5-5 Multiple-Angle and Product-to-Sum Identities
Find the values of sin 2 , cos 2 , and tan 2
1. cos
= for the given value and interval.
, (270 , 360 )
SOLUTION: Since
on the interval (270 , 360 ), one point on the terminal side of θ has x-coordinate 3 and a distance
of 5 units from the origin as shown. The y-coordinate of this point is therefore −
Using this point, we find that
and or −4.
Now use the double-angle identities for sine,
cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
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5-5 Multiple-Angle and Product-to-Sum Identities
2. tan
= , (180 , 270 )
SOLUTION: Since tan
= on the interval (180°, 270°), one point on the terminal side of θ has x-coordinate −15 and y-
coordinate −8 as shown. The distance from the point to the origin is
Using this point, we find that
and or 17.
. Now use the double-angle identities for
sine, cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
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5-5 Multiple-Angle and Product-to-Sum Identities
3. cos
= , (90 , 180 )
SOLUTION: Since
on the interval (90°, 180°), one point on the terminal side of θ has x-coordinate −9 and a
distance of 41 units from the origin as shown. The y-coordinate of this point is therefore
Using this point, we find that
and or 40.
Now use the double-angle identities for sine,
cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
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Using this point, we and
find that
and 5-5 Multiple-Angle
Product-to-Sum
Identities
Now use the double-angle identities for sine,
cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
4. sin
=
,
SOLUTION: Since
on the interval , one point on the terminal side of
has y-coordinate −7 and a
distance of 12 units from the origin as shown. The x-coordinate of this point is therefore
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Using this point, we find that
or .
Page 4
and Now use the double-angle identities for
SOLUTION: Since
on the interval , one point on the terminal side of
5-5 Multiple-Angle and Product-to-Sum Identities
has y-coordinate −7 and a
distance of 12 units from the origin as shown. The x-coordinate of this point is therefore
and Using this point, we find that
sine and cosine to find sin 2
or .
Now use the double-angle identities for
and cos 2 .
Use the definition of tangent to find tan 2 .
5. tanManual
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by Cognero
SOLUTION: Page 5
5-5 Multiple-Angle and Product-to-Sum Identities
5. tan
=
,
SOLUTION: Since
on the interval , one point on the terminal side of
coordinate −1 as shown. The distance from the point to the origin is
Using this point, we find that
and has x-coordinate 2 and y or .
. Now use the double-angle identities for
sine, cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
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5-5 Multiple-Angle and Product-to-Sum Identities
6. tan
= ,
SOLUTION: If tan
= , then tan
= has x-coordinate 1 and y-coordinate
Using this point, we find that
on the interval . Since
, one point on the terminal side of θ
as shown. The distance from the point to the origin is 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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5-5 Multiple-Angle and Product-to-Sum Identities
6. tan
= ,
SOLUTION: = If tan
, then tan
= has x-coordinate 1 and y-coordinate
Using this point, we find that
on the interval . Since
, one point on the terminal side of θ
as shown. The distance from the point to the origin is and or 2.
. Now use the double-angle identities for sine,
cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
7. sin
= ,
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Page 8
SOLUTION: Since
on the interval , one point on the terminal side of
has y-coordinate 4 and a distance of 5
5-57. Multiple-Angle and Product-to-Sum Identities
sin
= ,
SOLUTION: Since
on the interval , one point on the terminal side of
has y-coordinate 4 and a distance of 5
units from the origin as shown. The x-coordinate of this point is therefore −
Using this point, we find that
and cosine to find sin 2
and or −3.
Now use the double-angle identities for sine
and cos 2 .
Use the definition of tangent to find tan 2 .
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5-5 Multiple-Angle and Product-to-Sum Identities
Use the definition of tangent to find tan 2 .
8. cos
= ,
SOLUTION: Since
on the interval , one point on the terminal side of
has x-coordinate −5 and a distance
of 13 units from the origin as shown. The y-coordinate of this point is therefore −
Using this point, we find that
and or −12.
