3 3
3
3
3
3
θ
3
34. cos θ = −
2
⎧
5π
7π
⎫
+ 2k π or θ =
+ 2k π ⎬ , k is any
⎨θ θ =
6
6
⎩
⎭
integer. Six solutions are
5π 7 π 17 π 19π 29π 31π
θ= , ,
.
,
,
,
6 6 6
6
6
6
291
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Chapter 3: Analytic Trigonometry
40. tan
θ
47. sec θ = −4
1
cos θ = −
4
= −1
2
3π
=
+ kπ , k is any integer
2
4
⎧
3π
⎫
+ 2kπ ⎬ , k is any integer
⎨θ θ =
2
⎩
⎭
Six solutions are
3π 7π 11π 15π 19π 23π
θ= , ,
.
,
,
,
2 2 2
2
2
2
θ
⎛ 1⎞
⎝
⎠
θ ≈ 1.82 or θ ≈ 2π − 1.82 ≈ 4.46 .
The solution set is {1.82, 4.46} .
θ = cos −1 ⎜ − ⎟ ≈ 1.82
4
48. csc θ = −3
1
sin θ = −
3
41. sin θ = 0.4
θ = sin −1 ( 0.4 ) ≈ 0.41
θ ≈ 0.41 or θ ≈ π − 0.41 ≈ 2.73 .
⎛ 1⎞
⎝
⎠
θ ≈ −0.34 + 2π or θ ≈ π − ( −0.34 ) .
≈ 5.94
≈ 3.48
The solution set is {3.48, 5.94} .
θ = sin −1 ⎜ − ⎟ ≈ −0.34
3
The solution set is {0.41, 2.73} .
42. cos θ = 0.6
θ = cos−1 ( 0.6 ) ≈ 0.93
θ ≈ 0.93 or θ ≈ 2π − 0.93 ≈ 5.36 .
49. 5 tan θ + 9 = 0
5 tan θ = −9
9
tan θ = −
5
The solution set is {0.93, 5.36} .
43. tan θ = 5
θ = tan −1 ( 5 ) ≈ 1.37
θ ≈ 1.37 or θ ≈ π + 1.37 ≈ 4.51 .
⎛ 9⎞
⎝
⎠
θ ≈ −1.064 + π or θ ≈ −1.064 + 2π
≈ 2.08
≈ 5.22
The solution set is {2.08, 5.22} .
θ = tan −1 ⎜ − ⎟ ≈ −1.064
5
The solution set is {1.37, 4.51} .
44. cot θ = 2
1
tan θ =
2
50. 4 cot θ = −5
5
cot θ = −
4
4
tan θ = −
5
⎛1⎞
⎝ ⎠
θ ≈ 0.46 or θ ≈ π + 0.46 ≈ 3.61 .
The solution set is {0.46, 3.61} .
θ = tan −1 ⎜ ⎟ ≈ 0.46
2
45. cos θ = −0.9
⎛ 4⎞
⎝
⎠
θ ≈ −0.675 + π or θ ≈ −0.675 + 2π .
≈ 2.47
≈ 5.61
The solution set is {2.47, 5.61} .
θ = tan −1 ⎜ − ⎟ ≈ −0.675
5
θ = cos ( −0.9 ) ≈ 2.69
θ ≈ 2.69 or θ ≈ 2π − 2.69 ≈ 3.59 .
−1
The solution set is {2.69, 3.59} .
46. sin θ = −0.2
θ = sin −1 ( −0.2 ) ≈ −0.20
θ ≈ −0.20 + 2π or θ ≈ π − ( −0.20 ) .
≈ 6.08
≈ 3.34
The solution set is {3.34, 6.08} .
292
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Section 3.7: Trigonometric Equations (I)
51. 3sin θ − 2 = 0
3sin θ = 2
2
sin θ =
3
55. a.
f ( x) = 0
3sin x = 0
sin x = 0
x = 0 + 2kπ or x = π + 2kπ , k is any integer
On the interval [ −2π , 4π] , the x-intercepts
⎛2⎞
⎝ ⎠
θ ≈ 0.73 or θ ≈ π − 0.73 ≈ 2.41 .
The solution set is {0.73, 2.41} .
θ = sin −1 ⎜ ⎟ ≈ 0.73
3
of the graph of f are
−2π, −π, 0, π, 2π, 3π, 4π .
b.
52. 4 cos θ + 3 = 0
4 cos θ = −3
f ( x ) = 3sin x
3
4
cos θ = −
⎛ 3⎞
⎝
⎠
θ ≈ 2.42 or θ ≈ 2π − 2.42 ≈ 3.86 .
The solution set is {2.42, 3.86} .
θ = cos −1 ⎜ − ⎟ ≈ 2.42
4
f ( x) = 0
53.
c.
4sin 2 x − 3 = 0
4sin 2 x = 3
sin 2 x =
3
4
5π
+ 2kπ , k is any integer
6
6
On the interval [ −2π , 4π] , the solution set is
x=
3
3
=±
sin x = ±
4
2
π
2π
x = + kπ or x =
+ kπ , k is any integer
3
3
On the interval [ 0, 2π] , the x-intercepts of the
graph of f are
+ 2kπ or x =
d. From the graph in part (b) and the results of
3
part (c), the solutions of f ( x ) > on the
2
⎧ 11π
7π
interval [ −2π , 4π] is ⎨ x −
<x<−
6
6
⎩
π 2π 4π 5π
, ,
,
.
3 3 3 3
2 cos ( 3 x ) + 1 = 0
2 cos ( 3 x ) = −1
cos ( 3 x ) = −
π
⎧ 11π 7π π 5π 13π 17 π ⎫
,− , ,
,
,
⎨−
⎬.
6 6 6 6
6 ⎭
⎩ 6
f ( x) = 0
54.
3
2
3
3sin x =
2
1
sin x =
2
f ( x) =
or
1
2
π
5π
13π
17π ⎫
<x<
or
<x<
⎬.
6
6
6
6 ⎭
2π
4π
+ 2kπ or 3x =
+ 2kπ
3
3
2π 2kπ
4π 2kπ
+
+
x=
or x =
,
9
3
9
3
k is any integer
On the interval [ 0, π] , the x-intercepts of the
3x =
graph of f are
2π 4π 8π
.
,
,
9 9 9
293
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Chapter 3: Analytic Trigonometry
56. a.
f ( x) = 0
4 tan x < −4
tan x < −1
Graphing y1 = tan x and y2 = −1 on the
2 cos x = 0
cos x = 0
3π
+ 2kπ or x =
+ 2kπ , k is any
2
2
integer
On the interval [ −2π , 4π] , the x-intercepts
x=
π
⎛ π π⎞
interval ⎜ − , ⎟ , we see that y1 < y2 for
⎝ 2 2⎠
π
π
⎛ π π⎞
− < x < − or ⎜ − , − ⎟ .
2
4
⎝ 2 4⎠
of the graph of f are
3π π π 3π 5π 7π
− ,− , , ,
,
.
2
2 2 2 2 2
b.
f ( x ) < −4
b.
6
f ( x ) = 2 cos x
π
_
2
π
−_
2
−6
58.
f ( x ) = cot x
a.
f ( x) = − 3
cot x = − 3
c.
⎧
5π
⎫
+ kπ ⎬ , k is any integer
⎨x x =
6
⎩
⎭
f ( x) = − 3
2 cos x = − 3
cos x = −
b.
3
2
f ( x) > − 3
cot x > − 3
1
and y2 = − 3 on the
tan x
interval ( 0, π ) , we see that y1 > y2 for
Graphing y1 =
5π
7π
x=
+ 2kπ or x =
+ 2kπ , k is any
6
6
integer
On the interval [ −2π , 4π] , the solution set is
0< x<
⎧ 7π 5π 5π 7π 17π 19π ⎫
,
,
,
⎨− , − ,
⎬.
6 6 6
6
6 ⎭
⎩ 6
6
d. From the graph in part (b) and the results of
part (c), the solutions of f ( x ) < − 3 on the
57.
⎞
⎟.
⎠
π
0
⎧
7π
5π
interval [ −2π , 4π] is ⎨ x −
<x<−
6
6
⎩
5π
7π
17 π
19π ⎫
or
<x<
or
<x<
⎬.
6
6
6
6 ⎭
5π
⎛ 5π
or ⎜ 0,
6
⎝ 6
−6
59. a, d.
f ( x ) = 3sin ( 2 x ) + 2 ; g ( x ) =
7
2
f ( x ) = 4 tan x
a.
f ( x ) = −4
4 tan x = −4
tan x = −1
⎧
π
⎫
⎨ x x = − + kπ ⎬ , k is any integer
4
⎩
⎭
294
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Section 3.7: Trigonometric Equations (I)
f ( x) = g ( x)
b.
7
2
3
3sin ( 2 x ) =
2
1
sin ( 2 x ) =
2
c.
3sin ( 2 x ) + 2 =
2x =
π
6
[0, 4π] is
61. a, d.
⎧ 2π
10π ⎫
⎛ 2π 10π ⎞
<x<
⎨x
⎬ or ⎜ ,
⎟.
3
3 ⎭
⎝ 3 3 ⎠
⎩
f ( x ) = −4 cos x ; g ( x ) = 2 cos x + 3
5π
+ 2kπ
6
5π
+ kπ ,
x=
12
+ 2kπ or 2 x =
π
+ kπ or
12
k is any integer
x=
From the graph in part (a) and the results of
part (b), the solution of f ( x ) < g ( x ) on
⎧ π 5π ⎫
On [ 0, π] , the solution set is ⎨ , ⎬ .
⎩12 12 ⎭
c.
From the graph in part (a) and the results of
part (b), the solution of f ( x ) > g ( x ) on
[0, π]
60. a, d.
⎧ π
5π ⎫
⎛ π 5π ⎞
is ⎨ x
< x < ⎬ or ⎜ ,
⎟.
12 ⎭
⎝ 12 12 ⎠
⎩ 12
−4 cos x = 2 cos x + 3
−6 cos x = 3
3
1
cos x =
=−
2
−6
2π
4π
x=
+ 2kπ or x =
+ 2kπ ,
3
3
k is any integer
⎧ 2π 4π ⎫
On [ 0, 2π ] , the solution set is ⎨ ,
⎬.
⎩ 3 3 ⎭
x
f ( x ) = 2 cos + 3 ; g ( x ) = 4
2
c.
b.
f ( x) = g ( x)
b.
f ( x) = g ( x)
x
2 cos + 3 = 4
2
x
2 cos = 1
2
x 1
cos =
2 2
x π
x 5π
= + 2kπ or
=
+ 2kπ
2 3
2
3
2π
10π
x=
+ 4kπ or x =
+ 4kπ ,
3
3
k is any integer
⎧ 2π 10π ⎫
On [ 0, 4π ] , the solution set is ⎨ ,
⎬.
⎩3 3 ⎭
From the graph in part (a) and the results of
part (b), the solution of f ( x ) > g ( x ) on
[0, 2π ] is
62. a, d.
⎧ 2π
4π ⎫
⎛ 2π 4π ⎞
<x<
,
⎨x
⎬ or ⎜
⎟.
3
3
⎝ 3 3 ⎠
⎩
⎭
f ( x ) = 2sin x ; g ( x ) = −2sin x + 2
295
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Chapter 3: Analytic Trigonometry
b.
f ( x) = g ( x)
π⎞
⎛
b. Solve h ( t ) = 125sin ⎜ 0.157t − ⎟ + 125 = 250
2⎠
⎝
on the interval [ 0,80] .
2sin x = −2sin x + 2
4sin x = 2
2 1
sin x = =
4 2
x=
π
+ 2kπ or x =
6
k is any integer
π⎞
⎛
125sin ⎜ 0.157t − ⎟ + 125 = 250
2⎠
⎝
π⎞
⎛
125sin ⎜ 0.157t − ⎟ = 125
2⎠
⎝
π⎞
⎛
sin ⎜ 0.157t − ⎟ = 1
2⎠
⎝
5π
+ 2kπ ,
6
⎧ π 5π ⎫
On [ 0, 2π ] , the solution set is ⎨ , ⎬ .
⎩6 6 ⎭
c.
[0,
⎧ π
5π ⎫
⎛ π 5π ⎞
2π] is ⎨ x
< x < ⎬ or ⎜ , ⎟ .
6⎭
⎝6 6 ⎠
⎩ 6
π⎞
⎛
sin ⎜ 0.157t − ⎟ = 0
2⎠
⎝
2
c.
= kπ , k is any integer
0.157t = kπ +
t=
kπ +
For k = 0, t =
For k = 1, t =
π
2
, k is any integer
π
π
π
2 ≈ 30 seconds .
0.157
2π +
π⎞
⎛
Solve h ( t ) = 125sin ⎜ 0.157t − ⎟ + 125 > 125
2⎠
⎝
on the interval [ 0, 40] .
π⎞
⎛
Graphing y1 = sin ⎜ 0.157 x − ⎟ and y2 = 0
2⎠
⎝
on the interval [ 0, 40] , we see that y1 > y2 for
2 ≈ 10 seconds .
0.157
π+
π
π⎞
⎛
125sin ⎜ 0.157t − ⎟ + 125 > 125
2⎠
⎝
π⎞
⎛
125sin ⎜ 0.157t − ⎟ > 0
2⎠
⎝
π⎞
⎛
sin ⎜ 0.157t − ⎟ > 0
2⎠
⎝
2 , k is any integer
0.157
0+
π
≈ 20 seconds .
0.157
π + 2π
For k = 1, t =
≈ 60 seconds .
0.157
π + 4π
For k = 2, t =
≈ 100 seconds .
0.157
So during the first 80 seconds, an individual
on the Ferris Wheel is exactly 250 feet above
the ground when t ≈ 20 seconds and again
when t ≈ 60 seconds .
π⎞
⎛
125sin ⎜ 0.157t − ⎟ + 125 = 125
2⎠
⎝
π⎞
⎛
125sin ⎜ 0.157t − ⎟ = 0
2⎠
⎝
π
=
For k = 0, t =
π⎞
⎛
63. h ( t ) = 125sin ⎜ 0.157t − ⎟ + 125
2⎠
⎝
π⎞
⎛
a. Solve h ( t ) = 125sin ⎜ 0.157t − ⎟ + 125 = 125
2⎠
⎝
on the interval [ 0, 40] .
0.157t −
π
+ 2kπ , k is any integer
2 2
0.157t = π + 2kπ , k is any integer
π + 2kπ
t=
, k is any integer
0.157
0.157t −
From the graph in part (a) and the results of
part (b), the solution of f ( x ) > g ( x ) on
10 < x < 30 .
π
2 ≈ 50 seconds .
0.157
So during the first 40 seconds, an individual
on the Ferris Wheel is exactly 125 feet above
the ground when t ≈ 10 seconds and again
when t ≈ 30 seconds .
For k = 2, t =
296
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Section 3.7: Trigonometric Equations (I)
1.5
0
π⎞
⎛
b. Solve h ( t ) = 13sin ⎜ 4π t − ⎟ + 13 = 6.5 on
2⎠
⎝
the interval [ 0,1] .
40
π⎞
⎛
13sin ⎜ 4π t − ⎟ + 13 = 6.5
2⎠
⎝
π⎞
⎛
13sin ⎜ 4π t − ⎟ = −6.5
2⎠
⎝
1
π⎞
⎛
sin ⎜ 4π t − ⎟ = −
2⎠
2
⎝
π
π
π
5π
+ 2kπ
4π t − = − + 2kπ or 4π t − = −
2
6
2
6
π
π
4π t = 2kπ +
or 4π t = 2kπ −
3
3
π
π
2 kπ +
2 kπ −
3 or t =
3
t=
4π
4π
6k + 1
6k − 1
or
,
=
=
12
12
k is any integer
−1.5
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 < x < 30 , or (10, 30 ) .
π⎞
⎛
64. h ( t ) = 13sin ⎜ 4π t − ⎟ + 13
2⎠
⎝
π⎞
⎛
a. Solve h ( t ) = 13sin ⎜ 4π t − ⎟ + 13 = 13 on
2⎠
⎝
the interval [ 0,1] .
π⎞
⎛
13sin ⎜ 4π t − ⎟ + 13 = 13
2⎠
⎝
0 +1
0 −1
or t =
12
12
≈ 0.083 sec
≈ −0.083 sec
For k = 0, t =
π⎞
⎛
13sin ⎜ 4π t − ⎟ = 0
2⎠
⎝
π⎞
⎛
sin ⎜ 4π t − ⎟ = 0
2⎠
⎝
4π t −
π
2
6 +1
6 −1
or t =
12
12
≈ 0.583 sec
≈ 0.417 sec
For k = 1, t =
= kπ , k is any integer
4π t = kπ +
kπ +
π
2
12 + 1
12 − 1
or t =
12
12
≈ 1.083 sec
≈ 0.917 sec
For k = 2, t =
, k is any integer
π
18 − 1
12
≈ 1.417 sec
So during the first second, the point on the
tire is exactly 6.5 inches above the ground
when t ≈ 0.083 second , t ≈ 0.417 second ,
t ≈ 0.583 second , and t ≈ 0.917 second .
2 = 2k + 1 , k is any integer
t=
4π
8
0 +1
For k = 0, t =
= 0.125 second .
8
2 +1
For k = 1, t =
= 0.375 second .
8
4 +1
For k = 2, t =
= 0.625 second .
8
6 +1
For k = 3, t =
= 0.875 second .
8
8 +1
For k = 4, t =
= 1.125 second .
8
So during the first second, the point on the tire
is exactly 13 inches above the ground when
t = 0.125 second , t = 0.375 second ,
t = 0.625 second , and t = 0.875 second .
For k = 3,
t=
297
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Chapter 3: Analytic Trigonometry
c.
3.94 + 2π k
5.94 + 2π k
or x ≈
,
0.65
0.65
k is any integer
3.94 + 0
5.94 + 0
or x ≈
For k = 0 , x ≈
0.65
0.65
≈ 6.06 min
≈ 8.44 min
π⎞
⎛
Solve h ( t ) = 13sin ⎜ 4π t − ⎟ + 13 > 13 on
2⎠
⎝
the interval [ 0,1] .
x≈
π⎞
⎛
13sin ⎜ 4π t − ⎟ + 13 > 13
2⎠
⎝
π⎞
⎛
13sin ⎜ 4π t − ⎟ > 0
2⎠
⎝
π⎞
⎛
sin ⎜ 4π t − ⎟ > 0
2⎠
⎝
3.94 + 2π
5.94 + 2π
or x ≈
0.65
0.65
≈ 15.72 min
≈ 18.11 min
For k = 1 , x ≈
For k = 2 ,
3.94 + 4π
5.94 + 4π
x≈
or x ≈
0.65
0.65
≈ 25.39 min
≈ 27.78 min
π⎞
⎛
Graphing y1 = sin ⎜ 4π x − ⎟ and y2 = 0
2⎠
⎝
on the interval [ 0,1] , we see that y1 > y2 for
0.125 < x < 0.375 and again for
0.625 < x < 0.875 .
So during the first 20 minutes in the holding
pattern, the plane is exactly 100 miles from
the airport when x ≈ 6.06 minutes ,
x ≈ 8.44 minutes , x ≈ 15.72 minutes , and
x ≈ 18.11 minutes .
1.5
0
1
c.
So during the first second, the point on the
tire is more than 13 inches above the ground
when 0.125 < t < 0.375 second and for
0.625 < t < 0.875 second.
70sin ( 0.65 x ) + 150 > 100
70sin ( 0.65 x ) > −50
sin ( 0.65 x ) > −
65. d ( x ) = 70sin ( 0.65 x ) + 150
a.
Solve d ( x ) = 70sin ( 0.65 x ) + 150 > 100 on
the interval [ 0, 20] .
−1.5
5
on
7
the interval [ 0, 20] , we see that y1 > y2 for
d ( 0 ) = 70sin ( 0.65 ( 0 ) ) + 150
Graphing y1 = sin ( 0.65 x ) and y2 = −
= 70sin ( 0 ) + 150
= 150 miles
0 < x < 6.06 , 8.44 < x < 15.72 , and
18.11 < x < 20 .
b. Solve d ( x ) = 70sin ( 0.65 x ) + 150 = 100 on
1.5
the interval [ 0, 20] .
70sin ( 0.65 x ) + 150 = 100
0
70sin ( 0.65 x ) = −50
sin ( 0.65 x ) = −
5
7
5
7
20
−1.5
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.
⎛ 5⎞
0.65 x = sin −1 ⎜ − ⎟ + 2π k
⎝ 7⎠
⎛ 5⎞
sin −1 ⎜ − ⎟ + 2π k
⎝ 7⎠
x=
0.65
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.
298
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Section 3.7: Trigonometric Equations (I)
66. R (θ ) = 672sin ( 2θ )
a.
0.9330 + 0
2.2083 + 0
or θ =
2
2
≈ 1.10415
≈ 0.46665
For k = 0 , θ =
Solve R (θ ) = 672sin ( 2θ ) = 450 on the
⎡ π⎞
interval ⎢ 0, ⎟ .
