SOLUTION: The half-angle identity states cos
Study Guide and Review - Chapter 5
±
Complete each identity by filling in the blank.
Then name the identity.
1. sec = _______
.
9. =_______
SOLUTION: SOLUTION: = The power-reducing identity states
= The reciprocal identity states sec
2
.
sin
.
10. _______ =
2. _______ =
[cos (
–
) + cos (
+
)]
SOLUTION: SOLUTION: The quotient identity states tan
= The product-to-sum identity states cos α cos β =
.
[cos (α – β) + cos (α + β)].
3. _______ + 1 = sec2
Find the value of each expression using the
given information.
11. sec and cos ; tan = 3, cos > 0
SOLUTION: The Pythagorean identity states sec
4. cos (90° –
=
= .
SOLUTION: ) = _______
Use the Pythagorean Identity that involves tan
find sec .
SOLUTION: The cofunction identity states cos(90° −
) = sin
to .
5. tan (– ) = _______
SOLUTION: The odd-even identity states tan(− ) = −tan
6. sin (
+
) = sin
_______ + cos .
Since we are given that cos
_______
SOLUTION: must be positive. So, sec
is positive, sec = . Use the
to find cos .
reciprocal identity
The sum identity states sin (α + β) = sin α cos β +
cos α sin β.
7. _______ = cos2α – sin2 α
SOLUTION: 2
The double-angle identity states that cos 2α = cos α
2
– sin α.
8. _______= ±
12. cot
SOLUTION: The half-angle identity states cos
±
9. .
=
and sin ; cos
= , tan
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
find sin .
to =_______
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SOLUTION: Page 1
Study Guide and Review - Chapter 5
12. cot
and sin ; cos
= , tan
13. csc
< 0
SOLUTION: = negative, sin
= , sin
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
find sin .
Since tan
and tan ; cos
to is negative and cos is must be positive. So, sin =
Use the quotient identity cot
= Since sin
.
to Use the Pythagorean Identity that involves cos
find sin .
identity csc
< 0, = . Use the reciprocal
to find csc .
to find cot .
Use the quotient identity tan
= to find tan .
13. csc
and tan ; cos
= , sin
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
find sin .
to 14. cot
and cos ; tan
= , csc
> 0
SOLUTION: Use the reciprocal function cot
cot
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= to find .
Page 2
15. sec
and cos ; tan
= , csc
= –2, csc
< 0
SOLUTION: Study Guide and Review - Chapter 5
14. cot
and sin ; cot
Use the Pythagorean identity that involves cot
find csc .
to > 0
SOLUTION: Use the reciprocal function cot
cot
= to find Since csc < 0, csc = −
. Use the reciprocal
identity sin = to find sin .
.
Use the Pythagorean identity that involves tan
find sec .
Since csc
= to
Use the Pythagorean identity that involves sin
find cos .
to is positive, sin is positive. Since tan is positive and sin is positive, cos has to be positive. Since cos is positive, sec is positive. So, sec = identity cos
= . Use the reciprocal
= Since cot
to find cos .
negative, cos
is negative and sin is must be positive. So, cos =
. Use the reciprocal identity sec
find sec
15. sec
and sin ; cot
= –2, csc
=
to
.
< 0
SOLUTION: Use the Pythagorean identity that involves cot
find csc .
to 16. cos
and sin ; cot
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= , sec
< 0
Page 3
SOLUTION: Use the Pythagorean identity that involves cot to find
Study Guide and Review - Chapter 5
It was determined that cos
16. cos
and sin ; cot
= , sec
is negative. So, < 0
SOLUTION: Use the Pythagorean identity that involves cot to find
csc .
Simplify each expression.
17. sin2 (–x) + cos2(–x)
SOLUTION: Since sec < 0, cos is negative. Since cot = is positive and cos is negative, sin is negative. Since sin
is negative, csc must be 18. sin2 x + cos2 x + cot2 x
SOLUTION: negative. So,
identity sin
. Use the reciprocal
= to find sin .