Now use the double-angle identities for sine,
cosine, and tangent to find sin 2 , cos 2 , and tan 2 .
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5-5 Multiple-Angle and Product-to-Sum Identities
Solve each equation on the interval [0, 2 ].
9. sin 2 = cos
SOLUTION: or On the interval [0, 2π), cos
10. cos 2
= 0 when = and = and sin = when = and = .
= cos
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Page 11
or On the interval [0, 2π), cos = 0 when = and = and sin 5-5 Multiple-Angle
and Product-to-Sum Identities
10. cos 2
= when = and = .
= cos
SOLUTION: or when
=
and θ =
= −1 when
= and sin = On the interval [0, 2π),
11. cos 2
– sin
and cos
= 1 when
= 0.
= 0
SOLUTION: or On the interval [0, 2π), sin
when = and = .
12. tan 2 − tan 2 tan2 θ = 2
SOLUTION: On the interval [0, 2π), tan
13. sin 2
csc
= 1 when = and = .
= 1
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On the interval [0, 2π), cos
Page 12
= when =
and = .
5-5 Multiple-Angle
and
Identities
On the interval [0, 2π),
tanProduct-to-Sum
= 1 when = and = .
13. sin 2
csc
= 1
SOLUTION: On the interval [0, 2π), cos
14. cos 2
+ 4 cos
= when =
and = .
= –3
SOLUTION: On the interval [0, 2π), cos
= −1 when
= .
15. GOLF A golf ball is hit with an initial velocity of 88 feet per second. The distance the ball travels is found by d =
, where v0 is the initial velocity,
in feet per second squared.
is the angle that the path of the ball makes with the ground, and 32 is
a. If the ball travels 242 feet, what is to the nearest degree?
b. Use a double-angle identity to rewrite the equation for d.
SOLUTION: a. Substitute d = 242 and v0 = 88 into the equation.
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5-5 Multiple-Angle and Product-to-Sum Identities
On the interval [0, 2π), cos
= −1 when
= .
15. GOLF A golf ball is hit with an initial velocity of 88 feet per second. The distance the ball travels is found by d =
, where v0 is the initial velocity,
in feet per second squared.
is the angle that the path of the ball makes with the ground, and 32 is
a. If the ball travels 242 feet, what is to the nearest degree?
b. Use a double-angle identity to rewrite the equation for d.
SOLUTION: a. Substitute d = 242 and v0 = 88 into the equation.
If the ball travels 242 feet, θ is 45 .
b.
Rewrite each expression in terms with no power greater than 1.
16. cos3
SOLUTION: 17. tan3
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5-5 Multiple-Angle and Product-to-Sum Identities
17. tan3
SOLUTION: 18. sec4
SOLUTION: 19. cot3
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4
20. cos
4
− sin
SOLUTION: Page 15
5-5 Multiple-Angle and Product-to-Sum Identities
19. cot3
SOLUTION: 20. cos4
4
− sin
SOLUTION: 21. sin2
3
cos
SOLUTION: 22. sin2
− cos
2
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5-5 Multiple-Angle and Product-to-Sum Identities
22. sin2
− cos
2
SOLUTION: 23. SOLUTION: Solve each equation.
24. 1 – sin2
− cos 2
=
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Page 17
5-5 Multiple-Angle and Product-to-Sum Identities
Solve each equation.
24. 1 – sin2
− cos 2
=
SOLUTION: The graph of y = cos 2
25. cos2
−
cos 2
has a period of
, so the solutions are
+ n ,
+ n , where n is an integer.
has a period of
, so the solutions are
+ n ,
+ n , where n is an integer.
=0
SOLUTION: The graph of y = cos 2
26. sin2
2
− 1 = cos
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Page 18
5-5 Multiple-Angle
Product-to-Sum
Identities
The graph of y = cosand
2 has
a period of , so the
solutions are
26. sin2
+ n ,
+ n , where n is an integer.
2
− 1 = cos
SOLUTION: The graph of y = cos 2
27. cos2
– sin
has a period of
, so the solutions are
+ n , where n is an integer.