⎣ 2⎠
672sin ( 2θ ) = 450
For k = 1 , θ =
450 225
=
672 336
⎛ 225 ⎞
2θ = sin −1 ⎜
⎟ + 2kπ
⎝ 336 ⎠
⎛ 225 ⎞
sin −1 ⎜
⎟ + 2kπ
336 ⎠
⎝
θ=
2
0.7337 + 2kπ
2.408 + 2kπ
θ≈
or θ ≈
,
2
2
k is any integer
sin ( 2θ ) =
0.7337 + 2π
2
c.
≈ 200.99°
≈ 248.98°
2
≈ 4.246
Solve R (θ ) = 672sin ( 2θ ) ≥ 480 on the
480
672
5
sin ( 2θ ) ≥
7
Graphing y1 = sin ( 2 x ) and y2 =
5
on the
7
⎡ π⎞
interval ⎢ 0, ⎟ and using INTERSECT, we
⎣ 2⎠
see that y1 ≥ y2 when 0.3978 ≤ x ≤ 1.1730
radians, or 22.79° ≤ x ≤ 67.21° .
2
≈ 4.3456
2.2083 + 2π
sin ( 2θ ) ≥
2.408 + 2π
≈ 3.508
or θ =
⎡ π⎞
interval ⎢ 0, ⎟ .
⎣ 2⎠
672sin ( 2θ ) ≥ 480
≈ 68.98°
or θ =
2
≈ 243.28°
≈ 206.72°
So the golfer should hit the ball at an angle
of either 26.74° or 63.26° .
0.7337 + 0
2.408 + 0
or θ =
2
2
≈ 1.204
≈ 0.36685
≈ 21.02°
0.9330 + 2π
≈ 3.608
For k = 0 , θ =
For k = 1 , θ =
≈ 63.26°
≈ 26.74°
So the golfer should hit the ball at an angle
of either 21.02° or 68.98° .
1.5
b. Solve R (θ ) = 672sin ( 2θ ) = 540 on the
0
⎡ π⎞
interval ⎢ 0, ⎟ .
⎣ 2⎠
672sin ( 2θ ) = 540
π
_
2
−1.5
1.5
540 135
=
672 168
⎛ 135 ⎞
2θ = sin −1 ⎜
⎟ + 2kπ
⎝ 168 ⎠
⎛ 135 ⎞
sin −1 ⎜
⎟ + 2kπ
168 ⎠
⎝
θ=
2
0.9333 + 2kπ
2.2083 + 2kπ
θ≈
or θ ≈
,
2
2
k is any integer
sin ( 2θ ) =
0
π
_
2
−1.5
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.
299
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Chapter 3: Analytic Trigonometry
67.
72. The index of refraction of crown glass is 1.52.
sin 30º
≈ 1.52
sin θ 2
sin 40°
= 1.33
sin θ 2
1.33sin θ 2 = sin 40°
1.52sin θ 2 = sin 30°
sin 40°
≈ 0.4833
1.33
θ 2 = sin −1 ( 0.4833) ≈ 28.90°
sin θ 2 =
68.
sin 30°
≈ 0.3289
1.52
θ 2 ≈ sin −1 ( 0.3352 ) ≈ 19.20°
sin θ 2 =
sin 50°
= 1.66
sin θ 2
The angle of refraction is about 19.20° .
73. If θ is the original angle of incidence and φ is
1.66sin θ 2 = sin 50°
sin θ
= n2 . The
sin φ
angle of incidence of the emerging beam is also
1
φ , and the index of refraction is
. Thus, θ is
n2
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.
sin 50°
≈ 0.4615
1.66
θ 2 = sin −1 ( 0.4615 ) ≈ 27.48°
the angle of refraction, then
sin θ 2 =
69. Calculate the index of refraction for each:
v1 sin θ1
θ1
θ2
=
v2 sin θ 2
sin10º
≈ 1.2477
10º
8º
sin 8º
sin 20º
≈ 1.2798
20º 15º 30 ' = 15.5º
sin15.5º
sin 30º
≈ 1.3066
30º 22º 30 ' = 22.5º
sin 22.5º
sin 40º
40º
29º 0 ' = 29º
≈ 1.3259
sin 29º
sin 50º
50º
35º 0 ' = 35º
≈ 1.3356
sin 35º
sin 60º
60º 40º 30 ' = 40.5º
≈ 1.3335
sin 40.5º
sin 70º
≈ 1.3175
70º 45º 30 ' = 45.5º
sin 45.5º
sin 80º
80º
50º 0 ' = 50º
≈ 1.2856
sin 50º
74. Here we have n1 = 1.33 and n2 = 1.52 .
n1 sin θ B = n2 cos B
sin θ B n2
=
cos θ B n1
Yes, these data values agree with Snell’s Law.
The results vary from about 1.25 to 1.34.
70.
tan θ B =
n2
n1
θ B = tan −1
v1 2.998 × 108
=
≈ 1.56
v2
1.92 × 108
The index of refraction for this liquid is about
1.56.
n2
⎛ 1.52 ⎞
= tan −1 ⎜
⎟ ≈ 48.8°
n1
⎝ 1.33 ⎠
75. Answers will vary.
71. Calculate the index of refraction:
sin θ1 sin 40º
θ1 = 40º , θ 2 = 26º ;
=
≈ 1.47
sin θ 2 sin 26º
300
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Section 3.8: Trigonometric Equations (II)
Section 3.8
5.
4 x2 − x − 5 = 0
1.
cos θ = 0
( 4 x − 5)( x + 1) = 0
( −1) − 4 (1)( −1)
2 (1)
2
cos θ = −
sin 2 θ − 1 = 0
6.
(sin θ + 1)(sin θ − 1) = 0
sin θ + 1 = 0
1± 1+ 4
2
1± 5
=
2
or
sin θ = −1
θ=
sin θ − 1 = 0
sin θ = 1
3π
2
θ=
π
2
⎧ π 3π ⎫
The solution set is ⎨ ,
⎬.
⎩2 2 ⎭
⎪⎧1 − 5 1 + 5 ⎪⎫
,
⎬.
2 ⎭⎪
⎩⎪ 2
The solution set is ⎨
7.
(2 x − 1) 2 − 3(2 x − 1) − 4 = 0
[(2 x − 1) + 1][(2 x − 1) − 4] = 0
2sin 2 θ − sin θ − 1 = 0
(2sin θ + 1)(sin θ − 1) = 0
2sin θ + 1 = 0
2 x(2 x − 5) = 0
2 x = 0 or 2 x − 5 = 0
5
x = 0 or
x=
2
⎧ 5⎫
The solution set is ⎨0, ⎬ .
⎩ 2⎭
or
2sin θ = −1
sin θ − 1 = 0
sin θ = 1
π
1
θ=
2
2
7 π 11π
θ= ,
6 6
⎧ π 7π 11π ⎫
,
The solution set is ⎨ ,
⎬.
6 ⎭
⎩2 6
sin θ = −
4. 5 x3 − 2 = x − x 2
Let y1 = 5 x3 − 2 and y2 = x − x 2 . Use
INTERSECT to find the solution(s):
8.
2 cos 2 θ + cos θ − 1 = 0
(cos θ + 1)(2 cos θ − 1) = 0
6
cos θ + 1 = 0
cos θ = −1
6
−6
2 cos θ = −1
1
2
2π 4π
θ= ,
3 3
⎧ π 2π 4π 3π ⎫
The solution set is ⎨ ,
,
,
⎬.
3
2 ⎭
⎩2 3
=
3.
2 cos θ + 1 = 0
π 3π
2 2
2. x 2 − x − 1 = 0
x=
or
θ= ,
4 x − 5 = 0 or x + 1 = 0
5
x=
or
x = −1
4
5⎫
⎧
The solution set is ⎨−1, ⎬ .
4⎭
⎩
− ( −1) ±
2 cos 2 θ + cos θ = 0
cos θ (2 cos θ + 1) = 0
θ =π
or
2 cos θ − 1 = 0
2 cos θ = 1
1
2
π 5π
θ= ,
3 3
5π ⎫
⎧π
The solution set is ⎨ , π ,
⎬.
3 ⎭
⎩3
−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} .
cos θ =
301
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Chapter 3: Analytic Trigonometry
9. (tan θ − 1)(sec θ − 1) = 0
tan θ − 1 = 0
or sec θ − 1 = 0
tan θ = 1
sec θ = 1
π 5π
θ =0
θ= ,
4 4
⎧ π 5π ⎫
The solution set is ⎨0, ,
⎬.
⎩ 4 4 ⎭
1 − cos 2 θ = 6 cos θ + 6
cos 2 θ + 6 cos θ + 5 = 0
( cos θ + 5 )( cos θ + 1) = 0
cos θ + 5 = 0
θ=
3π 7 π
,
4 4
(
2
2
2 − 2 cos θ = 3 − 3cos θ
2
2 cos θ − 3cos θ + 1 = 0
2
( 2 cos θ − 1)( cos θ − 1) = 0
2 cos θ − 1 = 0
1
cos θ =
2
θ = 1 + cos θ
θ=
1 − 2 cos θ = 1 + cos θ
2
( cos θ )( 2 cos θ + 1) = 0
θ=
or
2 cos θ + 1 = 0
π 3π
cos θ = −
,
2 2
15.
1
2
2
2
3
2
π 2π
θ= ,
3 3
⎧ π 2π
,
The solution set is ⎨ ,
⎩3 3
θ + sin θ = 0
1 − 2sin θ + sin θ = 0
2sin 2 θ − sin θ − 1 = 0
( 2sin θ + 1)( sin θ − 1) = 0
sin θ = −
or
1
2
16.
sin θ = 1
θ=
,
4π 5π
,
3 3
4π 5π ⎫
,
⎬.
3
3 ⎭
cos ( 2θ ) = 2 − 2sin 2 θ
1 − 2sin 2 θ = 2 − 2sin 2 θ
1 = 2 (not possible)
The equation has no real solution.
sin θ − 1 = 0
7π 11π
θ=
,
6 6
⎧ π 7π 11π ⎫
The solution set is ⎨ ,
,
⎬.
6 ⎭
⎩2 6
cos ( 2θ ) + 6sin 2 θ = 4
sin θ = ±
2
2sin θ + 1 = 0
,
3 3
4sin 2 θ = 3
3
sin 2 θ =
4
cos 2 θ − sin 2 θ + sin θ = 0
(1 − sin θ ) − sin
π 5π
cos θ − 1 = 0
cos θ = 1
θ =0
1 − 2sin 2 θ + 6sin 2 θ = 4
2π 4π
θ=
,
3 3
⎧ π 2π 4π 3π ⎫
,
,
The solution set is ⎨ ,
⎬.
3
2 ⎭
⎩2 3
12.
or
⎧ π 5π ⎫
The solution set is ⎨0, ,
⎬.
⎩ 3 3 ⎭
2 cos 2 θ + cos θ = 0
cos θ = 0
)
2 1 − cos 2 θ = 3 (1 − cos θ )
sin 2 θ − cos 2 θ = 1 + cos θ
(1 − cos θ ) − cos
2sin 2 θ = 3 (1 − cos θ )
14.
⎧ 3π 7π ⎫
The solution set is ⎨ ,
⎬.
4 ⎭
⎩ 4
11.
θ =π
The solution set is {π } .
1
=0
2
1
csc θ =
2
(not possible)
cot θ = −1
cos θ = −1
(not possible)
csc θ −
or
cos θ + 1 = 0
or
cos θ = −5
1⎞
⎛
10. (cot θ + 1) ⎜ csc θ − ⎟ = 0
2⎠
⎝
cot θ + 1 = 0
sin 2 θ = 6 ( cos θ + 1)
13.
π
2
302
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Section 3.8: Trigonometric Equations (II)
17. cos θ = sin θ
sin θ
=1
cos θ
tan θ = 1
π 5π
θ= ,
4 4
21.
⎧ π 5π ⎫
The solution set is ⎨ ,
⎬.
⎩4 4 ⎭
18. cos θ + sin θ
sin θ
sin θ
cos θ
tan θ
⎧ π 3π ⎫
The solution set is ⎨ ,
⎬.
⎩2 2 ⎭
=0
= − cos θ
22.
tan θ = cot θ
1
tan θ =
tan θ
tan 2 θ = 1
tan θ = ±1
π 3π 5π 7 π
θ= , , ,
4 4 4 4
⎧ π 3π 5π 7π ⎫
,
,
The solution set is ⎨ ,
⎬.
4
4 ⎭
⎩4 4
23.
cos(2θ ) = cos θ
= −1
= −1
3π 7 π
θ= ,
4 4
⎧ 3π 7π ⎫
The solution set is ⎨ ,
⎬.
4 ⎭
⎩4
19.
20.
sin θ = csc θ
1
sin θ =
sin θ
sin 2 θ = 1
sin θ = ±1
π 3π
θ= ,
2 2
tan θ = 2sin θ
sin θ
= 2sin θ
cos θ
sin θ = 2sin θ cos θ
0 = 2sin θ cos θ − sin θ
0 = sin θ (2 cos θ − 1)
2 cos θ − 1 = 0
or sin θ = 0
θ = 0, π
1
cos θ =
2
π 5π
θ= ,
3 3
5π ⎫
⎧ π
The solution set is ⎨0, , π ,
⎬.
3
3 ⎭
⎩
2 cos 2 θ − 1 = cos θ
2 cos 2 θ − cos θ − 1 = 0
(2 cos θ + 1)(cos θ − 1) = 0
2 cos θ + 1 = 0
or cos θ − 1 = 0
1
cos θ = 1
cos θ = −
2
θ =0
2π 4π
θ= ,
3 3
⎧ 2π 4π ⎫
,
The solution set is ⎨0,
⎬.
3
3 ⎭
⎩
24.
sin(2θ ) = cos θ
2sin θ cos θ = cos θ
2sin θ cos θ − cos θ = 0
(cos θ )(2sin θ − 1) = 0
cos θ = 0 or 2sin θ = 1
cos θ = 0
1
sin θ =
2
π 3π
θ= ,
π 5π
2 2
θ= ,
6 6
⎧ π π 5π 3π ⎫
The solution set is ⎨ , ,
,
⎬.
2 ⎭
⎩6 2 6
sin(2θ ) sin θ = cos θ
2sin θ cos θ sin θ = cos θ
2sin 2 θ cos θ − cos θ = 0
(2sin 2 θ − 1) cos θ = 0
2sin 2 θ − 1 = 0
1
sin 2 θ =
2
or
cos θ = 0
π 3π
θ= ,
2 2
2
2
π 3π 5π 7π
θ= , , ,
4 4 4 4
⎧ π π 3π 5π 3π 7π ⎫
The solution set is ⎨ , , , , , ⎬ .
⎩4 2 4 4 2 4 ⎭
sin θ = ±
303
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Chapter 3: Analytic Trigonometry
25.
sin(2θ ) + sin(4θ ) = 0
sin(2θ ) + 2sin(2θ ) cos(2θ ) = 0
sin(2θ ) (1 + 2 cos(2θ ) ) = 0
sin(2θ ) = 0
⎛ 4θ − 6θ
2sin ⎜
⎝
1 + 2 cos(2θ ) = 0
1
cos(2θ ) = −
2
2θ = 0 + 2k π or 2θ = π + 2k π or
θ = kπ
π
θ = + kπ
2
2π
4π
2θ =
+ 2k π or 2θ =
+ 2k π
3
3
π
2π
θ = + kπ
θ=
+ kπ
3
3
On the interval 0 ≤ θ < 2π , the solution set is
4π 3π 5π ⎫
⎧ π π 2π
, π,
,
,
⎨0, , ,
⎬.
3
2
3
3
2
3 ⎭
⎩
or
π
3π
+ 2k π
+ 2k π
or 5θ =
2
2
π 2k π
3π 2k π
θ= +
θ= +
10
5
10
5
On the interval 0 ≤ θ < 2π , the solution set is
11π 13π 3π 17π 19π ⎫
⎧ π 3π π 7π 9π
,
, ,
,
⎨0, , , , , , π ,
⎬.
5θ =
⎩ 10 10 2 10 10
29.
π
3π
+ 2k π or 3θ =
+ 2k π or
2
2
π 2k π
π 2k π
θ= +
θ= +
6
3
2
3
π
3π
θ = + 2k π or θ = + 2k π
2
2
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π π 5π 7π 3π 11π ⎫
,
,
,
⎨ , ,
⎬.
6
2
6 ⎭
⎩6 2 6
3θ =
30.
sin 2 θ = 2 cos θ + 2
2
=0
cos θ + 1 = 0
cos θ = −1
θ =π
The solution set is {π } .
2sin ( 5θ ) sin θ = 0
5
1 + sin θ = 2 cos 2 θ
( cos θ + 1)
sin(5θ ) = 0 or sin θ = 0
5θ = 0 + 2k π or 5θ = π + 2k π
or
2k π
π 2k π
θ=
θ= +
5
5
5
θ = 0 + 2k π or θ = π + 2k π
On the interval 0 ≤ θ < 2π , the solution set is
6π 7π 8π 9π ⎫
⎧ π 2π 3π 4π
⎨0, , , , , π , , , , ⎬ .
5
10 ⎭
cos 2 θ + 2 cos θ + 1 = 0
4θ + 6θ ⎞
⎛ 4θ − 6θ ⎞ = 0
−2sin ⎛⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
− 2sin(5θ ) sin(− θ ) = 0
5
10
1 − cos 2 θ = 2 cos θ + 2
cos(4θ ) − cos(6θ ) = 0
5
2
2sin 2 θ + sin θ − 1 = 0
(2sin θ − 1)(sin θ + 1) = 0
2sin θ − 1 = 0
or sin θ + 1 = 0
1
sin θ = −1
sin θ =
2
3π
θ=
π 5π
2
θ= ,
6 6
⎧ π 5π 3π ⎫
The solution set is ⎨ ,
,
⎬.
2 ⎭
⎩6 6
cos(3θ ) = 0 or cos θ = 0
5
10
1 + sin θ = 2 − 2sin 2 θ
2 cos ( 3θ ) cos θ = 0
⎩ 5 5
10
1 + sin θ = 2(1 − sin 2 θ )
⎛ 2θ + 4θ ⎞
⎛ 2θ − 4θ ⎞
2 cos ⎜
⎟ cos ⎜
⎟=0
⎝ 2 ⎠
⎝ 2 ⎠
2 cos(3θ ) cos(− θ ) = 0
27.
⎞ cos ⎛ 4θ + 6θ ⎞ = 0
⎟
⎜
⎟
2
⎠
⎝ 2 ⎠
2sin(− θ ) cos(5θ ) = 0
− 2sin θ cos(5θ ) = 0
cos(5θ ) = 0 or sin θ = 0
θ = 0 + 2k π or θ = π + 2k π or
cos(2θ ) + cos(4θ ) = 0
26.
sin(4θ ) − sin(6θ ) = 0
28.
31.
2sin 2 θ − 5sin θ + 3 = 0
( 2sin θ − 3)( sin θ + 1) = 0
2sin θ − 3 = 0
or sin θ − 1 = 0
3
π
sin θ = (not possible)
θ=
2
2
⎧π ⎫
The solution set is ⎨ ⎬ .
⎩2⎭
5 ⎭
304
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Section 3.8: Trigonometric Equations (II)
32.
2 cos 2 θ − 7 cos θ − 4 = 0
36.
( 2 cos θ + 1)( cos θ − 4 ) = 0
2 cos θ + 1 = 0
or
1 + cot 2 θ = cot θ + 1
cos θ − 4 = 0
cos θ = 4
(not possible)
cot 2 θ − cot θ = 0
cot θ (cot θ − 1) = 0
1
2
2π 4 π
θ= ,
3 3
⎧ 2π 4π ⎫
The solution set is ⎨ ,
⎬.
3 ⎭
⎩ 3
sin θ = −
33.
cot θ = 0
or cot θ = 1
3
π π
π 5π
θ= ,
θ= ,
2 2
4 4
π
π
5π 3π ⎫
⎧
The solution set is ⎨ , ,
,
⎬.
2 ⎭
⎩4 2 4
3(1 − cos θ ) = sin 2 θ
37.
3 − 3cos θ = 1 − cos θ
2
2sin 2 θ − sin θ + 2 = 0
This equation is quadratic in sin θ .
The discriminant is b 2` − 4ac = 1 − 16 = −15 < 0 .
The equation has no real solutions.
( cos θ − 1)( cos θ − 2 ) = 0
cos θ − 1 = 0 or cos θ − 2 = 0
cos θ = 1
cos θ = 2
θ =0
(not possible)
38.
The solution set is {0} .
2 cos 2 θ + 5cos θ + 2 = 0
(2 cos θ + 1)(cos θ + 2) = 0
4 + 4sin θ = 1 − sin 2 θ
2 cos θ = −1
1
cos θ = −
2
2π 4π
θ= ,
3 3
sin 2 θ + 4sin θ + 3 = 0
( sin θ + 1)( sin θ + 3) = 0
or
sin θ + 3 = 0
sin θ = −3
(not possible)
or
cos θ = − 2
(not possible)
⎧ 2π 4π ⎫
The solution set is ⎨ ,
⎬.