19. SOLUTION: Use the Pythagorean identity that involves sin
find cos .
to 20. SOLUTION: 21. SOLUTION: It was determined that cos
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Simplify each expression.
is negative. So, Page 4
SOLUTION: Study Guide and Review - Chapter 5
Verify each identity.
21. 23. SOLUTION: = 2 csc +
SOLUTION: 24. = 1
22. +
SOLUTION: SOLUTION: 25. = 2 sec +
SOLUTION: 26. =
SOLUTION: Verify each identity.
23. +
= 2 csc SOLUTION: eSolutions Manual - Powered by Cognero
27. = csc
SOLUTION: – 1
Page 5
Study Guide and Review - Chapter 5
27. = csc
31. – 1
=
SOLUTION: SOLUTION: 32. cos4
28. +
= sec + csc 4
– sin
= SOLUTION: SOLUTION: 29. = csc
SOLUTION: Find all solutions of each equation on the
interval [0, 2π].
33. 2 sin x =
SOLUTION: On the interval [0, 2π),
30. cot
2
csc + sec = csc
SOLUTION: sec x=
when x =
and
.
34. 4 cos2 x = 3
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31. On the interval [0, 2 ),
when x =
and6
Page
=
SOLUTION: x=
and when x =
and x =
when x =
On the interval [0, 2π),
and
and x=
x=
.
Study
Guide
and Review - Chapter 5
when x =
and x =
.
34. 4 cos2 x = 3
37. 2 sin2 x = sin x
SOLUTION: SOLUTION: when x =
On the interval [0, 2 ),
and x=
when x =
and
and x =
On the interval [0, 2π), sin x = 0 when x = 0 and
.
and when x =
and x =
.
2
35. tan x – 3 = 0
38. 3 cos x + 3 = sin2 x
SOLUTION: 2
SOLUTION: tan x – 3 = 0
when x =
On the interval [0, 2π),
and x=
when x =
and
and x =
.
36. 9 + cot2 x = 12
SOLUTION: On the interval [0, 2 ), cos x = −1 when x = . The
equation cos x = −2 has no solution since the
minimum value the cosine function can attain is −1.
Solve each equation for all values of x.
when x =
On the interval [0, 2π),
and x=
when x =
and and x =
39. sin2 x – sin x = 0
SOLUTION: .
37. 2 sin2 x = sin x
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Page 7
The period of sine is 2 , so you only need to find
solutions on the interval
. The solutions to sin
x = 0 on this interval are 0 and and the solution to On the interval [0, 2 ), cos x = −1 when x = . The
equation
cos and
x = −2Review
has no solution
since the5
Study
Guide
- Chapter
minimum value the cosine function can attain is −1.
Solve each equation for all values of x.
Solutions on the interval (– , ) are found by
adding integer multiples of π. Therefore, the general
form of the solutions is nπ, + n , where n is an
integer.
41. 3 cos x = cos x – 1
2
39. sin x – sin x = 0
SOLUTION: SOLUTION: The period of cosine is 2 , so you only need to find
solutions on the interval
. The solutions to
on this interval are
and .
The period of sine is 2 , so you only need to find
solutions on the interval
. The solutions to sin
x = 0 on this interval are 0 and and the solution to sin x = 1 on this interval is . Solutions on the interval (– , ) are found by
adding integer multiples of 2π. Therefore, the general
Solutions on the interval (– , ), are found by
adding integer multiples of 2π. The solutions x = 0 +
2n and x = + 2n can be combined to x = n .
Therefore, the general form of the solutions is n ,
+ 2n , where n is an integer.
where n is an integer.
40. tan2 x = tan x
form of the solutions is
+ 2n ,
+ 2n ,
42. sin2 x = sin x + 2
SOLUTION: SOLUTION: The period of tangent is , so you only need to find
solutions on the interval
. The solution to tan x
= 0 on this interval is 0 and the solution to tan x = 1
on this interval is . Solutions on the interval (– , ) are found by
adding integer multiples of π. Therefore, the general
form of the solutions is nπ, + n , where n is an
integer.