=1
SOLUTION: or The graph of y = sin
solutions are n ,
has a period of 2 . The solutions 0 + 2n
and
+ 2n
can be combined to n . So, the
+ 2n , where n is an integer.
28. MACH NUMBER The angle
at the vertex of the cone-shaped shock wave produced by a plane breaking the
sound barrier is related to the mach number M describing the plane's speed by the half-angle equation sin
= .
a. Express the mach number of the plane in terms of cosine.
b. Use the expression found in part a to find the mach number of a plane if cos
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SOLUTION: a.
= .
Page 19
The graph of y = sin
has a period of 2 . The solutions 0 + 2n
and
+ 2n
can be combined to n . So, the
5-5 Multiple-Angle
Identities
solutions are n , and
+ 2nProduct-to-Sum
, where n is an integer.
28. MACH NUMBER The angle
at the vertex of the cone-shaped shock wave produced by a plane breaking the
sound barrier is related to the mach number M describing the plane's speed by the half-angle equation sin
= .
a. Express the mach number of the plane in terms of cosine.
b. Use the expression found in part a to find the mach number of a plane if cos
= .
SOLUTION: a.
Since the mach number describing a plane’s speed cannot be negative,
b. Substitute cos
= .
into the expression found in part a.
The mach number of the plane is about 3.2.
Find the exact value of each expression.
29. sin 67.5
SOLUTION: Notice that 67.5 is half of 135 . Therefore, apply the half-angle identity for sine, noting that since 67.5 lies in
Quadrant I, its sine is positive.
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5-5 Multiple-Angle and Product-to-Sum Identities
The mach number of the plane is about 3.2.
Find the exact value of each expression.
29. sin 67.5
SOLUTION: Notice that 67.5 is half of 135 . Therefore, apply the half-angle identity for sine, noting that since 67.5 lies in
Quadrant I, its sine is positive.
30. cos
SOLUTION: Notice that
is half of . Therefore, apply the half-angle identity for cosine, noting that since
Quadrant I, its cosine is positive.
lies in 31. tan 157.5
SOLUTION: Notice that 157.5 is half of 315 . Therefore, apply the half-angle identity for tangent, noting that since 157.5 lies
in Quadrant II, its tangent is negative.
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5-5 Multiple-Angle and Product-to-Sum Identities
31. tan 157.5
SOLUTION: Notice that 157.5 is half of 315 . Therefore, apply the half-angle identity for tangent, noting that since 157.5 lies
in Quadrant II, its tangent is negative.
32. sin
SOLUTION: Notice that
is half of . Therefore, apply the half-angle identity for sine, noting that since
Quadrant II, its sine is positive.
lies in Solve each equation on the interval [0, 2 ].
33. sinManual
+ cos = 1by Cognero
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- Powered
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5-5 Multiple-Angle and Product-to-Sum Identities
Solve each equation on the interval [0, 2 ].
33. sin
+ cos = 1
SOLUTION: On the interval [0, 2 ),
when and
when
. Check each of these solutions in the original equation.
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5-5 Multiple-Angle and Product-to-Sum Identities
All solutions are valid. Therefore, the solutions to the equation on [0, 2 ) are 0,
34. tan
, and
.
= sin SOLUTION: On the interval [0, 2 ),
when
. Check this solution in the original equation.
The solution is valid. Therefore, the solution to the equation on [0, 2 ) is 0.
35. 2 sin
= sin SOLUTION: eSolutions Manual - Powered by Cognero
Page 24
5-5 Multiple-Angle and Product-to-Sum Identities
The solution is valid. Therefore, the solution to the equation on [0, 2 ) is 0.
35. 2 sin
= sin SOLUTION: On the interval [0, 2 ),
when
. Check this solution in the original equation.
The solution is valid. Therefore, the solution to the equation on [0, 2 ) is 0.
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Page 25
The solution is valid. Therefore, the solution to the equation on [0, 2 ) is 0.
36. 5-5 Multiple-Angle and Product-to-Sum Identities
SOLUTION: On the interval [0, 2 ),
when
. The equation
has no solution since
> 1. Check each of these solutions in the original equation.
