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Study Guide and Review - Chapter 5
Complete each identity by filling in the blank. Then name the identity.
= _______
1. sec
SOLUTION: = The reciprocal identity states sec
.
2. _______ =
SOLUTION: The quotient identity states tan
= .
3. _______ + 1 = sec2
SOLUTION: The Pythagorean identity states sec
4. cos (90° –
= .
) = _______
SOLUTION: The cofunction identity states cos(90° −
) = sin
.
5. tan (– ) = _______
SOLUTION: The odd-even identity states tan(− ) = −tan
6. sin (
+
) = sin
_______ + cos .
_______
SOLUTION: The sum identity states sin (α + β) = sin α cos β + cos α sin β.
7. _______ = cos2α – sin2 α
SOLUTION: 2
2
The double-angle identity states that cos 2α = cos α – sin α.
8. _______= ±
SOLUTION: The half-angle identity states cos
9. = ±
.
=_______
SOLUTION: 2
= sin
The power-reducing identity states
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10. _______ =
[cos (
–
.
Page 1
) + cos (
+
)]
SOLUTION: Study
and
Review
- Chapter
5
TheGuide
half-angle
identity
states cos
= ±
9. .
=_______
SOLUTION: 2
= sin
The power-reducing identity states
10. _______ =
[cos (
–
) + cos (
+
.
)]
SOLUTION: The product-to-sum identity states cos α cos β =
[cos (α – β) + cos (α + β)].
Find the value of each expression using the given information.
and cos ; tan = 3, cos > 0
11. sec
SOLUTION: Use the Pythagorean Identity that involves tan
Since we are given that cos
to find sec .
is positive, sec must be positive. So, sec
= . Use the reciprocal identity
to find cos .
12. cot
and sin ; cos
= , tan
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
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to find sin .
Page 2
Study Guide and Review - Chapter 5
12. cot
and sin ; cos
= , tan
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
Since tan
= quotient identity cot
13. csc
and tan ; cos
to find sin .
is negative and cos is negative, sin must be positive. So, sin =
= . Use the
to find cot .
= , sin
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
Since sin
< 0, to find sin .
. Use the reciprocal identity csc
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= to find csc .
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Study Guide and Review - Chapter 5
13. csc
and tan ; cos
= , sin
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
Since sin
< 0, and cos ; tan
= . Use the reciprocal identity csc
Use the quotient identity tan
14. cot
to find sin .
= = to find csc .
to find tan .
, csc
> 0
SOLUTION: Use the reciprocal function cot
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= to find cot .
Use the Pythagorean identity that involves tan
to find sec
.
Page 4
Study Guide and Review - Chapter 5
14. cot
and cos ; tan
= , csc
> 0
SOLUTION: Use the reciprocal function cot
= to find cot .
Use the Pythagorean identity that involves tan
Since csc
is positive, sin
positive. Since cos
to find sec
is positive. Since tan = .
is positive and sin is positive, cos has to be is positive, sec is positive. So, sec = . Use the reciprocal identity cos
= to find cos .
15. sec
and sin ; cot
= –2, csc
< 0
SOLUTION: Use the Pythagorean identity that involves cot
Since
csc - Powered
< 0, csc =
−
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Manual
by Cognero
to find csc .
. Use the reciprocal identity sin
= to find sin .
Page 5
Study Guide and Review - Chapter 5
15. sec
and sin ; cot
= –2, csc
< 0
SOLUTION: Use the Pythagorean identity that involves cot
Since csc
< 0, csc = −
. Use the reciprocal identity sin
Use the Pythagorean identity that involves sin
Since cot
= reciprocal identity sec
16. cos
and sin ; cot
to find csc .
= to find sin .
to find cos .
is negative and sin is negative, cos must be positive. So, cos =
to find sec .
=
= . Use the
, sec
< 0
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SOLUTION: Use the Pythagorean identity that involves cot to find csc
.
Study Guide and Review - Chapter 5
16. cos
and sin ; cot
= , sec
< 0
SOLUTION: Use the Pythagorean identity that involves cot to find csc
Since sec
sin
< 0, cos is negative. Since cot
is negative, csc must be negative. So, = .
is positive and cos is negative, sin is negative. Since
. Use the reciprocal identity sin
= to find sin
.
Use the Pythagorean identity that involves sin
It was determined that cos
to find cos .
is negative. So, Simplify each expression.
17. sin2 (–x) + cos2(–x)
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Study
Guide
and that
Review
- Chapter 5
It was
determined
cos is negative. So, Simplify each expression.
17. sin2 (–x) + cos2(–x)
SOLUTION: 18. sin2 x + cos2 x + cot2 x
SOLUTION: 19. SOLUTION: 20. SOLUTION: 21. SOLUTION: eSolutions Manual - Powered by Cognero
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SOLUTION: Study Guide and Review - Chapter 5
21. SOLUTION: 22. SOLUTION: Verify each identity.
23. +
= 2 csc SOLUTION: eSolutions Manual - Powered by Cognero
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Study Guide and Review - Chapter 5
Verify each identity.