The period of sine is 2 , so you only need to find
solutions on the interval
. The equation sin x
= 2 has no solution since the maximum value the sine
function can attain is 1. The solution to sin x = −1 on
this interval is
.
Solutions on the interval (– , ) are found by
adding integer multiples of 2 . Therefore, the
general form of the solutions is
+ 2n , where n
is an integer.
43. sin2 x = 1 – cos x
SOLUTION: 41. 3 cos x = cos x – 1
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Page 8
Solutions on the interval (– , ) are found by
adding integer multiples of 2 . Therefore, the
general form of the solutions is
+ 2n , where n
Study Guide and Review - Chapter 5
is an integer.
43. sin2 x = 1 – cos x
+ n . be combined to x =
Therefore, the general form of the solutions is 2n ,
+ n , where n is an integer.
44. sin x = cos x + 1
SOLUTION: SOLUTION: The period of cosine is 2 , so you only need to find
solutions on the interval
. The solutions to cos
x = 0 on this interval are
and and the solution to cos x = 1 on this interval is 0. Solutions on the interval (– , ), are found by
adding integer multiples of 2π. The period of cosine is 2 , so you only need to find
solutions on the interval
. The solutions to cos
be combined to x =
+ 2n
and x =
+ 2n
and x = 0 on this interval are
The solutions x =
can
+ n . and the solution to cos x = −1 on this interval is . Since each side of
the equation was squared, check for extraneous
solutions.
Therefore, the general form of the solutions is 2n ,
+ n , where n is an integer.
44. sin x = cos x + 1
SOLUTION: Solutions on the interval (– , ), are found by
adding integer multiples of 2 . Therefore, the
general form of the solutions is + 2n , + 2n ,
where n is an integer.
Find the exact value of each trigonometric
expression.
45. cos 15
The period of cosine is 2 , so you only need to find
solutions on the interval
. The solutions to cos
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x = 0 on this interval are
and to cos x = −1 on this interval is
and the solution . Since each side of
SOLUTION: Write 15° as the sum or difference of angle measures with cosines that you know.
Page 9
Solutions on the interval (– , ), are found by
adding integer multiples of 2 . Therefore, the
general form of the solutions is + 2n , + 2n ,
Study Guide and Review - Chapter 5
where n is an integer.
Find the exact value of each trigonometric
expression.
45. cos 15
47. tan
SOLUTION: SOLUTION: Write 15° as the sum or difference of angle measures with cosines that you know.
46. sin 345
SOLUTION: Write 345° as the sum or difference of angle measures with sines that you know.
48. sin
47. tan
SOLUTION: Write
SOLUTION: as the sum or difference of angle measures with sines that you know.
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Page 10
49. cos
Study Guide and Review - Chapter 5
48. sin
50. tan
SOLUTION: Write
SOLUTION: Write
as the sum or difference of angle measures with sines that you know.
as the sum or difference of angle measures with tangents that you know.
49. cos
SOLUTION: Write
as the sum or difference of angle measures with cosines that you know.
Simplify each expression.
51. SOLUTION: 50. tan
SOLUTION: Write
as the sum or difference of angle measures with tangents that you know.
52. cos 24 cos 36 − sin 24 sin 36
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Page 11
53. Study Guide and Review - Chapter 5
52. cos 24 cos 36 − sin 24 sin 36
57. SOLUTION: SOLUTION: 53. sin 95 cos 50 − cos 95 sin 50
SOLUTION: 58. SOLUTION: 54. cos
cos + sin sin SOLUTION: Verify each identity.
55. cos ( + 30 ) – sin (
Find the values of sin 2 , cos 2 , and tan 2
for the given value and interval.
+ 60 ) = –sin
59. cos
= , (0 , 90 )
SOLUTION: SOLUTION: on the interval (0°, 90°), one point Since
on the terminal side of θ has x-coordinate 1 and a
distance of 3 units from the origin as shown. The ycoordinate of this point is therefore
56. 2
or .