3 ⎭
⎩ 3
⎧ 3π ⎫
The solution set is ⎨ ⎬ .
⎩2 ⎭
35.
cos(2θ ) + 5cos θ + 3 = 0
2 cos 2 θ − 1 + 5cos θ + 3 = 0
4(1 + sin θ ) = cos 2 θ
sin θ + 1 = 0
sin θ = −1
3π
θ=
2
3 − sin θ = cos(2θ )
3 − sin θ = 1 − 2sin 2 θ
cos 2 θ − 3cos θ + 2 = 0
34.
csc2 θ = cot θ + 1
39.
sec 2 θ + tan θ = 0
tan 2 θ + 1 + tan θ = 0
This equation is quadratic in tan θ .
The discriminant is b 2` − 4ac = 1 − 4 = −3 < 0 .
The equation has no real solutions.
3
tan 2 θ = sec θ
2
3
2
sec θ − 1 = sec θ
2
2
2sec θ − 2 = 3sec θ
2sec2 θ − 3sec θ − 2 = 0
(2sec θ + 1)(sec θ − 2) = 0
2sec θ + 1 = 0
or sec θ − 2 = 0
1
sec θ = 2
sec θ = −
2
π 5π
θ= ,
(not possible)
3 3
⎧ π 5π ⎫
The solution set is ⎨ ,
⎬.
⎩3 3 ⎭
305
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Chapter 3: Analytic Trigonometry
40.
sec θ
1
cos θ
1
cos θ
1
cos θ
sin θ cos θ
cos θ
sin θ
= tan θ + cot θ
sin θ cos θ
=
+
cos θ sin θ
sin 2 θ + cos 2 θ
=
sin θ cos θ
1
=
sin θ cos θ
42.
=1
=1
π
2
⎛π⎞
⎛π⎞
Since sec ⎜ ⎟ and tan ⎜ ⎟ do not exist, the
2
⎝ ⎠
⎝2⎠
equation has no real solutions.
θ=
41. sin θ − 3 cos θ = 1
Divide each side by 2:
1
3
1
sin θ −
cos θ =
2
2
2
Rewrite in the difference of two angles form
π
1
3
using cos φ = , sin φ =
, and φ = :
3
2
2
1
sin θ cos φ − cos θ sin φ =
2
1
sin(θ − φ ) =
2
π
5π
or θ − φ =
θ −φ =
6
6
π π
π 5π
θ− =
θ− =
3 6
3 6
π
7π
θ=
θ=
2
6
3 sin θ + cos θ = 1
Divide each side by 2:
3
1
1
sin θ + cos θ =
2
2
2
Rewrite in the sum of two angles form using
π
1
3
, sin φ = , and φ = :
cos φ =
6
2
2
1
sin θ cos φ + cos θ sin φ =
2
1
sin(θ + φ ) =
2
π
5π
or θ + φ =
θ +φ =
6
6
π π
π 5π
or θ + =
θ+ =
6 6
6 6
2π
θ = 0 or
θ=
3
⎧ 2π ⎫
The solution set is ⎨0,
⎬.
3 ⎭
⎩
tan(2θ ) + 2sin θ = 0
43.
sin(2θ )
+ 2sin θ = 0
cos(2θ )
sin 2θ + 2sin θ cos 2θ
=0
cos 2θ
2sin θ cos θ + 2sin θ (2 cos 2 θ − 1) = 0
(
2sin θ ( 2 cos
)
θ + cos θ − 1) = 0
2sin θ cos θ + 2 cos 2 θ − 1 = 0
2
2sin θ (2 cos θ − 1)(cos θ + 1) = 0
2 cos θ − 1 = 0
or 2sin θ = 0
1
2
π 5π
θ= ,
3 3
cos θ =
⎧ π 7π ⎫
The solution set is ⎨ ,
⎬.
⎩2 6 ⎭
or
sin θ = 0
θ = 0, π
cos θ + 1 = 0
cos θ = −1
θ =π
5π ⎫
⎧ π
The solution set is ⎨0, , π ,
⎬.
3
3 ⎭
⎩
306
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Section 3.8: Trigonometric Equations (II)
tan(2θ ) + 2 cos θ = 0
44.
Rewrite in the sum of two angles form using
1
1
π
, sin φ =
, and φ = :
cos φ =
4
2
2
sin θ cos φ + cos θ sin φ = −1
sin(θ + φ ) = −1
3π
θ +φ =
2
π 3π
θ+ =
4 2
5π
θ=
4
⎧ 5π ⎫
The solution set is ⎨ ⎬ .
⎩ 4 ⎭
sin(2θ )
+ 2 cos θ = 0
cos(2θ )
sin ( 2θ ) + 2 cos θ cos 2θ
cos ( 2θ )
=0
2sin θ cos θ + 2 cos θ (1 − 2sin 2 θ ) = 0
(
− 2 cos θ ( 2sin
)
θ − sin θ − 1) = 0
2 cos θ sin θ + 1 − 2sin 2 θ = 0
2
− 2 cos θ (2sin θ + 1)(sin θ − 1) = 0
−2 cos θ = 0
or 2sin θ + 1 = 0
cos θ = 0
π 3π
2 2
θ= ,
or
1
2
7π 11π
θ= ,
6 6
sin θ = −
47.
sin θ − 1 = 0
sin θ = 1
θ=
f ( x) = 0
4 cos x − 1 = 0
1
cos 2 x =
4
2
π
2
cos x = ±
⎧ π 7π 3π 11π ⎫
,
,
The solution set is ⎨ ,
⎬.
2
6 ⎭
⎩2 6
1
1
=±
4
2
The zeros on 0 ≤ x < 2π are
48.
45. sin θ + cos θ = 2
Divide each side by 2 :
1
1
sin θ +
cos θ = 1
2
2
Rewrite in the sum of two angles form using
1
1
π
cos φ =
, sin φ =
, and φ = :
4
2
2
sin θ cos φ + cos θ sin φ = 1
sin(θ + φ ) = 1
π
θ +φ =
2
π π
θ+ =
4 2
π
θ=
4
⎧π ⎫
The solution set is ⎨ ⎬ .
⎩4⎭
π 2π 4π 5π
,
3
3
,
3
,
3
.
f ( x) = 0
4sin x − 3 = 0
4sin 2 x = 3
3
sin 2 x =
4
2
cos x = ±
3
3
=±
4
2
The zeros on 0 ≤ x < 2π are
49.
π 2π 4π 5π
,
3
,
3
.
or 2 cos x − 1 = 0
1
cos x =
2
x=
π 5π
3
The zeros on 0 ≤ x < 2π are 0,
Divide each side by 2 :
1
1
sin θ +
cos θ = −1
2
2
3
f ( x) = 0
sin ( 2 x ) − sin x = 0
2sin x cos x − sin x = 0
sin x ( 2 cos x − 1) = 0
sin x = 0
x = 0, π
46. sin θ + cos θ = − 2
,
3
,
π
3
3
,π ,
5π
.
3
307
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Chapter 3: Analytic Trigonometry
50.
f ( x) = 0
cos ( 2 x ) + cos x = 0
2 cos 2 x − 1 + cos x = 0
2 cos 2 x + cos x − 1 = 0
( 2 cos x − 1)( cos x + 1) = 0
2 cos x − 1 = 0
1
cos x =
2
x=
π 5π
3
,
10
3π −3π
−10
10
π
3
,π ,
−10
f ( x) = 0
2
sin x + 2 cos x + 2 = 0
1 − cos 2 x + 2 cos x + 2 = 0
− cos 2 x + 2 cos x + 3 = 0
cos 2 x − 2 cos x − 3 = 0
( cos x − 3)( cos x + 1) = 0
x ≈ −1.31, 1.98, 3.84
54. x − 4sin x = 0
Find the zeros (x-intercepts) of Y1 = x − 4sin x :
10
3π −3π
−10
10
3π
−10
3π
−3π
−10
π 3π
x ≈ − 2.47, 0, 2.47
,
2 2
The zeros on 0 ≤ x < 2π are
10
−3π
f ( x) = 0
cos ( 2 x ) + sin 2 x = 0
cos 2 x − sin 2 x + sin 2 x = 0
cos 2 x = 0
cos x = 0
x=
3π
−3π
5π
.
3
3π
−10
3
cos x − 3 = 0
or cos x + 1 = 0
cos x = 3
cos x = −1
(not possible)
x =π
The only zero on 0 ≤ x < 2π is π .
52.
10
−3π
or cos x + 1 = 0
cos x = −1
x =π
The zeros on 0 ≤ x < 2π are
51.
53. x + 5cos x = 0
Find the zeros (x-intercepts) of Y1 = x + 5cos x :
π 3π
,
2 2
55. 22 x − 17 sin x = 3
Find the intersection of Y1 = 22 x − 17 sin x and
Y2 = 3 :
.
5
π
−π
−5
x ≈ 0.52
308
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Section 3.8: Trigonometric Equations (II)
56. 19 x + 8cos x = 2
Find the intersection of Y1 = 19 x + 8cos x and
Y2 = 2 :
60. x 2 + 3sin x = 0
Find the zeros (x-intercepts) of Y1 = x 2 + 3sin x :
3
5
π −π
−π
π
−π
3
π
−3
−5
−3
x ≈ −1.72, 0
x ≈ − 0.30
61. x 2 − 2sin ( 2 x ) = 3 x
57. sin x + cos x = x
Find the intersection of Y1 = sin x + cos x and
Y2 = x :
Find the intersection of Y1 = x 2 − 2sin ( 2 x ) and
Y2 = 3 x :
3
12
12
π
−π
__
3π −π
−π
__
3π
2
−3
2
−3
x ≈ 1.26
−3
x ≈ 0, 2.15
58. sin x − cos x = x
Find the intersection of Y1 = sin x − cos x and
Y2 = x :
62. x 2 = x + 3cos(2 x)
Find the intersection of Y1 = x 2 and
Y2 = x + 3cos(2 x) :
3
10
10
π
−π
2π −2π
−2π
2π
−3
x ≈ −1.26
−5
x ≈ − 0.62, 0.81
59. x 2 − 2 cos x = 0
Find the zeros (x-intercepts) of Y1 = x 2 − 2 cos x :
3
−3
63. 6sin x − e x = 2, x > 0
3
π −π
−π
−5
Find the intersection of Y1 = 6sin x − e x and
Y2 = 2 :
π
6
−3
6
2π 0
0
2π
x ≈ −1.02, 1.02
−6
−6
x ≈ 0.76, 1.35
309
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
d. Graph Y1 = 16sin x ( cos x + 1) and use the
64. 4 cos(3 x) − e x = 1, x > 0
MAXIMUM feature:
Find the intersection of Y1 = 4 cos(3 x) − e and
Y2 = 1 :
x
25
6
π
0
0
The maximum area is approximately
20.78 in.2 when the angle is 60˚.
−6
x ≈ 0.31
65. a.
66. a.
cos(2θ ) + cos θ = 0 , 0º < θ < 90º
2 cos 2 θ + cos θ − 1 = 0
(2 cos θ − 1)(cos θ + 1) = 0
2 cos θ − 1 = 0
or cos θ + 1 = 0
1
cos θ = −1
2
θ = 180º
θ = 60º , 300º
On the interval 0º < θ < 90º , the solution is
60˚.
cos θ =
cos(2θ ) + cos θ = 0, 0º < θ < 90º
⎛ 3θ ⎞
⎛θ ⎞
2 cos ⎜ ⎟ cos ⎜ ⎟ = 0
⎝ 2 ⎠
⎝2⎠
⎛ 3θ ⎞
⎛θ ⎞
cos ⎜ ⎟ = 0 or cos ⎜ ⎟ = 0
2
⎝ ⎠
⎝2⎠
3θ
3θ
θ
θ
= 90º ;
= 270º ;
= 90º ;
= 270º ;
2
2
2
2
θ =60º
θ =180º
θ = 180º θ = 540º
On the interval 0º < θ < 90º , the solution is
60˚.
c.
b.
sin(2θ ) = − cos(2θ )
3⎛1 ⎞
⎜ + 1⎟
2 ⎝2 ⎠
= 12 3 in ≈ 20.78 in
2
sin(2θ ) + cos(2θ ) = 0
sin(2θ )
= −1
cos(2θ )
tan(2θ ) = −1
A(60º ) = 16sin ( 60º ) ⎡⎣cos ( 60º ) + 1⎤⎦
= 16 ⋅
sin(2θ ) + cos(2θ ) = 0
Divide each side by 2 :
1
1
sin(2θ ) +
cos(2θ ) = 0
2
2
Rewrite in the sum of two angles form using
1
1
π
cos φ =
and sin φ =
and φ = :
4
2
2
sin(2θ ) cos φ + cos(2θ ) sin φ = 0
sin(2θ + φ ) = 0
2θ + φ = 0 + k π
π
2θ + = 0 + k π
4
π
2θ = − + k π
4
π kπ
θ =− +
8 2
3π
θ=
= 67.5º
8
2 cos 2 θ − 1 + cos θ = 0
b.
90
0
3π
4
3π
θ=
= 67.5º
8
2θ =
2
310
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Section 3.8: Trigonometric Equations (II)
c.
R=
b.
322 2
( sin(2 ⋅ 67.5º ) − cos(2 ⋅ 67.5º ) − 1)
32
3sec θ tan θ − 4 csc θ cot θ = 0
3sec θ tan θ = 4 csc θ cot θ
sec θ tan θ 4
=
csc θ cot θ 3
4
tan 3 θ =
3
= 32 2 ( sin (135º ) − cos (135º ) − 1)
⎛ 2 ⎛
2⎞ ⎞
= 32 2 ⎜
− ⎜⎜ −
⎟⎟ − 1⎟⎟
⎜ 2
⎝ 2 ⎠ ⎠
⎝
= 32 2
(
(
)
4
≈ 1.10064
3
θ ≈ 47.74º
2 −1
)
tan θ =
= 32 2 − 2 feet ≈ 18.75 feet
322 2
( sin(2 x) − cos(2 x) − 1)
32
and use the MAXIMUM feature:
c.
d. Graph Y1 =
L ( 47.74º ) =
3
3
4
+
cos ( 47.74º ) sin ( 47.74º )
≈ 9.87 feet
20
3
4
+
and use the
cos x sin x
MINIMUM feature:
d. Graph Y1 =
20
45
90
0
The angle that maximizes the distance is
67.5˚, and the maximum distance is 18.75
feet.
0
An angle of θ ≈ 47.74° minimizes the
length at L ≈ 9.87 feet .
67. Find the first two positive intersection points of
Y1 = − x and Y2 = tan x .
2
e.
2
2π 0
0
−12
2π
−12
The first two positive solutions are x ≈ 2.03 and
x ≈ 4.91 .
68. a.
69. 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
3
cos θ =
x
x=
L(θ ) =
3
cos θ
For this problem, only one minimum length
exists. This minimum length is 9.87 feet,
and it occurs when θ ≈ 47.74° . No matter
if we find the minimum algebraically (using
calculus) or graphically, the minimum will
be the same.
107 =
(34.8) 2 sin ( 2θ )
9.8
107(9.8)
≈ 0.8659
sin ( 2θ ) =
(34.8) 2
2θ ≈ sin −1 ( 0.8659 )
2θ ≈ 60º or 120º
4
sin θ =
y
y=
90
0
θ ≈ 30º or 60º
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° :
(34.8) 2 sin ⎡⎣ 2 ( 45° ) ⎤⎦
≈ 123.6
R ( 45° ) =
9.8
The maximum distance is 123.6 meters.
4
sin θ
3
4
+
= 3sec θ + 4 csc θ
cos θ sin θ
311
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Chapter 3: Analytic Trigonometry
c.
Let Y1 =
125
Chapter 3 Review Exercises
(34.8) 2 sin(2 x)
9.8
1. sin −1 1
Find the angle θ , −
π
π
≤ θ ≤ , whose sine
2
2
equals 1.
0
90
0
sin θ = 1,
d.
θ=
−
π
2
π
π
≤θ ≤
2
2
Thus, sin −1 (1) =
70. a.
2. cos −1 0
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
equals 0.
cos θ = 0, 0 ≤ θ ≤ π
π
θ=
2
π
Thus, cos −1 ( 0 ) = .
2
(40) 2 sin(2θ )
9.8
110 ⋅ 9.8
sin(2θ ) =
≈ 0.67375
402
2θ ≈ sin −1 ( 0.67375 )
110 =
3. tan −1 1
2θ ≈ 42.4º or 137.6º
θ ≈ 21.2º or 68.8º
Find the angle θ , −
Let Y1 =
170
tan θ = 1,
θ=
π
4
−
π
π
<θ <
2
2
Thus, tan −1 (1) =
(40) 2 sin(2 x)
:
9.8
π
.
4
⎛ 1⎞
4. sin −1 ⎜ − ⎟
⎝ 2⎠
Find the angle θ , −
0
0
π
π
< θ < , whose tangent
2
2
equals 1.
b. The maximum distance will occur when the
angle of elevation is 45° :
(40) 2 sin [ 2(45°) ]
R ( 45° ) =
≈ 163.3
9.8
The maximum distance is approximately
163.3 meter
c.
π
.
2
π
π
≤ θ ≤ , whose sine
2
2
1
equals − .
2
1
π
π
sin θ = − , − ≤ θ ≤
2
2
2
π
θ =−
6
π
⎛ 1⎞
Thus, sin −1 ⎜ − ⎟ = − .
6
⎝ 2⎠
90
d.
312
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Chapter 3 Review Exercises
⎛
3⎞
5. cos −1 ⎜⎜ −
⎟⎟
⎝ 2 ⎠
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
⎛ ⎛ 3π
9. sin −1 ⎜⎜ sin ⎜
⎝ ⎝ 8
(
(
θ =−
π
3
(
(
)
π
.
3
(
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
(
)
cannot use the formula directly since
2π
is not
3
⎡ π π⎤
in the interval ⎢ − , ⎥ . We need to find an
⎣ 2 2⎦
⎡ π π⎤
angle θ in the interval ⎢ − , ⎥ for which
⎣ 2 2⎦
2π
⎛ 2π ⎞
is in quadrant
tan ⎜
⎟ = tan θ . The angle
3
3
⎝
⎠
II so tangent is negative. Therefore, we want θ
to be in quadrant IV so tangent will still be
⎡ π π⎤
negative and in the interval ⎢ − , ⎥ . Thus, we
⎣ 2 2⎦
0 ≤θ ≤ π
Thus, sec −1 2 =
)
equation f −1 f ( x ) = tan −1 tan ( x ) = x but we
2.
sec θ = 2,
π
θ=
4
(
⎛
⎛ 2π
11. tan −1 ⎜⎜ tan ⎜
⎝ 3
⎝
7. sec −1 2
Find the angle θ , 0 ≤ θ ≤ π, whose secant
equals
)
3π ⎞ 3π
⎛
directly and get cos −1 ⎜ cos ⎟ =
.
4 ⎠ 4
⎝
π
π
<θ <
2
2
Thus, tan −1 − 3 = −
)
3π ⎞
⎛
10. cos −1 ⎜ cos ⎟ follows the form of the equation
4 ⎠
⎝
3π
f −1 f ( x ) = cos −1 cos ( x ) = x . Since
is
4
in the interval ⎡⎣ 0, π ⎤⎦ , we can apply the equation
π
π
Find the angle θ , − < θ < , whose tangent
2
2
equals − 3 .
−
(
3π
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply
8
⎣ 2 2⎦
the equation directly and get
⎛ ⎛ 3π ⎞ ⎞ 3π
.
sin −1 ⎜⎜ sin ⎜ ⎟ ⎟⎟ =
⎝ ⎝ 8 ⎠⎠ 8
)
tan θ = − 3,
)
equation f −1 f ( x ) = sin −1 sin ( x ) = x . Since
3
equals −
.
2
3
cos θ = −
, 0 ≤θ ≤ π
2
5π
θ=
6
⎛
3 ⎞ 5π
Thus, cos −1 ⎜⎜ −
.
⎟⎟ =
⎝ 2 ⎠ 6
6. tan −1 − 3
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
π
.
4
8. cot −1 ( −1)
Find the angle θ , 0 < θ < π , whose cotangent
equals −1 .
cot θ = −1, 0 < θ < π
3π
θ=
4
3π
Thus, cot −1 ( −1) =
.
4
π
⎛ 2π ⎞
⎛ π⎞
have tan ⎜
⎟ = tan ⎜ − ⎟ . Since − is in the
3
⎝ 3 ⎠
⎝ 3⎠
⎡ π π⎤
interval ⎢ − , ⎥ , we can apply the equation
⎣ 2 2⎦
above and get
⎛
π
⎛ 2π ⎞ ⎞
⎛ π ⎞⎞
−1 ⎛
tan −1 ⎜⎜ tan ⎜
⎟ ⎟⎟ = tan ⎜⎜ tan ⎜ − ⎟ ⎟⎟ = − .