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5-5 Multiple-Angle and Product-to-Sum Identities
The only valid solution is
. Therefore, the solution to the equation on [0, 2 ) is
.
Rewrite each product as a sum or difference.
cos
37. cos 3
SOLUTION: 38. cos 12x sin 5x
SOLUTION: 39. sin 3x cos 2x
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40. sin 8θ sin θ
Page 27
5-5 Multiple-Angle and Product-to-Sum Identities
39. sin 3x cos 2x
SOLUTION: 40. sin 8θ sin θ
SOLUTION: Find the exact value of each expression.
41. 2 sin 135 sin 75
SOLUTION: 42. cos
+ cos
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43. sin 172.5 sin 127.5
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5-5 Multiple-Angle and Product-to-Sum Identities
43. sin 172.5 sin 127.5
SOLUTION: 44. sin 142.5 cos 352.5
SOLUTION: 45. sin 75 + sin 195
SOLUTION: 46. 2 cos 105 + 2 cos 195
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5-5 Multiple-Angle and Product-to-Sum Identities
46. 2 cos 105 + 2 cos 195
SOLUTION: 47. 3 sin
− 3 sin
SOLUTION: 48. cos
+ cos SOLUTION: Solve each equation.
49. cos – cos 3 = 0
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Page 30
5-5 Multiple-Angle and Product-to-Sum Identities
Solve each equation.
– cos 3 = 0
49. cos
SOLUTION: The graph of y = sin 2
solutions are n ,
50. 2 cos 4
has a period of π. The solutions 0 + 2n
and
+ 2n
can be combined to n . So, the
+ 2nπ, where n is an integer.
+ 2 cos 2 = 0
SOLUTION: The graph of y = cos 3 has a period of n . So, the solutions are
51. sin 3
+ sin 5
+ n,
+ . The solutions
n, and + 2n
and
+ 2n
can be combined to
+ + n , where n is an integer.
=0
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Page 31
The graph of y = cos 3 has a period of . The solutions
5-5 Multiple-Angle
Identities
n . So, the solutionsand
are Product-to-Sum
n, + n, and + + n
51. sin 3
+ sin 5
+ 2n
and
+ 2n
+ can be combined to
, where n is an integer.
=0
SOLUTION: The graph of y = sin 4 has a period of . So, the solutions are
52. sin 2 – sin
n and
+ n, where n is an integer.
= 0
SOLUTION: The graph of y = cos
has a period of . The solutions 0 + 2n
and
+ 2n
can be combined to nπ. + The solutions are
53. 3 cos 6 – 3 cos 4
n,
+ n, and n , where n is an integer.
=0
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Page 32
The graph of y = cos
has a period of . The solutions 0 + 2n
and
+ 2n
can be combined to nπ. 5-5 Multiple-Angle
and Product-to-Sum
The solutions are + n, + n, and n , Identities
where n is an integer.
53. 3 cos 6 – 3 cos 4
=0
SOLUTION: The graph of y = sin 5 has a period of ,
54. 4 sin
+ and the graph of y = sin 2
has a period of
. So, the solutions are
, and n , where n is an integer.
+ 4 sin 3 = 0
SOLUTION: The graph of y = sin 2
has a period of
. The solutions are n
and
+ n , where n is an integer.
Simplify each expression.
55. SOLUTION: If
, then
. Use this form of the Cosine Half-Angle Identity to simplify the
given expression.
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5-5 Multiple-Angle
Product-to-Sum
Identities
The graph of y = sinand
2 has
a period of . The solutions
are n
and
+ n , where n is an integer.
Simplify each expression.
55. SOLUTION: If
, then
. Use this form of the Cosine Half-Angle Identity to simplify the
, then
. Use this form of the Sine Half-Angle Identity to simplify the
given expression.
56. SOLUTION: If
given expression.
Write each expression as a sum or difference.