23. = 2 csc +
SOLUTION: 24. = 1
+
SOLUTION: 25. +
= 2 sec SOLUTION: eSolutions Manual - Powered by Cognero
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Study Guide and Review - Chapter 5
25. = 2 sec +
SOLUTION: 26. =
SOLUTION: 27. = csc
– 1
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Study Guide and Review - Chapter 5
27. = csc
– 1
SOLUTION: 28. +
= sec + csc SOLUTION: 29. = csc
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Study Guide and Review - Chapter 5
29. = csc
SOLUTION: 30. cot
2
csc + sec = csc
sec SOLUTION: 31. =
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Study Guide and Review - Chapter 5
31. =
SOLUTION: 32. cos4
4
– sin
= SOLUTION: Find all solutions of each equation on the interval [0, 2π].
33. 2 sin x =
SOLUTION: On the interval [0, 2π),
when x =
and x =
.
34. 4 cos2 x = 3
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Page 14
On the interval [0, 2π),
when x =
Study
Guide and Review - Chapter
5
and x =
.
34. 4 cos2 x = 3
SOLUTION: On the interval [0, 2 ),
when x =
and x =
and when x =
and x =
.
35. tan2 x – 3 = 0
SOLUTION: 2
tan x – 3 = 0
On the interval [0, 2π),
when x =
and x =
and when x =
and x =
.
when x =
and x =
and when x =
and x =
.
36. 9 + cot2 x = 12
SOLUTION: On the interval [0, 2π),
37. 2 sin2 x = sin x
SOLUTION: On the interval [0, 2π), sin x = 0 when x = 0 and
38. 3 cos x + 3 = sin2 x
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SOLUTION: and when x =
and x =
.
Page 15
Study
Guide
and
Review
On the
interval
[0, 2π),
sin x =-0Chapter
when x = 0 5and
and when x =
and x =
.
38. 3 cos x + 3 = sin2 x
SOLUTION: On the interval [0, 2 ), cos x = −1 when x =
the cosine function can attain is −1.
. The equation cos x = −2 has no solution since the minimum value
Solve each equation for all values of x.
39. sin2 x – sin x = 0
SOLUTION: The period of sine is 2 , so you only need to find solutions on the interval
interval are 0 and and the solution to sin x = 1 on this interval is . . The solutions to sin x = 0 on this
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.
40. tan2 x = tan x
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The period of tangent is
, so you only need to find solutions on the interval
Page 16
. The solution to tan x = 0 on this
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
Study
Guide and Review - Chapter 5
integer.
40. tan2 x = tan x
SOLUTION: The period of tangent is , so you only need to find solutions on the interval
interval is 0 and the solution to tan x = 1 on this interval is . . The solution to tan x = 0 on this
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
SOLUTION: The period of cosine is 2 , so you only need to find solutions on the interval
on this interval are
Solutions on the interval (–
+ 2n ,
solutions is
,
and . The solutions to
.
) are found by adding integer multiples of 2π. Therefore, the general form of the
+ 2n , where n is an integer.
42. sin2 x = sin x + 2
SOLUTION: 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
.
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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.
on this interval are
Solutions on the interval (–
,
and .
) are found by adding integer multiples of 2π. Therefore, the general form of the
Study
Guide
Review
- Chapter
solutions
is and
+ 2n
,
+ 2n
, where n5is an integer.
42. sin2 x = sin x + 2
SOLUTION: 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: The period of cosine is 2 , so you only need to find solutions on the interval
this interval are
and . The solutions to cos x = 0 on
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 solutions x =
+ 2n
and x =
+ 2n
can be combined to x =
Therefore, the general form of the solutions is 2n ,
+ n . + n , where n is an integer.
44. sin x = cos x + 1
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The solutions x =
+ 2n
and x =
+ 2n
can be combined to x =
+ n . Study
Guidetheand
Review
Chapter
5 2n
Therefore,
general
form of-the
solutions is
,
+ n , where n is an integer.
44. sin x = cos x + 1
SOLUTION: The period of cosine is 2 , so you only need to find solutions on the interval
this interval are
and and the solution to cos x = −1 on this interval is
. The solutions to cos x = 0 on
. Since each side of the equation was
squared, check for extraneous solutions.
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
SOLUTION: Write 15° as the sum or difference of angle measures with cosines that you know.
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Solutions on the interval (–
,
), are found by adding integer multiples of 2 . Therefore, the general form of the
Study
Guide
5 integer.
solutions
is and
+ 2n Review
, + 2n - ,Chapter
where n is an
Find the exact value of each trigonometric expression.
45. cos 15
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.
47. tan
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Study Guide and Review - Chapter 5
47. tan
SOLUTION: 48. sin
SOLUTION: Write
as the sum or difference of angle measures with sines that you know.