SOLUTION: 57. Using this point, we find that
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Now use the double-angle
and Page 12
identities for sine, cosine, and tangent to find sin 2 ,
cos 2 , and tan 2 .
the interval (180°, 270°), one point on the terminal side of has x-coordinate −1 and y-coordinate −2
as shown. The distance from the point to the origin is
Study Guide and Review - Chapter 5
and Using this point, we find that
or .
Now use the double-angle
identities for sine, cosine, and tangent to find sin 2 ,
cos 2 , and tan 2 .
Using this point, we find that
and . Now
use the double-angle identities for sine, cosine, and
tangent to find sin 2 , cos 2 , and tan 2 .
60. tan
= 2, (180 , 270 )
SOLUTION: If tan
= 2, then tan = . Since tan
= on the interval (180°, 270°), one point on the terminal side of has x-coordinate −1 and y-coordinate −2
as shown. The distance from the point to the origin is
or .
61. SOLUTION: Since
on the interval , one point on
the terminal side of θ has y-coordinate 4 and a
distance of 5 units from the origin as shown. The xcoordinate of this point is therefore −
−3.
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Using this point, we find that
and . Now
or 13
Page
SOLUTION: Since
on the interval , one point on
Study Guide and Review - Chapter 5
the terminal side of θ has y-coordinate 4 and a
distance of 5 units from the origin as shown. The xcoordinate of this point is therefore −
−3.
or
62. SOLUTION: If
, then
the interval
. Since
on , one point on the terminal side
of θ has x-coordinate 5 and a distance of 13 units
from the origin as shown. The y-coordinate of this
point is therefore –
or –12.
and Using this point, we find that
Now use the double-angle
identities for sine and cosine to find sin 2θ and cos
2θ.
Using this point, we find that
and Now use the double-angle
identities for sine and cosine to find sin 2θ and cos
2θ.
Use the definition of tangent to find tan 2θ.
Use the definition of tangent to find tan 2θ.
62. eSolutions
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SOLUTION: If
, then
Page 14
. Since
on Find the exact value of each expression.
63. sin 75
Study Guide and Review - Chapter 5
Use the definition of tangent to find tan 2θ.
64. cos
SOLUTION: Notice that
is half of . Therefore, apply
the half-angle identity for cosine, noting that since
lies in Quadrant II, its cosine is negative.
Find the exact value of each expression.
63. sin 75
SOLUTION: Notice that 75° is half of 150°. Therefore, apply the half-angle identity for sine, noting that since 75° lies in Quadrant I, its sine is positive.
65. tan 67.5
SOLUTION: Notice that 67.5° is half of 135°. Therefore, apply the half-angle identity for tangent, noting that since
67.5° lies in Quadrant I, its tangent is positive.
64. cos
SOLUTION: Notice that
is half of . Therefore, apply
the half-angle identity for cosine, noting that since
lies in Quadrant II, its cosine is negative.
66. cos
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Notice that
is half of . Therefore, applyPage
the 15
half-angle identity for cosine, noting that since
lies in Quadrant I, its cosine is positive.
Study Guide and Review - Chapter 5
66. cos
68. tan
SOLUTION: Notice that
SOLUTION: is half of . Therefore, apply the
half-angle identity for cosine, noting that since
Notice that
is half of . Therefore, apply
the half-angle identity for tangent, noting that since
lies in Quadrant III, its tangent is positive.
lies in Quadrant I, its cosine is positive.
69. CONSTRUCTION Find the tangent of the angle
that the ramp makes with the building if sin
and cos = 67. sin
= .
SOLUTION: Notice that
is half of
. Therefore, apply
the half-angle identity for sine, noting that since
lies in Quadrant IV, its sine is negative.
SOLUTION: Use the Quotient Identity tan θ =
Therefore, the tangent of the angle the ramp makes
with the building is
.