3
⎝ 3 ⎠⎠
⎝ 3 ⎠⎠
⎝
⎝
313
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Chapter 3: Analytic Trigonometry
⎡ π π⎤
negative and in the interval ⎢ − , ⎥ . Thus, we
⎣ 2 2⎦
π
⎛ 8π ⎞
⎛ π⎞
have sin ⎜ − ⎟ = sin ⎜ − ⎟ . Since − is in
9
9
9
⎝
⎠
⎝
⎠
⎛ ⎛ π ⎞⎞
12. sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ follows the form of the
⎝ ⎝ 8 ⎠⎠
(
)
(
)
equation f −1 f ( x ) = sin −1 sin ( x ) = x . Since
π
⎡ π π⎤
is in the interval ⎢ − , ⎥ , we can apply
8
⎣ 2 2⎦
the equation directly and get
⎛ ⎛ π ⎞⎞
π
sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ = − .
8
⎝ ⎝ 8 ⎠⎠
−
⎛
⎛ 15π
13. cos −1 ⎜⎜ cos ⎜
⎝ 7
⎝
(
⎡ π π⎤
the interval ⎢ − , ⎥ , we can apply the equation
⎣ 2 2⎦
above and get
⎛ ⎛ 8π ⎞ ⎞
⎛ ⎛ π ⎞⎞
π
sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ = sin −1 ⎜⎜ sin ⎜ − ⎟ ⎟⎟ = − .
9
9
9
⎠⎠
⎠⎠
⎝ ⎝
⎝ ⎝
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
(
(
7
⎛ ⎛ 8π
14. sin −1 ⎜⎜ sin ⎜ −
⎝ ⎝ 9
(
(
)
(
)
(
)
directly and get cos cos −1 0.6 = 0.6 .
(
)
17. cos cos −1 ( −0.3) follows the form of the
(
)
(
)
equation f f −1 ( x ) = cos cos −1 ( x ) = x .
π⎞ π
⎞⎞
−1 ⎛
⎟ ⎟⎟ = cos ⎜ cos ⎟ = .
7⎠ 7
⎠⎠
⎝
(
)
in the interval ⎡⎣ −1,1⎤⎦ , we can apply the equation
. Thus, we
Since −0.3 is in the interval ⎡⎣ −1,1⎤⎦ , we can
apply the equation directly and get
(
⎞⎞
⎟ ⎟⎟ follows the form of the
⎠⎠
)
(
f f −1 ( x ) = cos cos −1 ( x ) = x . Since 0.6 is
interval ⎡⎣ 0, π ⎤⎦ , we can apply the equation above
⎛ 15π
and get cos ⎜⎜ cos ⎜
⎝ 7
⎝
)
16. cos cos −1 0.6 follows the form of the equation
π
π
⎛ 15π ⎞
is in the
have cos ⎜
⎟ = cos . Since
7
7
⎝ 7 ⎠
−1 ⎛
)
(
15π
⎞
is in
⎟ = cos θ . The angle
7
⎠
π
(
directly and get sin sin −1 0.9 = 0.9 .
angle θ in the interval ⎣⎡ 0, π ⎦⎤ for which
quadrant I and coterminal with
)
the interval ⎡⎣ −1,1⎤⎦ , we can apply the equation
15π
is
7
not in the interval ⎡⎣0, π ⎤⎦ . We need to find an
⎛ 15π
cos ⎜
⎝ 7
)
f f −1 ( x ) = sin sin −1 ( x ) = x . Since 0.9 is in
)
equation f −1 f ( x ) = cos −1 cos ( x ) = x , but
we cannot use the formula directly since
(
15. sin sin −1 0.9 follows the form of the equation
)
cos cos −1 ( −0.3) = −0.3 .
)
(
)
equation f −1 f ( x ) = sin −1 sin ( x ) = x , but we
18. tan tan −1 5 follows the form of the equation
8π
is not
9
f f −1 ( x ) = tan tan −1 ( x ) = x . Since 5 is a
cannot use the formula directly since −
(
)
(
)
real number, we can apply the equation directly
⎡ π π⎤
in the interval ⎢ − , ⎥ . We need to find an
⎣ 2 2⎦
⎡ π π⎤
angle θ in the interval ⎢ − , ⎥ for which
⎣ 2 2⎦
8π
⎛ 8π ⎞
is in
sin ⎜ − ⎟ = sin θ . The angle −
9
⎝ 9 ⎠
quadrant III so sine is negative. Therefore, we
want θ to be in quadrant IV so sine will still be
(
)
and get tan tan −1 5 = 5 .
19. Since there is no angle θ such that cos θ = −1.6 ,
the quantity cos −1 ( −1.6 ) is not defined. Thus,
(
)
cos cos −1 ( −1.6 ) is not defined.
314
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Chapter 3 Review Exercises
20. Since there is no angle θ such that sin θ = 1.6 ,
the quantity sin −1 1.6 is not defined. Thus,
(
⎛
3⎞
27. sec ⎜⎜ tan −1
⎟
3 ⎟⎠
⎝
)
sin sin −1 1.6 is not defined.
Find the angle θ , −
2π ⎞
π
−1 ⎛
−1 ⎛ 1 ⎞
21. sin ⎜ cos ⎟ = sin ⎜ − ⎟ = −
3
2
6
⎝
⎠
⎝
⎠
π
π
< θ < , whose tangent is
2
2
3
3
3
π
π
, − <θ <
3
2
2
π
θ=
6
3 π
So, tan −1
= .
3
6
⎛
3⎞
⎛π⎞ 2 3
.
Thus, sec ⎜⎜ tan −1
⎟⎟ = sec ⎜ ⎟ =
3 ⎠
3
⎝6⎠
⎝
tan θ =
3π ⎞
⎛
−1
22. cos ⎜ tan ⎟ = cos ( −1) = π
4
⎝
⎠
−1
7π ⎞
π
−1 ⎛
−1
23. tan ⎜ tan ⎟ = tan ( −1) = −
4 ⎠
4
⎝
⎛
7π ⎞
3 ⎞ 5π
⎛
24. cos −1 ⎜ cos ⎟ = cos −1 ⎜⎜ −
⎟⎟ =
6 ⎠
⎝
⎝ 2 ⎠ 6
⎛
⎡
⎛
3 ⎞⎤
25. tan ⎢sin −1 ⎜⎜ −
⎟⎟ ⎥
⎝ 2 ⎠ ⎥⎦
⎣⎢
28. csc ⎜ sin −1
⎝
3⎞
⎟
2 ⎠
Find the angle θ , −
π
π
Find the angle θ , − ≤ θ ≤ , whose sine
2
2
3
.
equals −
2
3
π
π
sin θ = −
, − ≤θ ≤
2
2
2
π
θ =−
3
⎛
3⎞
π
So, sin −1 ⎜⎜ −
⎟⎟ = − .
3
⎝ 2 ⎠
π
π
≤ θ ≤ , whose sine equals
2
2
3
.
2
3
π
π
, − ≤θ ≤
2
2
2
π
θ=
3
3 π
= .
So, sin −1
2
3
⎛ −1 3 ⎞
⎛ π⎞ 2 3
Thus, csc ⎜ sin
.
⎟ = csc ⎜ ⎟ =
2
3
⎝3⎠
⎝
⎠
sin θ =
⎡
⎛
3 ⎞⎤
⎛ π⎞
Thus, tan ⎢sin −1 ⎜⎜ −
⎟⎟ ⎥ = tan ⎜ − ⎟ = − 3 .
2
⎝ 3⎠
⎝
⎠ ⎥⎦
⎣⎢
3
29. sin ⎛⎜ cot −1 ⎞⎟
4⎠
⎝
3
4
⎡
⎛ 1 ⎞⎤
26. tan ⎢cos −1 ⎜ − ⎟ ⎥
⎝ 2 ⎠⎦
⎣
Find the angle θ , 0 ≤ θ ≤ π, whose cosine
Since cot θ = , 0 < θ < π , θ is in quadrant I.
Let x = 3 and y = 4 . Solve for r: 9 + 16 = r 2
r 2 = 25
1
equals − .
2
1
cos θ = − , 0 ≤ θ ≤ π
2
2π
θ=
3
⎛ 1 ⎞ 2π
So, cos −1 ⎜ − ⎟ =
⎝ 2⎠ 3
⎡
⎛ 1 ⎞⎤
⎛ 2π ⎞
Thus, tan ⎢cos −1 ⎜ − ⎟ ⎥ = tan ⎜ ⎟ = − 3 .
2
⎝
⎠⎦
⎝ 3 ⎠
⎣
r =5
3
y 4
Thus, sin ⎛⎜ tan −1 ⎞⎟ = sin θ = = .
4⎠
r 5
⎝
315
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Chapter 3: Analytic Trigonometry
find the domain of f −1 ( x ) we note that the
30. cos ⎛⎜ csc−1 ⎞⎟
3
5
⎝
⎠
x
and
2
that it must lie in the interval ⎣⎡ −1,1⎦⎤ . That is,
5
π
π
≤ θ ≤ , let r = 5 and y = 3 .
3
2
2
Solve for x: x 2 + 9 = 25
argument of the inverse sine function is
Since cscθ = , −
x
≤1
2
−2 ≤ x ≤ 2
The domain of f −1 ( x ) is { x | −2 ≤ x ≤ 2} , or
x 2 = 16
−1 ≤
x = ±4
Since θ is in quadrant I, x = 4 .
x 4
5
Thus, cos ⎛⎜ csc−1 ⎞⎟ = cosθ = = .
3⎠
⎝
⎡
r
5
⎡⎣ −2, 2 ⎤⎦ in interval notation.
4 ⎤
31. tan ⎢sin −1 ⎛⎜ − ⎞⎟ ⎥
⎝ 5⎠
34.
⎣
⎦
4
π
π
Since sin θ = − , − ≤ θ ≤ , let y = − 4 and
5
2
2
2
r = 5 . Solve for x: x + 16 = 25
y = tan ( 2 x + 3) − 1
x = tan ( 2 y + 3) − 1
x + 1 = tan ( 2 y + 3)
x2 = 9
2 y + 3 = tan −1 ( x + 1)
x = ±3
Since θ is in quadrant IV, x = 3 .
⎡
4 ⎤
−4
2 y = tan −1 ( x + 1) − 3
Thus, tan ⎢sin −1 ⎛⎜ − ⎞⎟ ⎥ = tan θ = =
=−
x
3
3
⎝ 5 ⎠⎦
⎣
y
4
1
3
tan −1 ( x + 1) − = f −1 ( x )
2
2
The domain of f ( x ) is all real numbers such
y=
⎡
⎛ 3 ⎞⎤
32. tan ⎢cos −1 ⎜ − ⎟ ⎥
⎝ 5 ⎠⎦
⎣
3
Since cos θ = − , 0 ≤ θ ≤ π , let x = − 3 and
5
that x ≠
( 2k + 1) π − 3
where k is an integer. To
4
2
find the domain of f −1 ( x ) we note that the
argument of the inverse tangent function can be
any real number. Thus, the domain of f −1 ( x ) is
r = 5 . Solve for y: 9 + y 2 = 25
y 2 = 16
y = ±4
Since θ is in quadrant II, y = 4 .
all real numbers, or ( −∞, ∞ ) in interval notation.
⎡
y 4
4
⎛ 3 ⎞⎤
=−
Thus, tan ⎢cos −1 ⎜ − ⎟ ⎥ = tan θ = =
x
5
−
3
3
⎝
⎠
⎣
⎦
33.
f ( x ) = tan ( 2 x + 3) − 1
35.
f ( x ) = − cos x + 3
y = − cos x + 3
x = − cos y + 3
f ( x ) = 2sin ( 3 x )
x − 3 = − cos y
y = 2sin ( 3x )
3 − x = cos y
x = 2sin ( 3 y )
y = cos −1 ( 3 − x ) = f −1 ( x )
x
= sin ( 3 y )
2
⎛ x⎞
3 y = sin −1 ⎜ ⎟
⎝2⎠
The domain of f ( x ) is the set of all real
numbers, or ( −∞, ∞ ) in interval notation. To
find the domain of f −1 ( x ) we note that the
1
⎛x⎞
y = sin −1 ⎜ ⎟ = f −1 ( x )
3
⎝2⎠
argument of the inverse cosine function is 3 − x
and that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That
The domain of f ( x ) is the set of all real
is,
numbers, or ( −∞, ∞ ) in interval notation. To
316
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Chapter 3 Review Exercises
−1 ≤ 3 − x ≤ 1
38. Let θ = csc −1 u so that csc θ = u , −
−4 ≤ − x ≤ −2
4≥ x≥2
(
)
cos csc−1 u = cos θ = cos θ ⋅
The domain of f −1 ( x ) is { x | 2 ≤ x ≤ 4} , or
⎡⎣ 2, 4 ⎤⎦ in interval notation.
=
f ( x ) = 2sin ( − x + 1)
=
y = 2sin ( − x + 1)
x = 2sin ( − y + 1)
≤θ ≤
π
2
u2 −1
u
(
)
sin csc −1 u = sin θ =
π
2
≤θ ≤
π
2
1
1
=
csc θ u
40. Let θ = csc −1 u so that csc θ = u , −
(
x
and
2
that it must lie in the interval ⎡⎣ −1,1⎤⎦ . That is,
=
x
≤1
2
−2 ≤ x ≤ 2
The domain of f −1 ( x ) is { x | −2 ≤ x ≤ 2} , or
2
u2 −1
42. sin θ csc θ − sin 2 θ = sin θ ⋅
2
1
− sin 2 θ
tan θ
= 1 − sin 2 θ
= cos 2 θ
⎡⎣ −2, 2 ⎤⎦ in interval notation.
π
1
csc 2 θ − 1
1
41. tan θ cot θ − sin 2 θ = tan θ ⋅
−1 ≤
2
π
)
= sec 2 θ − 1 =
argument of the inverse sine function is
≤θ ≤
2
≤θ ≤
tan csc−1 u = tan θ = tan 2 θ
find the domain of f −1 ( x ) we note that the
π
π
and θ ≠ 0 , u ≥ 1 . Then,
numbers, or ( −∞, ∞ ) in interval notation. To
−1 ≤ u ≤ 1 . Then,
cot θ
cot 2 θ
csc 2 θ − 1
=
=
csc θ
csc θ
csc θ
and θ ≠ 0 , u ≥ 1 . Then,
⎛ x⎞
y = 1 − sin −1 ⎜ ⎟
⎝2⎠
The domain of f ( x ) is the set of all real
37. Let θ = sin −1 u so that sin θ = u , −
sin θ
= cot θ sin θ
sin θ
39. Let θ = csc −1 u so that csc θ = u , −
x
= sin ( − y + 1)
2
⎛ x⎞
− y + 1 = sin −1 ⎜ ⎟
⎝2⎠
⎛ x⎞
− y = sin −1 ⎜ ⎟ − 1
⎝2⎠
(
2
u ≥ 1 . Then,
2≤ x≤4
36.
π
1
− sin 2 θ
sin θ
= 1 − sin 2 θ
= cos 2 θ
,
)
43. sin 2 θ (1 + cot 2 θ ) = sin 2 θ ⋅ csc2 θ
1
= sin 2 θ ⋅ 2
sin θ
=1
cos sin −1 u = cos θ = cos 2 θ
= 1 − sin 2 θ = 1 − u 2
44. (1 − sin 2 θ )(1 + tan 2 θ ) = cos 2 θ ⋅ sec2 θ
1
= cos 2 θ ⋅
cos 2 θ
=1
317
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,
Chapter 3: Analytic Trigonometry
45. 5cos 2 θ + 3sin 2 θ = 2 cos 2 θ + 3cos 2 θ + 3sin 2 θ
= 2 cos 2 θ + 3 cos 2 θ + sin 2 θ
(
50. 1 −
)
(
= 2 cos θ + 3 ⋅1
= 3 + 2 cos 2 θ
2
(
)
46. 4sin 2 θ + 2 cos 2 θ = 4 1 − cos 2 θ + 2 cos 2 θ
= 4 − 4 cos θ + 2 cos 2 θ
= 4 − 2 cos 2 θ
2
47.
)
1
csc θ
θ ⋅ sin θ
sin
51.
=
1 + csc θ 1 + 1 sin θ
sin θ
1
=
sin θ + 1
1
1 − sin θ
=
⋅
1 + sin θ 1 − sin θ
1 − sin θ
=
1 − sin 2 θ
1 − sin θ
=
cos 2 θ
1 − cosθ
sin θ
(1 − cosθ ) 2 + sin 2 θ
+
=
sin θ
1 − cosθ
sin θ (1 − cosθ )
1 − 2cosθ + cos 2 θ + sin 2 θ
=
sin θ (1 − cosθ )
1 − 2cosθ + 1
sin θ (1 − cosθ )
2 − 2cosθ
=
sin θ (1 − cosθ )
2(1 − cosθ )
=
sin θ (1 − cosθ )
2
=
sin θ
= 2cscθ
=
48.
sin 2 θ
1 + cosθ − sin 2 θ
=
1 + cosθ
1 + cosθ
1 + cosθ − 1 − cos 2 θ
=
1 + cosθ
cosθ + cos 2 θ
=
1 + cosθ
cosθ (1 + cosθ )
=
1 + cosθ
= cosθ
1
cos θ ⋅ cos θ
1
cos θ
cos θ
1 + cos θ
=
1
1 + cos θ 1 − cos θ
=
⋅
1
1 − cos θ
1 − cos 2 θ
=
1 − cos θ
sin 2 θ
=
1 − cos θ
1 + sec θ
52.
=
sec θ
sin θ
1 + cos θ sin 2 θ + (1 + cos θ )2
+
=
1 + cos θ
sin θ
sin θ (1 + cos θ )
sin 2 θ + 1 + 2 cos θ + cos 2 θ
=
sin θ (1 + cos θ )
1 + 2 cos θ + 1
sin θ (1 + cos θ )
2 + 2 cos θ
=
sin θ (1 + cos θ )
2(1 + cos θ )
=
sin θ (1 + cos θ )
2
=
sin θ
= 2 csc θ
=
1+
1
− sin θ
sin θ
1 − sin 2 θ
=
sin θ
cos 2 θ
=
sin θ
cos θ
= cos θ ⋅
sin θ
= cos θ cot θ
53. csc θ − sin θ =
1
cosθ
cosθ
cos
θ
49.
=
⋅
1
cosθ − sin θ cosθ − sin θ
cosθ
1
=
sin θ
1−
cosθ
1
=
1 − tan θ
318
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Chapter 3 Review Exercises
( 2sin
1
csc θ
1 + cos θ
54.
= sin θ ⋅
1 − cos θ 1 − cos θ 1 + cos θ
1 1 + cos θ
=
⋅
sin θ 1 − cos 2 θ
1 + cos θ
=
sin θ sin 2 θ
1 + cos θ
=
sin 3 θ
θ − 1)
(
=
2
)(
[ − cos(2θ )]
(
)
−1 cos 2 θ − sin 2 θ ⋅1
cos ( 2θ )
2
=
− cos ( 2θ )
= − cos ( 2θ )
(
)
= −1 2 cos 2 θ − 1
= 1 − 2 cos θ
2
1 + sin θ
= cos θ (1 − sin θ ) ⋅
1 + sin θ
cos θ 1 − sin 2 θ
=
1 + sin θ
cos θ cos 2 θ
=
1 + sin θ
cos3 θ
=
1 + sin θ
(
)
2
1 − sin θ
55.
= cos θ (1 − sin θ )
sec θ
(
(
2
⎡ − 1 − 2sin 2 θ ⎤
⎣
⎦
=
58.
sin 4 θ − cos 4 θ
sin 2 θ − cos 2 θ sin 2 θ + cos 2 θ
2
59.
)
)
1 − cos θ
1 − cos θ
56.
= sin θ
1 + cos θ 1 + cos θ
sin θ
csc θ − cot θ csc θ − cot θ
=
⋅
csc θ + cot θ csc θ − cot θ
(csc θ − cot θ ) 2
=
csc2 θ − cot 2 θ
(csc θ − cot θ ) 2
=
1
= (csc θ − cot θ )2
60.
61.
cos θ sin θ
−
sin θ cos θ
cos 2 θ − sin 2 θ
=
sin θ cos θ
1 − sin 2 θ − sin 2 θ
=
sin θ cos θ
1 − 2sin 2 θ
=
sin θ cos θ
62.
57. cot θ − tan θ =
cos(α + β ) cos α cos β − sin α sin β
=
cos α sin β
cos α sin β
cos α cos β sin α sin β
=
−
cos α sin β cos α sin β
cos β sin α
=
−
sin β cos α
= cot β − tan α
sin(α − β ) sin α cos β − cos α sin β
=
sin α cos β
sin α cos β
sin α cos β cos α sin β
=
−
sin α cos β sin α cos β
= 1 − cot α tan β
cos(α − β ) cos α cos β + sin α sin β
=
cos α cos β
cos α cos β
cos α cos β sin α sin β
=
+
cos α cos β cos α cos β
= 1 + tan α tan β
cos(α + β ) cos α cos β − sin α sin β
=
sin α cos β
sin α cos β
cos α cos β sin α sin β
=
−
sin α cos β sin α cos β
cos α sin β
=
−
sin α cos β
= cot α − tan β
63. (1 + cos θ ) tan
64. sin θ tan
θ
2
θ
2
= (1 + cos θ ) ⋅
= sin θ ⋅
sin θ
= sin θ
1 + cos θ
1 − cos θ
= 1 − cos θ
sin θ
319
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
)
Chapter 3: Analytic Trigonometry
65. 2 cot θ cot ( 2θ ) = 2 ⋅
=
cos θ cos ( 2θ )
⋅
sin θ sin ( 2θ )
70.