57. cos (a + b) cos (a – b)
SOLUTION: 58. sin (
−
) sin (
+
)
SOLUTION: 59. sin (b +
) cos (b + π)
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5-5 Multiple-Angle and Product-to-Sum Identities
59. sin (b +
) cos (b + π)
SOLUTION: 60. cos (a − b) sin (b − a)
SOLUTION: 61. MAPS A Mercator projection is a flat projection of the globe in which the distance between the lines of latitude
increases with their distance from the equator.
The calculation of a point on a Mercator projection contains the expression
, where
is the latitude of
the point.
a. Write the expression in terms of sin and cos .
b. Find the value of this expression if = 60 .
SOLUTION: a.
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Page 35
the point.
a. Write the expression in terms of sin and cos .
b. Find the value of this expression if = 60 .
5-5 Multiple-Angle
and Product-to-Sum Identities
SOLUTION: a.
b.
62. BEAN BAG TOSS Ivan constructed a bean bag tossing game as shown in the figure below.
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5-5 Multiple-Angle
and Product-to-Sum Identities
62. BEAN BAG TOSS Ivan constructed a bean bag tossing game as shown in the figure below.
a. Exactly how far will the back edge of the board be from the ground?
b. Exactly how long is the entire setup?
SOLUTION: a. Use the right triangle formed by the distance from the back edge to the ground y, the entire length of the setup x
and the length of the board, 46.5 in. to find y.
Since length cannot be negative, the back edge is
b. Use the same right triangle to find x.
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in. from the ground.
Page 37
5-5 Multiple-Angle
Product-to-Sum
Since length cannot and
be negative,
the back edge is Identities
in. from the ground.
b. Use the same right triangle to find x.
Since length cannot be negative, the entire setup is
in. long.
PROOF Prove each identity.
63. cos 2
2
− sin
= cos
2
SOLUTION: 64. cos 2
= 2 cos
2
− 1
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5-5 Multiple-Angle and Product-to-Sum Identities
65. SOLUTION: 66. SOLUTION: Let x =
.
67. SOLUTION: Let x =
.
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Let x =
.
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5-5 Multiple-Angle and Product-to-Sum Identities
68. SOLUTION: Let x =
.
Verify each identity by first using the power-reducing identities and then again by using the product-tosum identities.
69. 2 cos2 5
− 1 = cos 10
SOLUTION: 70. cos2 2
2
− sin 2
= cos 4
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5-5 Multiple-Angle and Product-to-Sum Identities
70. cos2 2
2
− sin 2
= cos 4
SOLUTION: Rewrite each expression in terms of cosines of multiple angles with no power greater than 1.
71. sin6
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5-5 Multiple-Angle and Product-to-Sum Identities
Rewrite each expression in terms of cosines of multiple angles with no power greater than 1.
71. sin6
SOLUTION: 72. sin8
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5-5 Multiple-Angle and Product-to-Sum Identities
72. sin8
SOLUTION: 73. cos7
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5-5 Multiple-Angle and Product-to-Sum Identities
73. cos7
SOLUTION: 74. sin4
4
cos
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5-5 Multiple-Angle and Product-to-Sum Identities
74. sin4
4
cos
SOLUTION: 75. MULTIPLE REPRESENTATIONS In this problem, you will investigate how graphs of functions can be used
to find identities.
a. GRAPHICAL Use a graphing calculator to graph f (x) =
on the interval [–2 , 2 ].
b. ANALYTICAL Write a sine function h(x) that models the graph of f (x). Then verify that f (x) = h(x)
algebraically.
c. GRAPHICAL Use a graphing calculator to graph g(x) =
on the interval [–2 , 2
].
d. ANALYTICAL Write a cosine function k(x) that models the graph of g(x). Then verify that g(x) = k(x)
algebraically.
SOLUTION: a.
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Page 45
d. ANALYTICAL Write a cosine function k(x) that models the graph of g(x). Then verify that g(x) = k(x)
algebraically.