70. LIGHT The intensity of light that emerges from a
68. tan
system of two polarizing lenses can be calculated by
SOLUTION: eSolutions Manual - Powered by Cognero
Notice that
is half of . Therefore, apply
the half-angle identity for tangent, noting that since
I = I0 −
Page 16
, where I0 is the intensity of light
entering the system and
is the angle of the axis of Therefore, the tangent of the angle the ramp makes
Study
andis Review
- Chapter 5
withGuide
the building
.
70. LIGHT The intensity of light that emerges from a
system of two polarizing lenses can be calculated by
I = I0 −
, where I0 is the intensity of light
entering the system and is the angle of the axis of the second lens with the first lens. Write the equation
for the light intensity using only tan .
71. MAP PROJECTIONS Stereographic projection is
used to project the contours of a three-dimensional
sphere onto a two-dimensional map. Points on the
sphere are related to points on the map using
r = . Verify that r =
.
SOLUTION: SOLUTION: Verify that
= .
71. MAP PROJECTIONS Stereographic projection is
used to project the contours of a three-dimensional
sphere onto a two-dimensional map. Points on the
sphere are related to points on the map using
r = . Verify that r =
.
By substitution, r =
.
72. PROJECTILE MOTION A ball thrown with an
initial speed of v0 at an angle that travels a
horizontal distance d will remain in the air t seconds,
where t =
. Suppose a ball is thrown with
an initial speed of 50 feet per second, travels 100
feet, and is in the air for 4 seconds. Find the angle at
which the ball was thrown.
SOLUTION: Verify that
= .
SOLUTION: Let t = 4, d = 100, v0 = 50, and solve for θ.
By substitution, r =
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.
72. PROJECTILE MOTION A ball thrown with an
Page 17
Study
Guide and
5
By substitution,
r =Review - Chapter
.
72. PROJECTILE MOTION A ball thrown with an
300º. Because the inverse cosine function is
restricted to acute angles of θ on the interval [0,
180º], θ = 60º. Therefore, the ball was thrown at an
angle of 60º.
73. BROADCASTING Interference occurs when two
an initial speed of 50 feet per second, travels 100
feet, and is in the air for 4 seconds. Find the angle at
which the ball was thrown.
waves pass through the same space at the same
time. It is destructive if the amplitude of the sum of
the waves is less than the amplitudes of the
individual waves. Determine whether the
interference is destructive when signals modeled by
y = 20 sin (3t + 45 ) and y = 20 sin (3t + 225 ) are
combined.
SOLUTION: SOLUTION: initial speed of v0 at an angle that travels a
horizontal distance d will remain in the air t seconds,
where t =
. Suppose a ball is thrown with
Let t = 4, d = 100, v0 = 50, and solve for θ.
The sum of the two functions is zero, which means
that the amplitude is 0. Because the amplitude of each of the original functions was 20 and 0 < 20, the
combination of the two waves can be characterized
as producing destructive interference.
74. TRIANGULATION Triangulation is the process
of measuring a distance d using the angles α and β
On the unit circle, cos θ =
when θ = 60º and θ =
300º. Because the inverse cosine function is
restricted to acute angles of θ on the interval [0,
180º], θ = 60º. Therefore, the ball was thrown at an
angle of 60º.
and the distance ℓ using
.
a. Solve the formula for d.
73. BROADCASTING Interference occurs when two
waves pass through the same space at the same
time. It is destructive if the amplitude of the sum of
the waves is less than the amplitudes of the
individual waves. Determine whether the
interference is destructive when signals modeled by
y = 20 sin (3t + 45 ) and y = 20 sin (3t + 225 ) are
combined.
b. Verify that d =
c. Verify that d =
d. Show that if = .
.
, then d = 0.5ℓ tan
SOLUTION: a.
SOLUTION: The sum of the two functions is zero, which means
that the amplitude is 0. Because the amplitude of each of the original functions was 20 and 0 < 20, the
combination of the two waves can be characterized
as producing
destructive
interference.
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74. TRIANGULATION Triangulation is the process
of measuring a distance d using the angles α and β
b.
Page 18
Study Guide and Review - Chapter 5
b.
c.
d.
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