(
2 cos θ cos 2 θ − sin 2 θ
sin θ ( 2sin θ cos θ )
cos 2 θ − sin 2 θ
=
sin 2 θ
2
cos θ sin 2 θ
=
−
sin 2 θ sin 2 θ
2
= cot θ − 1
(
)
)
68.
sin ( 2θ )
=−
= sin ( 2 ⋅ 2θ )
= sin ( 4θ )
sin ( 3θ ) cos θ − sin θ cos ( 3θ )
=
=
tan ( 3θ )
+
sin ( 2θ ) − sin ( 4θ )
tan θ
2θ + 4θ ⎞
⎛ 2θ − 4θ ⎞
2sin ⎛⎜
⎟ cos ⎜
⎟
2
⎝
⎠ ⎝ 2 ⎠ + tan ( 3θ )
=
2θ − 4θ ⎞
tan θ
⎛ 2θ + 4θ ⎞
2sin ⎛⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
2sin ( 3θ ) cos ( −θ ) tan ( 3θ )
=
+
2sin ( −θ ) cos ( 3θ )
tan θ
= tan ( 3θ ) cot ( − θ ) +
66. 2sin ( 2θ ) 1 − 2sin 2 θ = 2sin ( 2θ ) cos ( 2θ )
67. 1 − 8sin 2 θ cos 2 θ = 1 − 2 ( 2sin θ cos θ )
= 1 − 2sin 2 ( 2θ )
= cos ( 2 ⋅ 2θ )
= cos ( 4θ )
sin ( 2θ ) + sin ( 4θ )
=0
71.
2
sin ( 3θ − θ )
sin ( 2θ )
sin ( 2θ )
72.
sin ( 2θ )
=1
2θ + 4θ ⎞
⎛ 2θ − 4θ ⎞
2sin ⎛⎜
⎟ cos ⎜
⎟
sin ( 2θ ) + sin ( 4θ )
2
⎝
⎠
⎝ 2 ⎠
=
69.
cos ( 2θ ) + cos ( 4θ ) 2cos ⎛ 2θ + 4θ ⎞ cos ⎛ 2θ + 4θ ⎞
⎜
⎟ ⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎠
2sin ( 3θ ) cos ( −θ )
=
2cos ( 3θ ) cos ( −θ )
sin ( 3θ )
=
cos ( 3θ )
= tan ( 3θ )
tan ( 3θ )
tan θ
+
tan ( 3θ )
tan ( 3θ )
tan θ
tan θ
cos(2θ ) − cos(4θ )
− tan θ tan(3θ )
cos(2θ ) + cos(4θ )
− 2sin(3θ ) sin(− θ )
=
− tan θ tan(3θ )
2 cos(3θ ) cos(− θ )
2sin(3θ )sin θ
=
− tan θ tan(3θ )
2 cos(3θ ) cos θ
= tan(3θ ) tan θ − tan θ tan(3θ )
=0
cos(2θ ) − cos(10θ )
⎛ 2θ + 10θ ⎞ ⎛ 2θ − 10θ ⎞
= − 2sin ⎜
⎟ sin ⎜
⎟
2
2
⎝
⎠ ⎝
⎠
= − 2sin(6θ ) sin(− 4θ )
= 2sin(6θ ) sin(4θ )
=
=
sin ( 4θ )
cos ( 4θ )
sin ( 4θ )
cos ( 4θ )
⋅ 2sin ( 6θ ) cos ( 4θ )
1
⋅ 2 ⋅ ⎡⎣sin ( 6θ + 4θ ) sin ( 6θ − 4θ ) ⎤⎦
2
= tan ( 4θ ) ⎡⎣sin (10θ ) + sin ( 2θ ) ⎤⎦
= tan ( 4θ ) ⎡⎣sin ( 2θ ) + sin (10θ ) ⎤⎦
73. sin165º = sin (120º + 45º )
= sin120º ⋅ cos 45º + cos120º ⋅ sin 45º
⎛ 3⎞ ⎛ 2⎞ ⎛ 1⎞ ⎛ 2⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟ + ⎜ − ⎟ ⋅ ⎜⎜
⎟⎟
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2 ⎠
6
2
=
−
4
4
1
6− 2
=
4
(
)
320
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Chapter 3 Review Exercises
74. tan105º = tan ( 60º + 45º )
tan 60º + tan 45º
=
1 − tan 60º tan 45º
3 +1
=
1 − 3 ⋅1
3 +1 1+ 3
=
⋅
1− 3 1+ 3
π
⎛π⎞
1 − cos
1−
⎜4⎟
π
4 =
79. tan = tan ⎜ ⎟ =
π
8
⎝2⎠
1 + cos
1+
4
=
=
1+ 2 3 + 3
1− 3
4+2 3
=
−2
= −2− 3
=
75. cos
=
2+ 2
2− 2 2− 2
⋅
2+ 2 2− 2
(2 − 2 )
2
⎛
2⎞
5π
⎛ 5π ⎞
1 − ⎜⎜ −
⎟⎟
1 − cos
2
⎜
⎟
5π
⎝
⎠
4 =
= sin ⎜ 4 ⎟ =
80. sin
⎝ 2 ⎠
2
8
2
)
⎛ π⎞
⎛ 2π 3π ⎞
− ⎟
76. sin ⎜ − ⎟ = sin ⎜
⎝ 12 ⎠
⎝ 12 12 ⎠
π
π
π
π
= sin ⋅ cos − cos ⋅ sin
6
4
6
4
1 2
3 2
= ⋅
−
⋅
2 2
2 2
2
6
=
−
4
4
1
=
2− 6
4
(
2− 2
4
2− 2 2
=
⋅
2
2
2 2 −2
=
2
= 2 −1
5π
⎛ 3π 2π ⎞
= cos ⎜ +
⎟
12
⎝ 12 12 ⎠
π
π
π
π
= cos ⋅ cos − sin ⋅ sin
4
6
4
6
2 3
2 1
=
⋅
−
⋅
2 2
2 2
6
2
=
−
4
4
1
=
6− 2
4
(
2
2
2
2
4
π
81. sin α = , 0 < α < ;
5
2
sin β =
=
2+ 2
4
=
2+ 2
2
5 π
, <β <π
13 2
For α : x 2 + y 2 = r 2
x 2 + 4 2 = 52
x 2 + 16 = 25
)
x2 = 9
x = 3 (since α is in quadrant I)
77. cos80º ⋅ cos 20º + sin 80º ⋅ sin 20º = cos ( 80º − 20º )
= cos 60º
1
=
2
For β : x 2 + y 2 = r 2
x 2 + 52 = 132
x 2 + 25 = 169
x 2 = 144
78. sin 70º ⋅ cos 40º − cos 70º ⋅ sin 40º = sin ( 70º − 40º )
= sin 30º
1
=
2
x = −12 (since β is in quadrant II)
3
4
12
5
cos α = , tan α = , cos β = − , tan β = − ,
5
3
13
12
α π π β π
0< < ,
< <
2 4 4 2 2
321
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Chapter 3: Analytic Trigonometry
a.
b.
c.
sin(α + β ) = sin α cos β + cos α sin β
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞
= ⎜ ⎟⋅⎜ − ⎟ + ⎜ ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
− 48 + 15
=
65
33
=−
65
h.
e.
f.
cos(2 β ) = cos 2 β − sin 2 β
2
16 + y 2 = 25
y2 = 9
y = 3 (since α is in quadrant I)
2
For β : x + y 2 = r 2
52 + y 2 = 132
25 + y 2 = 169
y 2 = 144
y = −12 (since β is in quad. IV)
3
3
12
12
sin α = , tan α = , sin β = − , tan β = − ,
5
4
13
5
α π
π β
0< < , − < <0
2 4
4 2
a. sin(α + β ) = sin α cos β + cos α sin β
⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 12 ⎞
= ⎜ ⎟⋅⎜ ⎟ + ⎜ ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
15 − 48
=
65
33
=−
65
b.
cos(α + β ) = cos α cos β − sin α sin β
⎛ 4 ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞ ⎛ 12 ⎞
= ⎜ ⎟⋅⎜ ⎟ − ⎜ ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
20 + 36
=
65
56
=
65
c.
sin(α − β ) = sin α cos β − cos α sin β
3 5 4 −12
= ⋅ − ⋅
5 13 5 13
15 + 48
=
65
63
=
65
2
⎛ 12 ⎞ ⎛ 5 ⎞ 144 25 119
= ⎜− ⎟ −⎜ ⎟ =
−
=
⎝ 13 ⎠ ⎝ 13 ⎠ 169 169 169
g.
sin
β
4
2
2 5
=
=
5
5
5
4 2 + y 2 = 52
tan α + tan β
1 − tan α tan β
4 ⎛ 5⎞
+⎜− ⎟
3 ⎝ 12 ⎠
=
⎛4⎞ ⎛ 5 ⎞
1− ⎜ ⎟ ⋅⎜ − ⎟
⎝ 3 ⎠ ⎝ 12 ⎠
11
11 9 33
12
=
= ⋅ =
14 12 14 56
9
4 3 24
sin(2α ) = 2sin α cos α = 2 ⋅ ⋅ =
5 5 25
1 + cos α
2
2
3
8
1+
5
=
= 5 =
2
2
=
For α : x 2 + y 2 = r 2
sin(α − β ) = sin α cos β − cos α sin β
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞
= ⎜ ⎟⋅⎜ − ⎟ − ⎜ ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
− 48 − 15
=
65
63
=−
65
tan(α + β ) =
α
4
π
5
π
82. cos α = , 0 < α < ; cos β = , − < β < 0
5
2
13 2
cos(α + β ) = cos α cos β − sin α sin β
⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞
= ⎜ ⎟⋅⎜ − ⎟ − ⎜ ⎟ ⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
−36 − 20
=
65
56
=−
65
d.
cos
1 − cos β
2
2
⎛ 12 ⎞
1− ⎜ − ⎟
⎝ 13 ⎠
=
2
25
25
5
5 26
13
=
=
=
=
2
26
26
26
=
322
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Chapter 3 Review Exercises
d.
tan(α + β ) =
tan α + tan β
1 − tan α tan β
For β :
122 + y 2 = 132
3 ⎛ 12 ⎞
+⎜− ⎟
4 ⎝ 5⎠
=
⎛ 3 ⎞ ⎛ 12 ⎞
1− ⎜ ⎟ ⋅⎜ − ⎟
⎝4⎠ ⎝ 5 ⎠
33
−
20
= 20 ⋅
56 20
20
33
=−
56
e.
⎛ 3 ⎞ ⎛ 4 ⎞ 24
sin(2α ) = 2sin α cos α = 2 ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ =
⎝ 5 ⎠ ⎝ 5 ⎠ 25
f.
cos(2 β ) = cos 2 β − sin 2 β
2
144 + y 2 = 169
y 2 = 25
y = −5 (since β is in quad. IV)
4
3
5
5
cos α = − , tan α = , sin β = − , tan β = − ,
5
4
13
12
π α 3π 3π β
< <
< <π
,
2 2
4
4
2
a. sin(α + β ) = sin α cos β + cos α sin β
⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞
= ⎜ − ⎟⋅⎜ ⎟ + ⎜ − ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
−36 + 20
=
65
16
=−
65
2
25 144
119
⎛ 5 ⎞ ⎛ 12 ⎞
= ⎜ ⎟ −⎜− ⎟ =
−
=−
169
⎝ 13 ⎠ ⎝ 13 ⎠ 169 169
g.
h.
sin
cos
β
1 − cos β
2
2
5
1−
13
=−
2
8
4
2
2 13
13
=−
=−
=−
=−
2
13
13
13
α
2
b.
cos(α + β ) = cos α cos β − sin α sin β
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞
= ⎜ − ⎟⋅⎜ ⎟ − ⎜ − ⎟ ⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
− 48 − 15
=
65
63
=−
65
c.
sin(α − β ) = sin α cos β − cos α sin β
⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞
= ⎜ − ⎟ ⋅⎜ ⎟ − ⎜ − ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
−36 − 20
=
65
56
=−
65
d.
tan(α + β ) =
=−
1 + cos α
2
4
9
1+
5 = 5 = 9 = 3 = 3 10
=
2
2
10
10
10
=
3
3π
12 3π
83. sin α = − , π < α < ; cos β = ,
< β < 2π
5
2
13 2
For α :
x2 + y2 = r 2
tan α + tan β
1 − tan α tan β
3 ⎛ 5⎞
1
+⎜− ⎟
1 16 16
4 ⎝ 12 ⎠
=
= 3 = ⋅ =
21
3⎛ 5 ⎞
3 21 63
1− ⎜ − ⎟
4 ⎝ 12 ⎠ 16
x2 + y2 = r 2
x 2 + ( −3) = 52
2
e.
x 2 + 9 = 25
x 2 = 16
x = −4 (since α is in quad. III)
sin(2α ) = 2sin α cos α
⎛ 3 ⎞ ⎛ 4 ⎞ 24
= 2⋅⎜ − ⎟⋅⎜ − ⎟ =
⎝ 5 ⎠ ⎝ 5 ⎠ 25
323
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Chapter 3: Analytic Trigonometry
f.
cos(2 β ) = cos 2 β − sin 2 β
2
b.
cos(α + β ) = cos α cos β − sin α sin β
⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 12 ⎞
= ⎜ ⎟ ⋅⎜ − ⎟ − ⎜ − ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
−15 48
=
+
65 65
33
=
65
c.
sin(α − β ) = sin α cos β − cos α sin β
⎛ 4 ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞ ⎛ 12 ⎞
= ⎜ − ⎟⋅⎜ − ⎟ − ⎜ ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
20 36
=
−
65 65
16
=−
65
d.
tan(α + β ) =
2
⎛ 12 ⎞ ⎛ 5 ⎞
= ⎜ ⎟ −⎜− ⎟
⎝ 13 ⎠ ⎝ 13 ⎠
144 25 119
=
−
=
169 169 169
g.
h.
sin
cos
β
1 − cos β
2
2
12
1
1−
13
13
=
=
=
2
2
α
2
=
1
1
26
=
=
26
26
26
1 + cos α
2
⎛ 4⎞
1+ ⎜ − ⎟
⎝ 5⎠
=−
2
1
1
1
10
=− 5 =−
=−
=−
2
10
10
10
=−
4 ⎛ 12 ⎞
− +⎜− ⎟
3 ⎝ 5⎠
=
⎛ 4 ⎞⎛ 12 ⎞
1 − ⎜ − ⎟⎜ − ⎟
⎝ 3 ⎠⎝ 5 ⎠
56
−
⎛ 56 ⎞ ⎛ 15 ⎞ 56
15
=
= −
−
=
33 ⎜⎝ 15 ⎟⎠ ⎜⎝ 33 ⎟⎠ 33
−
15
4 π
5 π
84. sin α = − , − < α < 0; cos β = − , < β < π
5 2
13 2
For α :
x2 + y2 = r 2
x 2 + ( −4 ) = 52
2
e.
x 2 + 16 = 25
x2 = 9
2
x + y2 = r 2
( −5)
2
24
⎛ 4 ⎞⎛ 3 ⎞
sin(2α ) = 2sin α cos α = 2 ⎜ − ⎟⎜ ⎟ = −
5
5
25
⎝
⎠⎝ ⎠
2
f.
x = 3 (since α is in quad. IV)
For β :
tan α + tan β
1 − tan α tan β
+ y 2 = 132
25 + y 2 = 169
y 2 = 144
g.
y = 12 (since β is in quad. II)
3
4
12
12
cos α = , tan α = − , sin β = , tan β = − ,
5
3
13
5
π α
π β π
− < < 0,
< <
4 2
4 2 2
a. sin(α + β ) = sin α cos β + cos α sin β
⎛ 4 ⎞ ⎛ 5 ⎞ 3 12
= ⎜ − ⎟⋅⎜ − ⎟ + ⋅
⎝ 5 ⎠ ⎝ 13 ⎠ 5 13
20 36
=
+
65 65
56
=
65
⎛ 5 ⎞ ⎛ 12 ⎞
cos(2 β ) = cos 2 β − sin 2 β = ⎜ − ⎟ − ⎜ ⎟
⎝ 13 ⎠ ⎝ 13 ⎠
25 144
=
−
169 169
119
=−
169
sin
β
2
1 − cos β
2
⎛ 5⎞
1− ⎜ − ⎟
⎝ 13 ⎠
=
2
18
9
3
3 13
= 13 =
=
=
2
13
13
13
=
324
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2
Chapter 3 Review Exercises
h.
cos
α
1 + cos α
2
2
3
1+
5
=
2
8
4
2
2 5
= 5 =
=
=
2
5
5
5
=
c.
sin(α − β ) = sin α cos β − cos α sin β
⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 12 ⎞
= ⎜ − ⎟⋅⎜ ⎟ − ⎜ − ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
15 48
=− +
65 65
33
=
65
d.
tan(α + β ) =
e.
⎛ 3 ⎞ ⎛ 4 ⎞ 24
sin(2α ) = 2sin α cos α = 2 ⎜ − ⎟ ⎜ − ⎟ =
⎝ 5 ⎠ ⎝ 5 ⎠ 25
f.
cos(2 β ) = cos 2 β − sin 2 β
tan α + tan β
1 − tan α tan β
3 12
+
4 5
=
⎛ 3 ⎞ ⎛ 12 ⎞
1− ⎜ ⎟⎜ ⎟
⎝ 4 ⎠⎝ 5 ⎠
63
63 ⎛ 5 ⎞
63
= 20 = ⎜ − ⎟ = −
4 20 ⎝ 4 ⎠
16
−
5
3
3π
12
π
85. tan α = , π < α < ; tan β = , 0 < β <
4
2
5
2
For α :
x2 + y2 = r 2
( − 3 ) + ( −4 )
2
2
= r 2 (α in quad. III)
9 + 16 = r 2
For β :
r 2 = 25
r =5
2
2
2
x +y =r
52 + 122 = r 2 (β in quad I)
25 + 144 = r 2
2
b.
2
25 144
119
⎛ 5 ⎞ ⎛ 12 ⎞
= ⎜ ⎟ −⎜ ⎟ =
−
=−
169
⎝ 13 ⎠ ⎝ 13 ⎠ 169 169
r 2 = 169
r = 13
3
4
12
5
sin α = − , cos α = − , sin β = , cos β = ,
5
5
13
13
π α 3π
β π
< <
, 0< <
2 2
4
2 4
sin(
α
+
β
)
=
sin
α cos β + cos α sin β
a.
⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 12 ⎞
= ⎜ − ⎟⋅⎜ ⎟ + ⎜ − ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
15 48
=− −
65 65
63
=−
65
cos(α + β ) = cos α cos β − sin α sin β
⎛ 4 ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞ ⎛ 12 ⎞
= ⎜ − ⎟⋅⎜ ⎟ − ⎜ − ⎟⋅⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
20 36
=− +
65 65
16
=
65
g.
sin
h.
cos
β
1 − cos β
2
2
5
8
1−
4
2
2 13
13
13
=
=
=
=
=
2
2
13
13
13
α
2
=
1 + cos α
2
⎛ 4⎞
1+ ⎜ − ⎟
⎝ 5⎠
=−
2
1
1
1
10
=− 5 =−
=−
=−
2
10
10
10
=−
325
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Chapter 3: Analytic Trigonometry
4 π
12
3π
86. tan α = − , < α < π; cot β = , π < β <
3 2
5
2
For α :
tan α + tan β
1 − tan α tan β
4 5
− +
3 12
=
⎛ 4⎞ ⎛ 5 ⎞
1− ⎜ − ⎟ ⋅⎜ ⎟
⎝ 3 ⎠ ⎝ 12 ⎠
−11
= 12
56
36
11 36
=− ⋅
12 56
33
=−
56
d.
tan(α + β ) =
e.