5-5 Multiple-Angle
and Product-to-Sum Identities
SOLUTION: a.
b. Using the CALC maximum feature on the graphing calculator, you can determine that the function has a
maximum of 4 at x
2.356197 or x =
. Since the maximum height of y = sin x is 1, a function h(x) = a sin(x +
c) that models the graph of f (x) has an amplitude of 4 times that of f (x) or b = 4. Also, since the first maximum that
occurs for y = sin x, x > 0, is at x =
, the phase shift of the graph is about
Therefore, a function in terms of sine that models this graph is h(x) = 4 sin
4 sin
= –
or –
, so c = –
.
.
Sine Difference Identity
c.
d. Using the CALC maximum feature on the graphing calculator, you can determine that the function has a
maximum of 1 at x
1.0471978 or x =
. Since the maximum height of y = cos x is 1, a function k(x) = a cos(bx
+ c) that models the graph of f (x ) also has an amplitude of 1, so a = 1. Since the graph completes 1 cycle on [0, π]
the frequency of the graph is
which is twice the frequency of the cosine function, so b = 2. Since the first
maximum for y = cos x, x > 0, is at x = π, the phase shift of the graph is about π –
Therefore, a function in terms of sine that models this graph is k(x) = cos
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or
, so c =
.
.
Page 46
maximum for y = cos x, x > 0, is at x = π, the phase shift of the graph is about π –
or
, so c =
.
5-5 Multiple-Angle and Product-to-Sum Identities
Therefore, a function in terms of sine that models this graph is k(x) = cos
.
76. CHALLENGE Verify the following identity.
– cos 2
sin 2 cos
sin
= sin SOLUTION: REASONING Consider an angle in the unit circle. Determine what quadrant a double angle and half
angle would lie in if the terminal side of the angle is in each quadrant.
77. I
SOLUTION: If an angle
lies in Quadrant I, then 0° < θ < 90°.
If 0° <
< 45 then 0 < 2 < 90 , which is Quadrant I. If 45 < θ < 90 then 90 < 2 < 180 , which is Quadrant II. If
= 45 , then 2
= 90 and the angle is quadrantal, falling between Quadrants I and II. If 0 <
< 90 , then 0 < < 45 , which is still in Quadrant I.
78. II
SOLUTION: If an angle
lies in Quadrant II, then 90 <
< 180 . If 90 <
< 135 then 180 < 2
< 270 , which is Quadrant III. If 135 <
< 180 then 270 < 2 < 360 , which is Quadrant IV. If θ = 135 , then 2θ = 270 and the angle is quadrantal, falling between Quadrants III and IV. If 90 <
< 180 , then 45 < < 90 , which is in Quadrant I.
79. III
SOLUTION: eSolutions
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If an angle
lies in Quadrant III, then 180 <
Page 47
< 270 . If θ = 135 , then 2θ = 270 and the angle is quadrantal, falling between Quadrants III and IV. 5-5 Multiple-Angle
and Product-to-Sum Identities
If 90 < < 180 , then 45 < < 90 , which is in Quadrant I.
79. III
SOLUTION: If an angle
lies in Quadrant III, then 180 <
< 270 . If 180 < < 225 then 360 < 2 < 450 . Using coterminal angles, this is equivalent to 0 <2
which is Quadrant I. < 90 ,
If 225 < < 270 then 450 < 2 < 540 . Using coterminal angles, this is equivalent to 90° < 2 < 180°
which is Quadrant II.
If
= 225 , then 2 = 450 or 90 and the angle is quadrantal, falling between Quadrants I and II. If 180 < θ < 270 , then 90 <
< 135 , which is in Quadrant II.
CHALLENGE Verify each identity.
80. sin 4
3
= 4sin cos – 8sin
cos
SOLUTION: 81. cos 4
SOLUTION: PROOF Prove each identity.