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49. cos
Page 21
Study Guide and Review - Chapter 5
49. cos
SOLUTION: Write
as the sum or difference of angle measures with cosines that you know.
50. tan
SOLUTION: Write
as the sum or difference of angle measures with tangents that you know.
Simplify each expression.
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51. Page 22
Study Guide and Review - Chapter 5
Simplify each expression.
51. SOLUTION: 52. cos 24 cos 36 − sin 24 sin 36
SOLUTION: 53. sin 95 cos 50 − cos 95 sin 50
SOLUTION: 54. cos
cos + sin sin SOLUTION: Verify each identity.
55. cos ( + 30 ) – sin (
+ 60 ) = –sin
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Page 23
Study Guide and Review - Chapter 5
Verify each identity.
55. cos ( + 30 ) – sin (
+ 60 ) = –sin
SOLUTION: 56. SOLUTION: 57. SOLUTION: 58. SOLUTION: eSolutions Manual - Powered by Cognero
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Study Guide and Review - Chapter 5
58. SOLUTION: Find the values of sin 2 , cos 2 , and tan 2
59. cos
= for the given value and interval.
, (0 , 90 )
SOLUTION: Since
on the interval (0°, 90°), one point on the terminal side of θ has x-coordinate 1 and a distance of 3
units from the origin as shown. The y-coordinate of this point is therefore
Using this point, we find that
and or 2
.
Now use the double-angle identities for sine,
cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
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Study Guide and Review - Chapter 5
Find the values of sin 2 , cos 2 , and tan 2
59. cos
= for the given value and interval.
, (0 , 90 )
SOLUTION: on the interval (0°, 90°), one point on the terminal side of θ has x-coordinate 1 and a distance of 3
Since
units from the origin as shown. The y-coordinate of this point is therefore
and Using this point, we find that
or 2
.
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 )
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SOLUTION: If tan
= 2, then tan = . Since tan
= on the interval (180°, 270°), one point on the terminal side of has
Study Guide and Review - Chapter 5
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
Using this point, we find that
and or .
. Now use the double-angle identities for
sine, cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
61. SOLUTION: eSolutions Manual - Powered by Cognero
Since
on the interval Page 27
, one point on the terminal side of θ has y-coordinate 4 and a distance of 5
Study Guide and Review - Chapter 5
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 x-coordinate of this point is therefore −
or −3.
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.
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SOLUTION: Page 28
Study Guide and Review - Chapter 5
62. SOLUTION: If
, then
. Since
on the interval , 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.
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.
Find the exact value of each expression.
63. sin 75
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SOLUTION: Page 29
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.
Study Guide and Review - Chapter 5
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.
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.
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.
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Study Guide and Review - Chapter 5
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.
66. cos
SOLUTION: Notice that
is half of . Therefore, apply the half-angle identity for cosine, noting that since
lies in Quadrant I, its cosine is positive.
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.
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Study Guide and Review - Chapter 5
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.
68. tan
SOLUTION: is half of Notice that
. Therefore, apply the half-angle identity for tangent, noting that since
lies in Quadrant III, its tangent is positive.
69. CONSTRUCTION Find the tangent of the angle that the ramp makes with the building if sin
= = and cos .
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SOLUTION: Page 32
Study Guide and Review - Chapter 5
69. CONSTRUCTION Find the tangent of the angle that the ramp makes with the building if sin
= = and cos .
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 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
.
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Study
Guidetheand
Review
- Chapter
Therefore,
tangent
of the angle
the ramp 5
makes with the building is
.
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
.
SOLUTION: 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: Verify that
= eSolutions Manual - Powered by Cognero
.
Page 34
Study Guide and Review - Chapter 5
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: Verify that
= .
By substitution, r =
.
72. PROJECTILE MOTION A ball thrown with an initial speed of v0 at an angle
d will remain in the air t seconds, where t =
that travels a horizontal distance
. 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: Let t = 4, d = 100, v0 = 50, and solve for θ.
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On the unit circle, cos θ =
when θ = 60º and θ = 300º. Because the inverse cosine function is restricted to acute
Study
Guide and
5
By substitution,
r =Review - Chapter
.
72. PROJECTILE MOTION A ball thrown with an initial speed of v0 at an angle
d will remain in the air t seconds, where t =
that travels a horizontal distance
. 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: Let t = 4, d = 100, v0 = 50, and solve for θ.
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.
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.
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.
74. TRIANGULATION Triangulation is the process of measuring a distance d using the angles α and β and the
distance ℓ using
.
a. Solve
the
formulabyfor
d.
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b. Verify that d =
Page 36
.
distance ℓ using
.
Study Guide and Review - Chapter 5
a. Solve the formula for d.
b. Verify that d =
c. Verify that d =
d. Show that if = .
.
, then d = 0.5ℓ tan
SOLUTION: a.
b.
c.
d.
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Study Guide and Review - Chapter 5
d.
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