24
⎛ 4 ⎞⎛ 3 ⎞
sin(2α ) = 2sin α cos α = 2 ⎜ ⎟ ⎜ − ⎟ = −
25
⎝ 5 ⎠⎝ 5 ⎠
f.
cos(2 β ) = cos 2 β − sin 2 β
x2 + y2 = r 2
( −3 ) + ( 4 )
2
2
= r 2 (α in quad. II)
9 + 16 = r 2
For β :
r 2 = 25
r =5
x2 + y2 = r 2
(−12) 2 + (−5) 2 = r 2 (β in quad III)
144 + 25 = r 2
r 2 = 169
r = 13
4
3
5
12
sin α = , cos α = − , sin β = − , cos β = − ,
5
5
13
13
π α π π β 3π
< < ,
< <
4 2 2 2 2
4
a. sin(α + β ) = sin α cos β + cos α sin β
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞
= ⎜ ⎟ ⋅⎜ − ⎟ + ⎜ − ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
− 48 + 15
=
65
33
=−
65
b.
cos(α + β ) = cos α cos β − sin α sin β
⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞
= ⎜ − ⎟⋅⎜ − ⎟ − ⎜ ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
36 + 20
=
65
56
=
65
c.
sin(α − β ) = sin α cos β − cos α sin β
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞
= ⎜ ⎟⋅⎜ − ⎟ − ⎜ − ⎟⋅⎜ − ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
− 48 − 15
=
65
63
=−
65
2
⎛ 12 ⎞ ⎛ 5 ⎞
= ⎜− ⎟ −⎜− ⎟
⎝ 13 ⎠ ⎝ 13 ⎠
144 25
=
−
169 169
119
=
169
g.
sin
h.
cos
β
2
1 − cos β
2
2
⎛ 12 ⎞
1− ⎜ − ⎟
⎝ 13 ⎠
=
2
25
25
5
5 26
= 13 =
=
=
2
26
26
26
α
=
1 + cos α
2
2
⎛ 3⎞
1+ ⎜ − ⎟
⎝ 5⎠ =
=
2
=
2
5 = 1= 1 = 5
2
5
5
5
326
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Chapter 3 Review Exercises
π
3π
< α < 0; sec β = 3,
< β < 2π
2
2
For α : x 2 + y 2 = r 2
87. sec α = 2, −
2
2
1 +y =2
d.
tan(α + β ) =
=
2
1 + y2 = 4
y =3
y = 3 (α in quad. IV)
−9 3 − 8 2
− 23
8 2 +9 3
=
23
=
2
For β : x + y 2 = r 2
12 + y 2 = 32
2 + y2 = 9
y2 = 8
y = −2 2 (β in quad IV)
3
1
, cos α = , tan α = − 3,
2
2
2 2
1
sin β = −
, cos β = , tan β = − 2 2,
3
3
3π β
π α
− < < 0,
< <π
4 2
4
2
a. sin(α + β ) = sin α cos β + cos α sin β
3 ⎛1⎞ 1⎛ 2 2 ⎞
=−
⎟
⎜ ⎟+ ⎜−
2 ⎝ 3 ⎠ 2 ⎜⎝
3 ⎟⎠
− 3−2 2
=
6
sin α = −
c.
(
)
1 − ( − 3 )( − 2 2 )
− 3 + −2 2
⎛ − 3 − 2 2 ⎞ ⎛ 1+ 2 6 ⎞
= ⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟
⎝ 1− 2 6 ⎠ ⎝ 1+ 2 6 ⎠
2
b.
tan α + tan β
1 − tan α tan β
e.
⎛
3 ⎞⎛ 1 ⎞
3
sin(2α ) = 2sin α cos α = 2 ⎜⎜ −
⎟⎟ ⎜ ⎟ = −
2
⎝ 2 ⎠⎝ 2 ⎠
f.
cos(2 β ) = cos 2 β − sin 2 β
2
2
1 8
7
⎛1⎞ ⎛ 2 2 ⎞
= ⎜ ⎟ − ⎜⎜ −
⎟⎟ = − = −
3 ⎠
9 9
9
⎝3⎠ ⎝
cos(α + β ) = cos α cos β − sin α sin β
1 1 ⎛
3 ⎞⎛ 2 2 ⎞
= ⋅ − ⎜⎜ −
⎟⎜ −
⎟
2 3 ⎝ 2 ⎟⎠ ⎜⎝
3 ⎟⎠
1− 2 6
=
6
g.
sin
h.
cos
β
2
α
1 − cos β
2
1
2
1−
3 = 3 = 1= 1 = 3
=
2
2
3
3
3
=
1 + cos α
2
2
1
3
1+
2 = 2 =
=
2
2
=
3
3
=
4
2
sin(α − β ) = sin α cos β − cos α sin β
3 1 1 ⎛ 2 2⎞
=−
⋅ − ⋅⎜ −
⎟
2 3 2 ⎜⎝
3 ⎟⎠
− 3+2 2
=
6
327
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Chapter 3: Analytic Trigonometry
π
π
< α < π; sec β = −3, < β < π
2
2
2
2
2
For α : x + y = r
88. csc α = 2,
d.
tan α + tan β
1 − tan α tan β
3
−
+ −2 2
3
=
⎛
3⎞
1 − ⎜⎜ −
⎟⎟ − 2 2
⎝ 3 ⎠
tan(α + β ) =
(
x 2 + 12 = 22
(
x2 + 1 = 4
2
x =3
For β : x 2 + y 2 = r 2
12 + y 2 = 32
− 3 −6 2 3+ 2 6
⋅
3− 2 6 3+ 2 6
− 27 3 − 24 2
=
−15
9 3 +8 2
=
5
2 + y2 = 9
=
y2 = 8
y = 2 2 (β in quad II)
1
3
3
, cos α = −
, tan α = −
,
2
2
3
2 2
1
sin β =
, cos β = − , tan β = − 2 2,
3
3
π α π π β π
< < ,
< <
4 2 2 4 2 2
a. sin(α + β ) = sin α cos β + cos α sin β
3⎞ ⎛2 2⎞
⎛ 1⎞ ⎛ 1⎞ ⎛
= ⎜ ⎟ ⋅ ⎜ − ⎟ + ⎜⎜ −
⎟⋅⎜
⎟
⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 3 ⎟⎠
−1 − 2 6
=
6
sin α =
c.
)
− 3 −6 2
3
=
3− 2 6
3
x = − 3 (α in quad. III)
b.
)
e.
sin(2α ) = 2sin α cos α
3⎞
3
⎛ 1 ⎞⎛
= 2 ⎜ ⎟ ⎜⎜ −
⎟⎟ = −
2
2
2
⎝ ⎠⎝
⎠
f.
cos(2 β ) = cos 2 β − sin 2 β
2
⎛ 1⎞ ⎛ 2 2 ⎞
= ⎜ − ⎟ − ⎜⎜
⎟
⎝ 3 ⎠ ⎝ 3 ⎟⎠
1 8
7
= − =−
9 9
9
cos(α + β ) = cos α cos β − sin α sin β
⎛
3 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 2 2 ⎞
= ⎜⎜ −
⎟⎟ ⎜ − ⎟ − ⎜ ⎟ ⎜⎜
⎟⎟
⎝ 2 ⎠⎝ 3 ⎠ ⎝ 2 ⎠⎝ 3 ⎠
3−2 2
=
6
g.
sin(α − β ) = sin α cos β − cos α sin β
1⎛ 1⎞ ⎛
3 ⎞⎛ 2 2 ⎞
= ⎜ − ⎟ − ⎜⎜ −
⎟⎜
⎟
2 ⎝ 3 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 3 ⎟⎠
−1 + 2 6
=
6
h.
sin
cos
β
1 − cos β
2
2
⎛ 1⎞
1− ⎜ − ⎟
⎝ 3⎠
=
2
4
2
= 3 =
=
2
3
α
2
2
=
2
3
⋅
3
3
=
6
3
1 + cos α
2
⎛
3⎞
1 + ⎜⎜ −
⎟
2 ⎟⎠
⎝
=
2
=
=
2− 3
2
=
2
2− 3
=
4
2− 3
2
328
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3 Review Exercises
2
3π
2
3π
89. sin α = − , π < α < ; cos β = − , π < β <
3
2
3
2
For α :
d.
tan(α + β ) =
x2 + y 2 = r 2
2
2
2 5
5
+
2
= 5
2 5 5
1−
⋅
5
2
4 5 +5 5
10
=
1−1
9 5
= 10 ; Undefined
0
2
x + (−2) = 3
x2 + 4 = 9
x2 = 5
x = − 5 (α in quad. III)
2
x + y2 = r 2
For β :
(−2) 2 + y 2 = 32
4 + y2 = 9
y2 = 5
y = − 5 (β in quad III)
5
2 5
5
, tan α =
, sin β = −
,
3
5
3
5 π α 3π π β 3π
tan β =
, < <
, < <
2 2 2
4 2 2
4
a. sin(α + β ) = sin α cos β + cos α sin β
5 ⎞⎛
5⎞
⎛ 2 ⎞⎛ 2 ⎞ ⎛
= ⎜ − ⎟ ⎜ − ⎟ + ⎜⎜ −
⎟⎜ −
⎟
⎟⎜
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠⎝ 3 ⎟⎠
4 5
= +
9 9
=1
e.
sin(2α ) = 2sin α cos α
5⎞ 4 5
⎛ 2 ⎞⎛
= 2 ⎜ − ⎟ ⎜⎜ −
⎟=
9
⎝ 3 ⎠ ⎝ 3 ⎟⎠
f.
cos(2 β ) = cos 2 β − sin 2 β
cos α = −
b.
c.
tan α + tan β
1 − tan α tan β
2
2
5⎞
4 5
1
⎛ 2⎞ ⎛
= ⎜ − ⎟ − ⎜⎜ −
⎟⎟ = − = −
3
3
9
9
9
⎝
⎠ ⎝
⎠
g.
cos(α + β ) = cos α cos β − sin α sin β
⎛
5 ⎞⎛ 2 ⎞ ⎛ 2 ⎞⎛
5⎞
= ⎜⎜ −
⎟ ⎜ − ⎟ − ⎜ − ⎟ ⎜⎜ −
⎟⎟
⎟
⎝ 3 ⎠⎝ 3 ⎠ ⎝ 3 ⎠⎝ 3 ⎠
2 5 2 5
=
−
9
9
=0
h.
sin
β
1 − cos β
2
2
⎛ 2⎞
1− ⎜ − ⎟
⎝ 3⎠ =
=
2
=
5
3 = 5 = 30
2
6
6
⎛
5⎞
1 + ⎜⎜ −
⎟⎟
3
α
1 + cos α
⎝
⎠
cos = −
=−
2
2
2
3− 5
3
=−
2
sin(α − β ) = sin α cos β − cos α sin β
5 ⎞⎛
5⎞
⎛ 2 ⎞⎛ 2 ⎞ ⎛
= ⎜ − ⎟ ⎜ − ⎟ − ⎜⎜ −
⎟⎟ ⎜⎜ −
⎟
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎟⎠
4 5
= −
9 9
1
=−
9
=−
=−
3− 5
6
(
6 3− 5
)
6
6 3− 5
=−
6
329
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
90. tan α = − 2,
For α :
π
π
< α < π; cot β = − 2, < β < π
2
2
2
2
2
x +y =r
tan α + tan β
1 − tan α tan β
⎛ 1⎞
−2+⎜ − ⎟
⎝ 2⎠
=
⎛ 1⎞
1 − (− 2) ⋅ ⎜ − ⎟
⎝ 2⎠
5
−
= 2 ; undefined
0
d.
tan(α + β ) =
e.
4
⎛ 2 ⎞⎛ 1 ⎞
sin(2α ) = 2sin α cos α = 2 ⎜
⎟⎜ −
⎟=−
5
5⎠
⎝ 5 ⎠⎝
f.
cos(2 β ) = cos 2 β − sin 2 β
(−1) 2 + (2) 2 = r 2 (α in quad. II)
1+ 4 = r
r2 = 5
r= 5
For β :
2
2
x + y = r2
(−2)2 + 12 = r 2 (β in quad II)
4 +1 = r2
r2 = 5
r= 5
1
2
1
1
tan β = − , sin α =
, cos α = −
,sin β =
,
2
5
5
5
π α π π β π
2
< < , < <
cos β = −
,
5 4 2 2 4 2 2
a. sin(α + β ) = sin α cos β + cos α sin β
⎛ 2 ⎞⎛ 2 ⎞ ⎛ 1 ⎞⎛ 1 ⎞
=⎜
⎟⎜ −
⎟+⎜−
⎟⎜
⎟
5⎠ ⎝
5 ⎠⎝ 5 ⎠
⎝ 5 ⎠⎝
4 1
=− −
5 5
5
=−
5
= −1
b.
c.
cos(α + β ) = cos α cos β − sin α sin β
⎛ 1 ⎞⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
= ⎜−
⎟⎜ −
⎟−⎜
⎟⎜
⎟
5 ⎠⎝
5 ⎠ ⎝ 5 ⎠⎝ 5 ⎠
⎝
2 2
= −
5 5
=0
2
g.
β
1 − cos β
=
sin =
2
2
=
=
h.
sin(α − β ) = sin α cos β − cos α sin β
⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
=⎜
⎟⋅⎜ −
⎟−⎜−
⎟⋅⎜
⎟
5⎠ ⎝
5⎠ ⎝ 5⎠
⎝ 5⎠ ⎝
4 1
=− +
5 5
3
=−
5
2
⎛ 2 ⎞ ⎛ 1 ⎞
4 1 3
= ⎜−
⎟ −⎜
⎟ = − =
5 5 5
5⎠ ⎝ 5⎠
⎝
⎛ 2 ⎞
1− ⎜ −
⎟
5⎠
⎝
2
5+2
5
2
5+2
2 5
=
5+2 5
10
=
10 5 + 2 5
10
α
1 + cos α
=
cos =
2
2
=
=
⎛ 1 ⎞
1+ ⎜ −
⎟
5⎠
⎝
2
5 −1
5
2
5 −1
2 5
=
5− 5
10
=
10 5 − 5
10
330
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3 Review Exercises
3
1⎞
⎛
91. cos ⎜ sin −1 − cos −1 ⎟
5
2⎠
⎝
3
1
Let α = sin −1 and β = cos −1 . α is in
5
2
quadrant I; β is in quadrant I. Then sin α =
⎡
3⎤
⎛ 1⎞
93. tan ⎢sin −1 ⎜ − ⎟ − tan −1 ⎥
2
4⎦
⎝
⎠
⎣
3
⎛ 1⎞
Let α = sin −1 ⎜ − ⎟ and β = tan −1 . α is in
4
⎝ 2⎠
quadrant IV; β is in quadrant I. Then,
3
,
5
1
π
3
, 0 ≤ α ≤ , and tan β = ,
2
4
2
π
0<β < .
2
sin α = −
π
1
π
0 ≤ α ≤ , and cos β = , 0 ≤ β ≤ .
2
2
2
cos α = 1 − sin 2 α
cos α = 1 − sin 2 α
2
9
16 4
⎛3⎞
= 1− ⎜ ⎟ = 1−
=
=
5
25
25 5
⎝ ⎠
2
1
⎛ 1⎞
= 1− ⎜ − ⎟ = 1− =
4
⎝ 2⎠
sin β = 1 − cos β
2
2
1
3
3
⎛1⎞
= 1− ⎜ ⎟ = 1− =
=
2
4
4
2
⎝ ⎠
1⎞
⎛ −1 3
cos ⎜ sin
− cos −1 ⎟ = cos (α − β )
5
2⎠
⎝
= cos α cos β + sin α sin β
4 1 3 3
= ⋅ + ⋅
5 2 5 2
4 3 3 4+3 3
= +
=
10 10
10
tan α = −
0≤α ≤
=−
3
3
3
⎡ −1 ⎛ 1 ⎞
⎛ 3 ⎞⎤
tan ⎢sin ⎜ − ⎟ − tan −1 ⎜ ⎟ ⎥ = tan (α − β )
⎝ 2⎠
⎝ 4 ⎠⎦
⎣
tan α − tan β
=
1 + tan α tan β
3 3
−
3
4
=
⎛
⎞
3 ⎛3⎞
1 + ⎜⎜ −
⎟⎟ ⎜⎝ 4 ⎟⎠
3
⎝
⎠
−
5
4⎞
⎛
92. sin ⎜ cos −1 − cos −1 ⎟
13
5⎠
⎝
5
4
and β = cos −1 . α is in
Let α = cos −1
13
5
quadrant I; β is in quadrant I. Then cos α =
1
3
3
=
4
2
−4 3 −9
12
=
3 3
1−
12
−9 − 4 3 12 + 3 3
=
⋅
12 − 3 3 12 + 3 3
5
,
13
4
π
π
, and cos β = , 0 ≤ β ≤ .
5
2
2
−144 − 75 3
117
− 48 − 25 3
=
39
48 + 25 3
=−
39
=
sin α = 1 − cos α
2
2
25
144 12
⎛5⎞
= 1− ⎜ ⎟ = 1−
=
=
13
169
169 13
⎝ ⎠
sin β = 1 − cos 2 β
2
16
9
3
⎛4⎞
= 1− ⎜ ⎟ = 1−
=
=
5
25
25
5
⎝ ⎠
4⎞
⎛ −1 5
sin ⎜ cos
− cos −1 ⎟ = sin (α − β )
13
5⎠
⎝
= sin α cos β − cos α sin β
12 4 5 3
= ⋅ − ⋅
13 5 13 5
48 15 33
=
−
=
65 65 65
331
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
4⎞
⎛
96. cos ⎜ 2 tan −1 ⎟
3⎠
⎝
4
Let α = tan −1 . α is in quadrant I. Then
3
4
π
tan α = , 0 < α < .
3
2
⎡
⎛ 4 ⎞⎤
94. cos ⎢ tan −1 (−1) + cos −1 ⎜ − ⎟ ⎥
⎝ 5 ⎠⎦
⎣
⎛ 4⎞
⎝
⎠
Let α = tan −1 (−1) and β = cos −1 ⎜ − ⎟ . α is in
5
quadrant IV; β is in quadrant II. Then
tan α = −1, −
π
2
4
π
< α < 0 , and cos β = − ,
5
2
sec α = tan 2 α + 1
2
16
⎛4⎞
= ⎜ ⎟ +1 =
+1 =
9
⎝3⎠
3
cos α =
5
4⎞
⎛
cos ⎜ 2 tan −1 ⎟ = cos ( 2α )
3⎠
⎝
= 2 cos 2 α − 1
≤β ≤π.
sec α = 1 + tan 2 α = 1 + (−1) 2 = 2
cos α =
1
2
=
2
2
sin α = − 1 − cos 2 α
2
⎛ 2⎞
1
1
2
= − 1 − ⎜⎜
=−
⎟⎟ = − 1 − = −
2
2
2
2
⎝
⎠
25 5
=
9
3
2
7
⎛3⎞
⎛ 9 ⎞
= 2 ⎜ ⎟ −1 = 2 ⎜ ⎟ −1 = −
25
⎝5⎠
⎝ 25 ⎠
sin β = 1 − cos 2 β
2
97. cos θ =
16
9 3
⎛ 4⎞
= 1− ⎜ − ⎟ = 1−
=
=
25
25 5
⎝ 5⎠
5π
+ 2kπ , k is any integer
3
⎧ π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ,
⎬.
⎩3 3 ⎭
θ=
⎡
⎛ 4 ⎞⎤
cos ⎢ tan −1 (−1) + cos −1 ⎜ − ⎟ ⎥ = cos (α + β )
⎝ 5 ⎠⎦
⎣
= cos α cos β − sin α sin β
⎛ 2 ⎞⎛ 4 ⎞ ⎛
2 ⎞⎛ 3 ⎞
= ⎜⎜
⎟⎟ ⎜ − ⎟ − ⎜⎜ −
⎟⎟ ⎜ ⎟
⎝ 2 ⎠⎝ 5 ⎠ ⎝ 2 ⎠⎝ 5 ⎠
−4 2 3 2
=
+
10
10
2
=−
10
π
1
2
3
+ 2kπ or θ =
98. sin θ = −
3
2
4π
5π
+ 2kπ or θ =
+ 2kπ , k is any integer
3
3
⎧ 4π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ,
⎬.
3 ⎭
⎩ 3
θ=
⎡
⎛ 3 ⎞⎤
95. sin ⎢ 2 cos −1 ⎜ − ⎟ ⎥
⎝ 5 ⎠⎦
⎣
⎛ 3⎞
Let α = cos −1 ⎜ − ⎟ . α is in quadrant II. Then
⎝ 5⎠
3 π
cos α = − , ≤ α ≤ π .
5 2
99. 2 cos θ + 2 = 0
2 cos θ = − 2
cos θ = −
2
2
3π
5π
+ 2kπ or θ =
+ 2kπ , k is any integer
4
4
⎧ 3π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ,
⎬.
4 ⎭
⎩ 4
sin α = 1 − cos 2 α
θ=
2
9
16 4
⎛ 3⎞
= 1− ⎜ − ⎟ = 1−
=
=
5
25
25 5
⎝
⎠
⎡
⎛ 3 ⎞⎤
sin ⎢ 2cos −1 ⎜ − ⎟ ⎥ = sin 2α
⎝ 5 ⎠⎦
⎣
= 2sin α cos α
24
⎛ 4 ⎞⎛ 3 ⎞
= 2 ⎜ ⎟⎜ − ⎟ = −
5
5
25
⎝ ⎠⎝
⎠
332
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Chapter 3 Review Exercises
100. tan θ + 3 = 0
105. sec 2 θ = 4
sec θ = ±2
1
cos θ = ±
2
tan θ = − 3
2π
θ=
+ k π , k is any integer
3
On the interval 0 ≤ θ < 2π , the solution set is
⎧ 2π 5π ⎫
⎨ ,
⎬.
3 ⎭
⎩ 3
θ=
π
+k π
or
θ=
2π
+kπ ,
3
3
where k is any integer
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 2π 4π 5π ⎫
,
, ⎬.
⎨ ,
⎩3 3 3 3 ⎭
101. sin(2θ ) + 1 = 0
sin(2θ ) = −1
3π
+ 2kπ
2
3π
θ=
+ kπ , k is any integer
4
On the interval 0 ≤ θ < 2π , the solution set is
⎧ 3π 7π ⎫
⎨ ,
⎬
4 ⎭
⎩ 4
2θ =
106. csc 2 θ = 1
csc θ = ±1
sin θ = ±1
θ=
π
+kπ , where k is any integer
2
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 3π ⎫
⎨ , ⎬.
⎩2 2 ⎭
102. cos ( 2θ ) = 0
π
3π
+ 2k π or 2θ =
+ 2k π
2
2
π
3π
θ = + kπ
θ = + k π,
4
4
where k is any integer
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 3π 5π 7π ⎫
, ⎬.
⎨ , ,
⎩4 4 4 4 ⎭
2θ =
sin θ = tan θ
107.
sin θ
cos θ
sin θ cos θ = sin θ
sin θ =
sin θ cos θ − sin θ = 0
sin θ (cos θ − 1) = 0
cos θ − 1 = 0 or sin θ = 0
cos θ = 1
θ =0
θ =π
On the interval 0 ≤ θ < 2π , the solution set is
{0, π } .
103. tan ( 2θ ) = 0
2θ = 0 + k π
kπ
, where k is any integer
2
On the interval 0 ≤ θ < 2π , the solution set is
3π ⎫
⎧ π
⎨0, , π ,
⎬.
2 ⎭
⎩ 2
θ=
108.
cos θ = sec θ
1
cos θ
cos 2 θ = 1
cos θ =
104. sin ( 3θ ) = 1
cos θ = ±1
π
3θ = + 2k π
2
π 2k π
, where k is any integer
θ= +
6
3
On the interval 0 ≤ θ < 2π , the solution set is
⎧ π 5π 3π ⎫
, ⎬.
⎨ ,
⎩6 6 2 ⎭
θ = 0, π
On the interval 0 ≤ θ < 2π , the solution set is
{0, π } .
333
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Chapter 3: Analytic Trigonometry
109.
sin θ + sin(2θ ) = 0
113.
sin θ + 2sin θ cos θ = 0
sin θ (1 + 2 cos θ ) = 0
1 + 2 cos θ = 0
2sin θ − 1 = 0
or sin θ = 0
sin θ =
θ = 0, π
1
2
2π 4 π
θ= ,
3 3
cos θ = −
θ=
114.
(2sin θ − 1)(sin θ + 1) = 0
θ=
sin θ + 1 = 0
sin θ = −1
1
sin θ =
2
θ=
π 5π
6
,
3π
2
6
4sin 2 θ = 1 + 4 cos θ
2 cos θ − 1 = 0
cos θ =
θ=
θ =0
3
2
(not possible)
cos θ = −
π 5π
3
,
3
116. 8 − 12sin 2 θ = 4 cos 2 θ
8 − 12sin 2 θ = 4 1 − sin 2 θ
(
θ=
)
8 − 12sin θ = 4 − 4sin θ
2
2
4 = 8sin 2 θ
4 1
sin 2 θ = =
8 2
or sin θ = 1
3
1
2
⎧ π 5π ⎫
⎬.
⎩3 3 ⎭
(2 cos θ − 1)(sin θ − 1) = 0
3
or 2 cos θ + 3 = 0
On 0 ≤ θ < 2π , the solution set is ⎨ ,
sin θ (2 cos θ − 1) − 1(2 cos θ − 1) = 0
,
)
( 2 cos θ − 1)( 2 cos θ + 3) = 0
2sin θ cos θ − sin θ − 2 cos θ + 1 = 0
π 5π
5π ⎫
⎬.
3 ⎭
4 cos θ + 4 cos θ − 3 = 0
sin(2θ ) − sin θ − 2 cos θ + 1 = 0
θ=
3
4 − 4 cos θ = 1 + 4 cos θ
6
1
cos θ =
2
,
2
⎧ π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨0, , ⎬ .
⎩ 6 6 ⎭
112.
3
2
or cos θ = 1
,
θ =π
π 5π
(
(2sin θ − 1)(cos θ − 1) = 0
π 5π
cos θ = −1
1
2
4 1 − cos 2 θ = 1 + 4 cos θ
cos θ (2sin θ − 1) − 1(2sin θ − 1) = 0
θ=
or cos θ + 1 = 0
115.
6
2sin θ cos θ − cos θ − 2sin θ + 1 = 0
1
2
2
6
⎧π
⎩3
sin(2θ ) − cos θ − 2sin θ + 1 = 0
sin θ =
,
On 0 ≤ θ < 2π , the solution set is ⎨ , π ,
⎧ π 5π 3π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , , ⎬ .
⎩6 6 2 ⎭
111.
6
π
2 cos 2 θ + cos θ − 1 = 0
(2 cos θ − 1)(cos θ + 1) = 0
cos θ =
2sin 2 θ + sin θ − 1 = 0
θ=
θ=
π 5π
2 cos θ − 1 = 0
1 − 2sin 2 θ = sin θ
or
sin θ = 1
1
2
⎧ π π 5π ⎫
⎬.
⎩6 2 6 ⎭
cos ( 2θ ) = sin θ
2sin θ − 1 = 0
sin θ − 1 = 0
or
On 0 ≤ θ < 2π , the solution set is ⎨ , ,
4π ⎫
⎧ 2π
,π , ⎬ .
On 0 ≤ θ < 2π , the solution set is ⎨0,
3 ⎭
⎩ 3
110.
2sin 2 θ − 3sin θ + 1 = 0
(2sin θ − 1)(sin θ − 1) = 0
1
2
=±
2
2
π 3π 5π 7π
θ= , , ,
4 4 4 4
sin θ = ±
π
2
⎧ π π 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , , ⎬ .
⎩3 2 3 ⎭
⎧ π 3π 5π 7π ⎫
, , ⎬.
⎩4 4 4 4 ⎭
On 0 ≤ θ < 2π , the solution set is ⎨ ,
334
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Chapter 3 Review Exercises
sin ( 2θ ) = 2 cos θ
117.
120. sin θ − 3 cos θ = 2
Divide each side by
2sin θ cos θ = 2 cos θ
1
3
sin θ −
cos θ = 1
2
2
Rewrite using the difference of two angles
π
1
formula where φ = , so cos φ = and
3
2
2sin θ cos θ − 2 cos θ = 0
(
)
cos θ 2sin θ − 2 = 0
cos θ = 0
θ=
or 2sin θ − 2 = 0
π 3π
,
2 2
sin θ =
θ=
2
2
3
.
2
sin θ cos φ − cos θ sin φ = 1
sin (θ − φ ) = 1
sin φ =
π 3π
,
4 4
⎧ π π 3π 3π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , ,
⎩4 2 4
118.
,
⎬.
2 ⎭
π
2
π π
θ− =
3 2
5π
θ=
6
θ −φ =
1 + 3 cos θ + cos ( 2θ ) = 0
1 + 3 cos θ + 2 cos 2 (θ ) − 1 = 0
2 cos 2 (θ ) + 3 cos θ = 0
(
⎧ 5π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ ⎬ .
⎩6⎭
)
cos θ 2 cos θ + 3 = 0
cos θ = 0
or 2 cos θ + 3 = 0
121. sin −1 ( 0.7 ) ≈ 0.78
π 3π
3
cos θ = −
2 2
2
5π 7π
θ= ,
6 6
⎧ π 5π 7π 3π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , , , ⎬ .
θ=
,
⎩2 6
6
2 ⎭
⎛4⎞
122. cos −1 ⎜ ⎟ ≈ 0.64
⎝5⎠
119. sin θ − cos θ = 1
Divide each side by 2 :
1
1
1
sin θ −
cos θ =
2
2
2
Rewrite in the difference of two angles form
1
1
π
, sin φ =
, and φ = :
where cos φ =
4
2
2
1
sin θ cos φ − cos θ sin φ =
2
sin(θ − φ ) =
2:
123. tan −1 ( −2 ) ≈ −1.11
2
2
124. cos −1 ( −0.2 ) ≈ 1.77
π
3π
or θ − φ =
θ −φ =
4
4
π π
π 3π
θ − = or θ − =
4 4
4 4
π
θ=
or θ = π
2
⎧π ⎫
On 0 ≤ θ < 2π , the solution set is ⎨ , π ⎬ .
⎩2
⎭
335
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Chapter 3: Analytic Trigonometry
128. 2 x = 5sin x
Find the intersection of Y1 = 2 x and Y2 = 5sin x :
⎛1⎞
125. sec−1 ( 3) = cos −1 ⎜ ⎟
⎝3⎠
We seek the angle θ , 0 ≤ θ ≤ π , whose cosine
6
1
1
equals . Now cos θ = , so θ lies in
3
3
1
quadrant I. The calculator yields cos −1 ≈ 1.23 ,
3
which is an angle in quadrant I, so
sec−1 ( 3) ≈ 1.23 .
6
2π −2π
−2π
−6
2π
−6
On the interval 0 ≤ x ≤ 2π , x ≈ 0 or x ≈ 2.13 .
The solution set is {0, 2.13} .
129. 2sin x + 3cos x = 4 x
Find the intersection of Y1 = 2sin x + 3cos x and
Y2 = 4 x :
6
⎛ 1⎞
126. cot −1 ( −4 ) = tan −1 ⎜ − ⎟
⎝ 4⎠
We seek the angle θ , 0 ≤ θ ≤ π , whose tangent
2π
−2π
1
1
equals − . Now tan θ = − , so θ lies in
4
4
quadrant II. The calculator yields
−6
x ≈ 0.87 .
The solution set is {0.87} .
⎛ 1⎞
tan −1 ⎜ − ⎟ ≈ −0.24 , which is an angle in
⎝ 4⎠
quadrant IV. Since θ lies in quadrant II,
θ ≈ −0.24 + π ≈ 2.90 . Therefore,
cot −1 ( −4 ) ≈ 2.90 .
130. 3cos x + x = sin x
Find the intersection of Y1 = 3cos x + x and
Y2 = sin x :
6
6
2π −2π
−2π
127. 2 x = 5cos x
Find the intersection of Y1 = 2 x and
Y2 = 5cos x :
−6
−6
On the interval 0 ≤ x ≤ 2π , x ≈ 1.89 or
x ≈ 3.07 .
The solution set is {1.89,3.07} .
6
2π
−2π
2π
131. sin x = ln x
Find the intersection of Y1 = sin x and Y2 = ln x :
2
−6
x ≈ 1.11
The solution set is {1.11} .
2π
−2π
−2
x ≈ 2.22
The solution set is {2.22} .
336
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Chapter 3 Review Exercises
Using a difference formula:
sin15° = sin(45° − 30°)
= sin(45°) cos(30°) − cos(45°) sin(30°)
132. sin x = e − x
Find the intersection of Y1 = sin x and Y2 = e − x :
6
6
2 3
2 1
⋅
−
⋅
2 2
2 2
6
2
6− 2 1
=
−
=
=
4
4
4
4
=
2π −2π
−2π
−6
2π
−6
(
133. −3sin −1 x = π
π
3
=
−2 cos −1 x = −π
=
π
x = cos
π
=
=0
2
The solution set is {0}.
135. Using a half-angle formula:
⎛ 30° ⎞
sin15° = sin ⎜
⎟
⎝ 2 ⎠
=
1 − cos 30°
2
1−
(
)
3 −1
4
(
−2 cos −1 x + π = 0
2
2
) ⎞⎟
⎛ 2 3 −1
= ⎜
⎜
4
⎝
2 cos −1 x + π = 4 cos −1 x
cos −1 x =
)
)
=
3
⎛ π⎞
x = sin ⎜ − ⎟ = −
3
2
⎝
⎠
⎧⎪ 3 ⎫⎪
The solution set is ⎨−
⎬.
⎩⎪ 2 ⎭⎪
134.
6− 2
Verifying equality:
1
6− 2
6− 2 =
4
4
2⋅ 3− 2
=
4
On the interval 0 ≤ x ≤ 2π , x ≈ 0.59 or x ≈ 3.10 .
The solution set is {0.59,3.10} .
sin −1 x = −
(
2
⎟
⎠
(
)
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 cos θ , the most efficient
Double-angle Formula to use is
cos ( 2θ ) = 2 cos 2 θ − 1 .
3
2 =
2− 3
2− 3
=
2
4
2
Note: since 15º lies in quadrant I, we have
sin15° > 0 .
=
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Chapter 3: Analytic Trigonometry
Chapter 3 Test
7⎞
⎛
4. tan ⎜ tan −1 ⎟ follows the form
3⎠
⎝
⎛ 2 ⎞
1. Let θ = sec−1 ⎜
⎟ . We seek the angle θ , such
⎝ 3⎠
that 0 ≤ θ ≤ π and θ ≠
π
2
(
(
5. cot csc−1 10
2
. The
that − ≤ θ ≤ , whose sine equals −
2
2
2
only value in the restricted range with a sine of
⎛
2
π
2⎞
π
−
is − . Thus, sin −1 ⎜ −
⎟=− .
⎜
⎟
4
4
2
⎝ 2 ⎠
x 2 + 12 =
( 10 )
2
x 2 + 1 = 10
x2 = 9
x=3
θ is in quadrant I.
⎛ 11π ⎞
3. sin −1 ⎜ sin
⎟ follows the form of the equation
5 ⎠
⎝
(
r
π
π
= 10 , − ≤ θ ≤ , let
2
2
y
r = 10 and y = 1 . Solve for x:
π
)
)
)
Since csc −1 θ =
⎛
2⎞
2. Let θ = sin −1 ⎜ −
⎟ . We seek the angle θ , such
2
⎝
⎠
(
(
domain of the inverse tangent is all real numbers,
we can directly apply this equation to get
7⎞ 7
⎛
tan ⎜ tan −1 ⎟ = .
3⎠ 3
⎝
, whose secant equals
2
. The only value in the restricted range with
3
⎛ 2 ⎞ π
2
π
is . Thus, sec−1 ⎜
a secant of
⎟= .
6
3
⎝ 3⎠ 6
π
)
f f −1 ( x ) = tan tan −1 x = x . Since the
(
)
Thus, cot csc −1 10 = cot θ =
)
f −1 f ( x ) = sin −1 sin ( x ) = x , but because
x 3
= = 3.
y 1
⎛ 3⎞
6. Let θ = cos −1 ⎜ − ⎟ .
⎝ 4⎠
11π
⎡ π π⎤
is not in the interval ⎢ − , ⎥ , we cannot
5
⎣ 2 2⎦
directly use the equation.
We need to find an angle θ in the interval
11π
⎡ π π⎤
⎢ − 2 , 2 ⎥ for which sin 5 = sin θ . The angle
⎣
⎦
⎡
⎛ 3 ⎞⎤
sec ⎢cos −1 ⎜ − ⎟ ⎥ = sec θ
⎝ 4 ⎠⎦
⎣
1
=
cos θ
1
⎡ −1 ⎛ 3 ⎞ ⎤
cos ⎢ cos ⎜ − ⎟ ⎥
⎝ 4 ⎠⎦
⎣
1
= 3
−4
4
=−
3
=
11π
π
is in quadrant I and coterminal with , so
5
5
11π
π
π
sin
= sin . Since
is in the interval
5
5
5
⎡ π π⎤
⎢ − 2 , 2 ⎥ , we can apply the equation above and
⎣
⎦
7. sin −1 ( 0.382 ) ≈ 0.392 radian
11π ⎞ π
get sin ⎜ sin
⎟= .
5 ⎠ 5
⎝
−1 ⎛
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Chapter 3 Test
⎛ 1 ⎞
8. sec−1 1.4 = cos −1 ⎜
⎟ ≈ 0.775 radian
⎝ 1.4 ⎠
13. tan θ + cot θ =
sin θ cos θ
+
cos θ sin θ
sin 2 θ
cos 2 θ
+
sin θ cos θ sin θ cos θ
sin 2 θ + cos 2 θ
=
sin θ cos θ
1
=
sin θ cos θ
2
=
2sin θ cos θ
2
=
sin ( 2θ )
=
9. tan −1 3 ≈ 1.249 radians
= 2 csc ( 2θ )
⎛1⎞
10. cot −1 5 = tan −1 ⎜ ⎟ ≈ 0.197 radian
⎝5⎠
14.
sin (α + β )
tan α + tan β
sin α cos β + cos α sin β
sin α sin β
+
cos α cos β
sin α cos β + cos α sin β
=
sin α cos β cos α sin β
+
cos α cos β cos α cos β
sin α cos β + cos α sin β
=
sin α cos β + cos α sin β
cos α cos β
sin α cos β + cos α sin β
=
11.
csc θ + cot θ
sec θ + tan θ
csc θ + cot θ csc θ − cot θ
=
⋅
sec θ + tan θ csc θ − cot θ
csc2 θ − cot 2 θ
=
( sec θ + tan θ )( cscθ − cot θ )
=
=
=
=
1
( sec θ + tan θ )( cscθ − cot θ )
1
( sec θ + tan θ )( cscθ − cot θ )
(sec
⋅
1
cos α cos β
sin α cos β + cos α sin β
= cos α cos β
sec θ − tan θ
sec θ − tan θ
15.
sec θ − tan θ
2
⋅
)
θ − tan 2 θ ( csc θ − cot θ )
sin ( 3θ )
= sin (θ + 2θ )
= sin θ cos ( 2θ ) + cos θ sin ( 2θ )
(
)
= sin θ cos 2 θ − sin 2 θ + cos θ ⋅ 2sin θ cos θ
sec θ − tan θ
=
csc θ − cot θ
= sin θ cos θ − sin θ + 2sin θ cos 2 θ
= 3sin θ cos 2 θ − sin 3 θ
= 3sin θ 1 − sin 2 θ − sin 3 θ
2
3
(
sin θ
+ cos θ
cos θ
2
sin θ cos 2 θ
=
+
cos θ
cos θ
2
sin θ + cos 2 θ
=
cos θ
1
=
cos θ
= sec θ
12. sin θ tan θ + cos θ = sin θ ⋅
)
= 3sin θ − 3sin θ − sin 3 θ
= 3sin θ − 4sin 3 θ
3
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Chapter 3: Analytic Trigonometry
sin θ cos θ
tan θ − cot θ cos θ − sin θ
16.
=
tan θ + cot θ sin θ cos θ
+
cos θ sin θ
3⎞
⎛1
19. sin ⎜ cos −1 ⎟
5⎠
⎝2
3
π
(from the
Let θ = cos −1 . Since 0 < θ <
2
5
sin 2 θ − cos 2 θ
θ cos θ
= sin
2
sin θ + cos 2 θ
sin θ cos θ
sin 2 θ − cos 2 θ
=
sin 2 θ + cos 2 θ
− cos ( 2θ )
=
1
= − 2 cos 2 θ − 1
(
range of cos −1 x ),
1 − cos θ
⎛1 ⎞
sin ⎜ θ ⎟ =
2
⎝2 ⎠
3
1 − cos ⎛⎜ cos −1 ⎞⎟
5⎠
⎝
=
2
1 − 53
=
2
1
=
5
5
=
5
)
= 1 − 2 cos θ
2
17. cos15° = cos ( 45° − 30° )
= cos 45° cos 30° + sin 45° sin 30°
2 3
2 1
=
⋅
+
⋅
2 2
2 2
2
=
3 +1
4
6+ 2
1
=
or
6+ 2
4
4
(
6⎞
⎛
20. tan ⎜ 2sin −1 ⎟
11
⎝
⎠
6
−1 6
Let θ = sin
. Then sin θ =
and θ lies in
11
11
y 6
quadrant I. Since sin θ = = , let y = 6 and
r 11
r = 11 , and solve for x: x 2 + 62 = 112
x 2 = 85
x = 85
)
(
)
18. tan 75° = tan ( 45° + 30° )
tan 45° + tan 30°
1 − tan 45° tan 30°
3
1+
3
=
3
1 − 1⋅
3
3+ 3
=
3− 3
3+ 3 3+ 3
=
⋅
3− 3 3+ 3
9+6 3+3
=
32 − 3
12 + 6 3
=
6
= 2+ 3
=
y
6
6 85
=
=
x
85
85
2 tan θ
tan ( 2θ ) =
1 − tan 2 θ
⎛ 6 85 ⎞
2⎜
85 ⎟⎠
= ⎝
2
⎛ 6 85 ⎞
1− ⎜
⎟
⎝ 85 ⎠
tan θ =
12 85
= 85
36
1−
85
12 85 85
=
⋅
85 49
12 85
=
49
340
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Chapter 3 Test
23.
2
3⎞
⎛
21. cos ⎜ sin −1 + tan −1 ⎟
3
2⎠
⎝
2
3
Let α = sin −1 and β = tan −1 . Then
3
2
2
3
sin α = and tan β = , and both α and β
3
2
y
2
lie in quadrant I. Since sin α = 1 = , let
r1 3
⎛ 75° + 15° ⎞
⎛ 75° − 15° ⎞
= 2sin ⎜
cos ⎜
⎟
⎟
2
2
⎝
⎠
⎝
⎠
⎛ 2 ⎞⎛ 3 ⎞
6
= 2sin ( 45° ) cos ( 30° ) = 2 ⎜
⎟⎜
⎟=
2
2
2
⎝
⎠⎝
⎠
24. cos 65° cos 20° + sin 65° sin 20° = cos ( 65° − 20° )
= cos ( 45° )
y1 = 2 and r1 = 3 . Solve for x1 : x12 + 22 = 32
x12 + 4 = 9
x12 = 5
x1 = 5
Thus, cos α =
x1
5
=
.
r1
3
Since tan β =
y2 3
= , let x2 = 2 and y2 = 3 .
x2 2
=
4sin 2 θ = 3
3
sin 2 θ =
4
3
sin θ = ±
2
On the interval [ 0, 2π ) , the sine function takes
on a value of
sine takes on a value of −
{
5 2
2 3
=
⋅
− ⋅
3
13 3 13
2 13
(
5 −3
)
39
1
⎡sin (α + β ) + sin (α − β ) ⎤⎦ ,
2⎣
1
⎡sin ( 90° ) + sin ( 60° ) ⎤⎦
2⎣
1⎡
3⎤
= ⎢1 +
2⎣
2 ⎥⎦
1
= 2+ 3
4
2+ 3
=
4
sin 75° cos15° =
(
}
⎛π
⎞
26. −3cos ⎜ − θ ⎟ = tan θ
⎝2
⎠
−3sin θ = tan θ
sin θ
+ 3sin θ
0=
cos θ
⎛ 1
⎞
0 = sin θ ⎜
+ 3⎟
θ
cos
⎝
⎠
1
sin θ = 0 or
+3= 0
cos θ
1
cos θ = −
3
On the interval [ 0, 2π ) , the sine function takes on
22. Let α = 75° , β = 15° .
Since sin α cos β =
3
π
2π
when θ =
or θ =
. The
3
2
3
3
4π
when θ =
and
3
2
π 2π 4π 5π
5π
,
,
,
θ=
. The solution set is
.
3
3 3 3 3
y2
3
=
.
r2
13
Therefore, cos (α + β ) = cos α cos β − sin α sin β
Thus, sin β =
2 5 −6
=
3 13
2
2
25. 4sin 2 θ − 3 = 0
Solve for x2 : 22 + 32 = r2 2
4 + 9 = r2 2
r2 2 = 13
r2 = 13
=
sin 75° + sin15°
a value of 0 when θ = 0 or θ = π . The cosine
1
function takes on a value of − in the second and
3
1
third quadrants when θ = π − cos −1 and
3
−1 1
θ = π + cos
. That is θ ≈ 1.911 and θ ≈ 4.373 .
3
The solution set is {0,1.911, π , 4.373} .
)
341
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Chapter 3: Analytic Trigonometry
27.
The second equation has no solution since
−1 ≤ sin θ ≤ 1 for all values of θ .
Therefore, we only need to find values of θ
1
between 0 and 2π such that sin θ = . These will
4
occur in the first and second quadrants. Thus,
1
1
θ = sin −1 ≈ 0.253 and θ = π − sin −1 ≈ 2.889 .
4
4
The solution set is {0.253, 2.889} .
cos 2 θ + 2sin θ cos θ − sin 2 θ = 0
( cos
2
θ − sin 2 θ ) + 2sin θ cos θ = 0
cos ( 2θ ) + sin ( 2θ ) = 0
sin ( 2θ ) = − cos ( 2θ )
tan ( 2θ ) = −1
The tangent function takes on the value −1
3π
+ kπ . Thus, we need
when its argument is
4
3π
2θ =
+ kπ
4
3π
π
θ=
+k
8
2
30. Using the average grade, we approximate the
situation by considering the start of the route as
the center of a circle of radius r = 15 (the length
of the route). Then the end of the route is
located at the point P ( x, y ) on the circle
π
( 3 + 4k )
8
On the interval [ 0, 2π ) , the solution set is
θ=
{π
x 2 + y 2 = 152 = 225 . We have that
y
tan θ = = 0.079
x
θ = tan −1 0.079 ≈ 4.5°
y y
sin θ = =
r 15
y = 15sin θ = 15sin 4.5° ≈ 1.18
The change in elevation during the time trial was
about 1.18 kilometers.
}
3 7π 11π 15π
,
,
,
.
8 8 8
8
28. sin (θ + 1) = cos θ
sin θ cos1 + cos θ sin1 = cos θ
sin θ cos1 + cos θ sin1 cos θ
=
cos θ
cos θ
tan θ cos1 + sin1 = 1
tan θ cos1 = 1 − sin1
1 − sin1
tan θ =
cos1
⎛ 1 − sin1 ⎞
Therefore, θ = tan −1 ⎜
⎟ ≈ 0.285 or
⎝ cos1 ⎠
1 − sin1 ⎞
θ = π + tan −1 ⎛⎜
⎟ ≈ 3.427
⎝ cos1 ⎠
The solution set is {0.285,3.427} .
29. 4sin 2 θ + 7 sin θ = 2
4sin 2 θ + 7 sin θ − 2 = 0
Let u = sin θ . Then,
4u 2 + 7u − 2 = 0
( 4u − 1)( u + 2 ) = 0
4u − 1 = 0 or u + 2 = 0
u = −2
4u = 1
1
u=
4
Substituting back in terms of θ , we have
1
sin θ =
or sin θ = −2
4
342
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Chapter 3 Cumulative Review
Tests for symmetry:
Chapter 3 Cumulative Review
x-axis: Replace y with − y : 3x + ( − y ) = 9
2
1. 3x + x − 1 = 0
2
3x + y 2 = 9
Since we obtain the original equation, the graph
is symmetric with respect to the x-axis.
−b ± b − 4ac
2a
2
x=
−1 ± 12 − 4 ( 3)( −1)
=
y-axis: Replace x with − x : 3 ( − x ) + y 2 = 9
2 ( 3)
−3x + y 2 = 9
Since we do not obtain the original equation, the
graph is not symmetric with respect to the y-axis.
−1 ± 1 + 12
6
−1 ± 13
=
6
=
Origin: Replace x with − x and y with − y :
3( − x ) + ( − y ) = 9
2
⎧ −1 − 13 −1 + 13 ⎫
,
The solution set is ⎨
⎬.
6
6
⎩
⎭
−3x + y 2 = 9
Since we do not obtain the original equation, the
graph is not symmetric with respect to the origin.
2. Line containing points (−2,5) and (4, −1) :
m=
y2 − y1
−1 − 5
−6
=
=
= −1
x2 − x1 4 − ( −2 ) 6
4. y = x − 3 + 2
Using the graph of y = x , shift horizontally to
Using y − y1 = m( x − x1 ) with point (4, −1) ,
the right 3 units and vertically up 2 units.
y − (−1) = −1( x − 4 )
y + 1 = −1( x − 4 )
y +1 = −x + 4
y = − x + 3 or x + y = 3
Distance between points (−2,5) and (4, −1) :
d=
=
( x2 − x1 )
2
+ ( y2 − y1 )
2
( 4 − ( −2 ) ) + ( −1 − 5)2
2
= 62 + ( −6 ) = 72 = 36 ⋅ 2 = 6 2
π⎞
⎛
5. y = cos ⎜ x − ⎟ − 1
2⎠
⎝
Using the graph of y = cos x , horizontally shift
2
Midpoint of segment with endpoints (−2,5) and
(4, −1) :
to the right
⎛ x1 + x2 y1 + y2 ⎞ ⎛ −2 + 4 5 + ( −1) ⎞
⎜ 2 , 2 ⎟ = ⎜ 2 , 2 ⎟ = (1, 2 )
⎝
⎠ ⎝
⎠
π
2
units, and vertically shift down 1
unit.
3. 3x + y 2 = 9
x-intercept: 3x + 02 = 9 ; ( 3, 0 )
3x = 9
x=3
y-intercepts: 3 ( 0 ) + y 2 = 9 ; ( 0, −3) , ( 0,3)
y2 = 9
y = ±3
343
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Chapter 3: Analytic Trigonometry
6. a.
y = x3
y
5
(1, 1)
−5(−1,−1)
(0, 0)
x
5
−5
Inverse function: y = 3 x
c.
y
y = cos x , 0 ≤ x ≤ π
y
5
3
(1, 1)
(0, 0)
−5
x
5
⎛π
_ ⎞
⎜ 2 , 0⎟
⎝
⎠
(0, 1)
(−1,−1)
−1
−1
−5
x
3
(π,−1)
Inverse function: y = cos −1 x
y
(−1, π)
3
π⎞
⎛ _
⎜ 0, 2 ⎟
⎝
⎠
−1
−1
b.
y = sin x , −
y
π
≤x≤
2
2
x
π
2
2
x
−2
1
3π
7. sin θ = − , π < θ <
, so θ lies in Quadrant
3
2
III.
a. In Quadrant III, cos θ < 0
Inverse function: y = sin −1 x
y
2
_⎞
⎛ π
⎜ 1, 2 ⎟
⎠
⎝
(0, 0)
−2
π
⎛
_⎞
⎜ −1, − 2 ⎟
⎝
⎠
3
π ⎞
⎛_
⎜ 2 , 1⎟
⎠
⎝
(0, 0)
−2
π
⎛ _
⎞
⎜ − 2 , −1⎟
⎝
⎠
(1, 0)
2
⎛ 1⎞
cos θ = − 1 − sin 2 θ = − 1 − ⎜ − ⎟
⎝ 3⎠
x
= − 1−
−2
=−
2
1
8
=−
9
9
2 2
3
344
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Chapter 3 Cumulative Review
sin θ
=
tan θ =
cos θ
b.
−
1 π
1
3π
9. sin α = , < α < π ; cos β = − , π < β <
3 2
3
2
π
a. Since < α < π , we know that α lies in
2
Quadrant II and cos α < 0 .
1
3
2 2
3
1⎛
3 ⎞
1
2
= − ⎜−
=
=
3 ⎝ 2 2 ⎟⎠ 2 2
4
−
c.
sin(2θ ) = 2sin θ cos θ
⎛ 1 ⎞⎛ 2 2 ⎞ 4 2
= 2⎜ − ⎟⎜ −
⎟=
3 ⎠
9
⎝ 3 ⎠⎝
d.
cos(2θ ) = cos θ − sin θ
2
cos α = − 1 − sin 2 α
2
1
8
⎛1⎞
= − 1− ⎜ ⎟ = − 1− = −
9
9
⎝3⎠
=−
2
2
⎛ 2 2 ⎞ ⎛ 1 ⎞2 8 1 7
= ⎜−
⎟ −⎜ ⎟ = − =
3 ⎠ ⎝ 3⎠
9 9 9
⎝
e.
Since π < θ <
Thus,
3π
π θ 3π
, we have that < <
.
2
2 2 4
3π
, we know that β lies in
2
Quadrant III and sin β < 0 .
π <β <
sin β = − 1 − cos 2 β
( )
1
1
θ lies in Quadrant II and sin θ > 0 .
2
2
( )
1
1 − cos θ
sin θ =
=
2
2
=
f.
b.
Since
⎛ 2 2⎞
1− ⎜ −
⎟
3 ⎠
⎝
2
⎛ 1⎞
= − 1− ⎜ − ⎟
⎝ 3⎠
= − 1−
c.
3+ 2 2
3+ 2 2
3
=
2
6
1
8
2 2
=−
=−
9
9
3
2
⎛ 2 2 ⎞ ⎛ 1 ⎞2 8 1 7
= ⎜−
= − =
⎟ −
3 ⎠ ⎜⎝ 3 ⎟⎠
9 9 9
⎝
( )
1
1
θ lies in Quadrant II, cos θ < 0 .
2
2
d.
cos(α + β ) = cos α cos β − sin α sin β
=−
( )
=
3− 2 2
3−2 2
3
=−
=−
2
6
(
2
cos(2α ) = cos 2 α − sin 2 α
⎛ 2 2⎞
1+ ⎜ −
⎟
1
1 + cos θ
3 ⎠
⎝
cos θ = −
=−
2
2
2
8. cos tan −1 2
2 2
3
e.
2 2 ⎛ 1⎞ 1⎛ 2 2 ⎞
− − ⎜−
⎟
3 ⎜⎝ 3 ⎟⎠ 3 ⎝
3 ⎠
2 2 2 2 4 2
+
=
9
9
9
Since π < β <
3π
π β 3π
, we have that < <
.
2
2 2
4
β
β
lies in Quadrant II and sin > 0 .
2
2
1
1 − ⎛⎜ − ⎞⎟
β
1 − cos β
⎝ 3⎠
=
sin =
2
2
2
4
4
2
2 6
6
= 3 =
=
=
=
2
6
6
3
6
Thus,
)
Let θ = tan −1 2 . Then tan θ =
y 2
= ,
x 1
π
π
≤ θ ≤ . Let x = 1 and y = 2 .
2
2
Solve for r: r 2 = x 2 + y 2
−
r 2 = 12 + 22
r2 = 5
r= 5
θ is in quadrant I.
(
)
cos tan −1 2 = cos θ =
x
1
1
5
5
=
=
⋅
=
r
5
5
5 5
345
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Chapter 3: Analytic Trigonometry
⎛ 2kx − 2ωt + φ ⎞
⎛φ ⎞
y1 + y2 = 2 ym sin ⎜
⎟ cos ⎜ 2 ⎟
2
⎝
⎠
⎝ ⎠
⎛ 2 ⋅ 69.8 ⋅ 1 − 2 ⋅ 14.45t + 2.5 ⎞ ⎛ 2.5 ⎞
= 2 ⋅ 0.0045sin ⎜
⎟ cos ⎜ 2 ⎟
2
⎝
⎠ ⎝
⎠
⎛ 142.1 − 28.9t ⎞
= 0.009sin ⎜
⎟ cos(1.25)
2
⎝
⎠
= 0.009sin ( 71.05 − 14.45t ) cos(1.25)
Chapter 3 Projects
Project I
a. Amplitude = 0.00421 m
b. ω = 2.68 radians/sec
c.
f =
d. λ =
ω
2π
=
2.68
≈ 0.4265 vibrations/sec
2π
h. Let Y1 = 0.0045sin(69.8 − 14.45 x) ,
Y2 = 0.0045sin(72.3 − 14.45 x) , and
2π
2π
=
≈ 0.09199 m
68.3
k
Y3 = 0.009sin ( 71.05 − 14.45 x ) cos(1.25) .
e. If x = 1 , the resulting equation is
y = 0.00421sin(68.3 − 2.68t ) . To graph, let
Y1 = 0.00421sin(68.3 − 2.68 x) .
0.01
0.4
−0.4
0.005
y2
−0.01
4
−4
y1
y1 + y2
i. ym = 0.0045 , φ = 0.4 , λ = 0.09 , f = 2.3
Let x = 1 :
2π
ω
λ = 0.09 =
f = 2.3 =
k
2π
200π
ω = 4.6π = 14.45
= 69.8
k=
9
y1 = 0.0045sin(69.8 − 14.45t )
y2 = ym sin(kx − ωt + φ )
−0.005
f. Note: (kx − ωt ) + (kx − ωt + φ ) = 2kx − 2ωt + φ and
(kx − ωt ) − (kx − ωt + φ ) = −φ .
y1 + y2 = ym sin(kx − ωt ) + ym sin( kx − ωt + φ )
= ym [sin(kx − ωt ) + sin(kx − ωt + φ )]
⎡
⎛ 2kx − 2 wt + φ ⎞
⎛ −φ ⎞ ⎤
= ym ⎢ 2sin ⎜
cos ⎜ ⎟ ⎥
⎟
2
⎝
⎠
⎝ 2 ⎠⎦
⎣
⎛ 2kx − 2 wt + φ ⎞
⎛φ ⎞
= 2 ym sin ⎜
⎟ cos ⎜ 2 ⎟
2
⎝
⎠
⎝ ⎠
= 0.0045sin(69.8 ⋅1 − 14.45t + 0.4)
= 0.0045sin(70.2 − 14.45t )
⎛ 2kx − 2ωt + φ ⎞
⎛φ ⎞
y1 + y2 = 2 ym sin ⎜
⎟ cos ⎜ 2 ⎟
2
⎝
⎠
⎝ ⎠
⎛ 2 ⋅ 69.8 ⋅ 1 − 2 ⋅ 14.45t + 0.4 ⎞ ⎛ 0.4 ⎞
= 2 ⋅ 0.0045sin ⎜
⎟ cos ⎜ 2 ⎟
2
⎝
⎠ ⎝
⎠
⎛ 140 − 28.9t ⎞
= 0.009sin ⎜
⎟ cos(0.2)
2
⎝
⎠
= 0.009sin ( 70 − 14.45t ) cos(0.2)
g. ym = 0.0045 , φ = 2.5 , λ = 0.09 , f = 2.3
Let x = 1 :
2π
ω
λ = 0.09 =
f = 2.3 =
k
2π
200π
ω = 4.6π ≈ 14.45
≈ 69.8
k=
9
Let Y1 = 0.0045sin(69.8 − 14.45 x) ,
Y2 = 0.0045sin(70.2 − 14.45 x) , and
y1 = ym sin(kx − ωt )
= 0.0045sin(69.8 ⋅1 − 14.45t )
= 0.0045sin(69.8 − 14.45t )
Y3 = 0.009sin ( 70 − 14.45 x ) cos(0.2) .
y2 = ym sin(kx − ωt + φ )
= 0.0045sin(69.8 ⋅1 − 14.45t + 2.5)
= 0.0045sin(72.3 − 14.45t )
346
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3 Projects
0.01
Project III
y1 + y2
y1
a.
y2
0.4
−0.4
f ( x) = sin x (see table column 2)
f ( x)
0
1
2
2
2
3
2
x
0
−0.01
π
6
j. The phase shift causes the amplitude of y1 + y2
to increase from 0.009 cos(1.25) ≈ 0.003 to
0.009 cos(0.2) ≈ 0.009 .
π
4
π
3
π
Project II
y
a.
h
π 2π 3π 4π 5π x
−h
b. Let Y1 = 1 +
3
sin(17 x) ⎞
4 ⎛ sin x sin(3x )
c. Let Y1 = 1 + ⎜
+
+ ... +
3
17 ⎟⎠
π⎝ 1
3
5π
−3
d. Let Y1 = 1 +
3
0
h( x )
k ( x)
−0.311 −0.749
m( x )
6.085
0.791
−0.703 2.437
4.011
0.607
−1.341 1.387
−3.052
0.256
−0.978
0.588
−1.243
−0.256 −0.670 −0.063
0.413
−0.607 −0.703
0.153
8.507
−0.791 −0.623
2.380
−6.822
−0.954
0.594
−2.695
−0.954
−0.791
6
5π
−0.607
4
4π
−0.256
3
3π
−1 0.256
2
5π
3
0.607
−
3
2
7π
2
0.791
−
4
2
11π
1
0.954
−
6
2
2π
1
−3
0
3
2
2
2
1
2
0
1
−
2
2
−
2
3
−
2
π
7π
4 ⎛ sin x sin(3x ) sin(5 x) sin(7 x) ⎞
+
+
+
3
5
7 ⎟⎠
π ⎜⎝ 1
5π
0
1
2
2π
3
3π
4
5π
6
g ( x)
0.954
sin(37 x) ⎞
4 ⎛ sin x sin(3x )
+
+ ... +
3
37 ⎟⎠
π ⎜⎝ 1
b. g ( x) =
−0.117 −0.013 −5.248
1.341
−1.387
0.978
−0.588 1.243
0.670
0.063
0.703
−0.306
3.052
−0.705
0.623
f ( xi +1 ) − f ( xi )
(see table column 3)
xi +1 − xi
−0.5
−3
0.311 −0.817 1.536
1.2
c.
5π
0
6.5
−1.2
e. The best one is the one with the most terms.
347
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3: Analytic Trigonometry
The shape looks like a sinusoidal graph.
f. m( x) =
k ( xi +1 ) − k ( xi )
(see table column 6)
xi +1 − xi
12
Rounding a, b, c, and d to the nearest tenth, we
have that y = sin( x + 1.8) .
Barring error due to rounding and
approximation, this looks like y = cos x
5.5
−0.5
−10
The sinusoidal features are gone.
g ( xi +1 ) − g ( xi )
d. h( x) =
(see table column 4)
xi +1 − xi
1.8
Rounding a, b, c, and d to the nearest tenth, we
have that y = 2.1sin(5.1x − 1.5) + 0.6 .
6.5
−0.5
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.
−1.8
The shape is sinusoidal. It looks like an upsidedown sine wave.
Rounding a, b, c, and d to the nearest tenth, we
have that y = 0.5sin(6.4 x ) .
e. k ( x) =
h ( xi +1 ) − h ( xi )
(see table column 5)
xi +1 − xi
3
6
−0.5
−2
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 y = 0.8sin(1.1x) + 0.3 .
Note: The rounding error is getting greater and
greater.
348
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.