82. cos2
= SOLUTION: eSolutions Manual
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2
83. tan
=
SOLUTION: Page 48
5-5 Multiple-Angle and Product-to-Sum Identities
83. tan2
=
SOLUTION: 84. cos α cos
= [cos (α −
) + cos (α +
)]
SOLUTION: 85. sin
cos
=
[sin (
+
) + sin (
[sin (
+
) − sin (
−
)]
SOLUTION: 86. cos
sin
= −
)]
SOLUTION: eSolutions Manual - Powered by Cognero
87. cos
+ cos
= 2 cos
Page 49
cos
5-5 Multiple-Angle and Product-to-Sum Identities
86. cos
sin
= [sin (
+
) − sin (
−
)]
SOLUTION: 87. cos
+ cos
= 2 cos
cos
SOLUTION: and let y =
Let x =
88. sin
− sin
= 2 cos
.
sin
SOLUTION: Let x =
89. cos
− cos
and let y =
= –2 sin
.
sin
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SOLUTION: Let x =
and let y =
.
5-5 Multiple-Angle and Product-to-Sum Identities
89. cos
− cos
= –2 sin
sin
SOLUTION: and let y =
Let x =
.
90. Writing in Math Describe the steps that you would use to find the exact value of cos 8
if cos
= .
SOLUTION: 2
Sample answer: Since 8 = 2(4 ), use the identity cos 2 = 2cos – 1 to find an expression for cos 8 . Then,
since 4 = 2(2 ) use the same double-angle identity again to find an expression for cos 4 . Finally, use the same
double-angle identity to replace the remaining double-angle expression cos 2 . The result will be an expression in
terms of just cos
. Substitute
for cos in this expression and simplify.
Find the exact value of each trigonometric expression.
91. cos
SOLUTION: Write
as the sum or difference of angle measures with cosines that you know.
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92. cosManual - Powered by Cognero
SOLUTION: Page 51
5-5 Multiple-Angle and Product-to-Sum Identities
92. cos
SOLUTION: Write
as the sum or difference of angle measures with cosines that you know.
93. sin
SOLUTION: Write
as the sum or difference of angle measures with sines that you know.
94. sin
SOLUTION: Write
as the sum or difference of angle measures with sines that you know.
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Page 52
5-5 Multiple-Angle and Product-to-Sum Identities
94. sin
SOLUTION: Write
as the sum or difference of angle measures with sines that you know.
95. SOLUTION: Write
as the sum or difference of angle measures with cosines that you know.
96. sin
SOLUTION: Write
as the sum or difference of angle measures with sines that you know.
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Page 53
5-5 Multiple-Angle and Product-to-Sum Identities
96. sin
SOLUTION: as the sum or difference of angle measures with sines that you know.
Write
97. GARDENING Eliza is waiting for the first day of spring in which there will be 14 hours of daylight to start a
flower garden. The number of hours of daylight H in her town can be modeled by H = 11.45 + 6.5 sin (0.0168d –
1.333), where d is the day of the year, d = 1 represents January 1, d = 2 represents January 2, and so on. On what
day will Eliza begin gardening?
SOLUTION: Let H = 14, and solve for d.
Therefore, Eliza will begin gardening on the 104th day of the year. Because there are 31 days in January, 28 days
in February (on a non-leap year), and 31 days in March, the 104th day corresponds to April 14.
Find the exact value of each expression. If undefined, write undefined.
98. SOLUTION: Add
to
. eSolutions Manual - Powered by Cognero
Page 54
5-5
Therefore, Eliza willand
beginProduct-to-Sum
gardening on the 104th
day of the year. Because there are 31 days in January, 28 days
Multiple-Angle
Identities
in February (on a non-leap year), and 31 days in March, the 104th day corresponds to April 14.
Find the exact value of each expression. If undefined, write undefined.
98. SOLUTION: Add
to
. 99. tan 210
SOLUTION: 210 corresponds to the point (x, y) =
on the unit circle.
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Page 55
5-5 Multiple-Angle and Product-to-Sum Identities
99. tan 210
SOLUTION: 210 corresponds to the point (x, y) =
on the unit circle.
100. SOLUTION: Add an integer multiple of
to
. eSolutions Manual - Powered by Cognero
101. cos (–3780 )
Page 56
5-5 Multiple-Angle and Product-to-Sum Identities
100. SOLUTION: Add an integer multiple of
to
. 101. cos (–3780 )
SOLUTION: Add an integer multiple of
to –3780° .
Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and
where the function is increasing or decreasing.
102. SOLUTION: Evaluate the function for several x-values in its domain.
x - Powered
f(x)
eSolutions Manual
by Cognero
−6
−4
−0.66
−0.50
Page 57
5-5 Multiple-Angle and Product-to-Sum Identities
Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and
where the function is increasing or decreasing.
102. SOLUTION: Evaluate the function for several x-values in its domain.
x
−6
−4
−2
0
2
4
6
f(x)
−0.66
−0.50
−0.32
0
−0.32
−0.50
−0.66
Use these points to construct a graph.
All values of x are included in the graph, so the function exists for all values of x, and D = (−∞, ∞).
The only values of y that are included in the graph are negative infinity through 0, so R = (−∞, 0].
The only time the graph intersects the axes is when it goes through the origin, so the x- and y-intercepts are both 0.
The y-values approach negative infinity as x approaches negative or positive infinity, so
.
and
There are no breaks, holes, or gaps in the graph, so it is continuous for all real numbers.
As you read the graph from left to right, it is going up from negative infinity to 0, and then going down from 0 to
positive infinity, so the graph is increasing on (−∞, 0) and decreasing on (0, ∞).
103. SOLUTION: Evaluate the function for several x-values in its domain.
x
f(x)
0
0
1
4
2
9.51
3
15.8
4
22.6
eSolutions Manual - Powered by Cognero
5
29.9
6
37.6
Page 58
There are no breaks, holes, or gaps in the graph, so it is continuous for all real numbers.
As you read the graph from left to right, it is going up from negative infinity to 0, and then going down from 0 to
positive infinity, so the graph is increasing on (−∞, 0) and decreasing on (0, ∞).
5-5 Multiple-Angle and Product-to-Sum Identities
103. SOLUTION: Evaluate the function for several x-values in its domain.
x
0
1
2
3
4
5
6
f(x)
0
4
9.51
15.8
22.6
29.9
37.6
Use these points to construct a graph.
Since the denominator of the power is even, the domain must be restricted to nonnegative values, so D = (0, ∞).
The only values of y that are included in the graph are 0 through infinity, so R = [0, ∞).
The only time the graph intersects the axes is when it goes through the origin, so the x- and y-intercepts are both 0.
The y-values approach positive infinity as x approaches positive infinity, so .
There are no breaks, holes, or gaps in the graph, so it is continuous over the domain [0, ∞).
As you read the graph from left to right, it is going up from 0 to positive infinity, so the graph is increasing on (0,
∞).
104. f (x) = −3x6
SOLUTION: Evaluate the function for several x-values in its domain.
x
−1.5
−1
−0.5
0
0.5
1
1.5
f(x)
−34.2
−3
−0.05
0
−0.05
−3
−34.2
eSolutions Manual - Powered by Cognero
Use these points to construct a graph.
Page 59
There are no breaks, holes, or gaps in the graph, so it is continuous over the domain [0, ∞).
As you read the graph from left to right, it is going up from 0 to positive infinity, so the graph is increasing on (0,
5-5 Multiple-Angle
and Product-to-Sum Identities
∞).
104. f (x) = −3x6
SOLUTION: Evaluate the function for several x-values in its domain.
x
−1.5
−1
−0.5
0
0.5
1
1.5
f(x)
−34.2
−3
−0.05
0
−0.05
−3
−34.2
Use these points to construct a graph.
The function is a monomial with an even degree and a negative value for a.
All values of x are included in the graph, so the function exists for all values of x, and D = (−∞, ∞).
The only values of y that are included in the graph are negative infinity through 0, so R = (−∞, 0].
The only time the graph intersects the axes is when it goes through the origin, so the x- and y-intercepts are both 0.
The y-values approach negative infinity as x approaches negative or positive infinity, so
and
.
There are no breaks, holes, or gaps in the graph, so it is continuous for all real numbers.
As you read the graph from left to right, it is going up from negative infinity to 0, and then going down from 0 to
positive infinity, so the graph is increasing on (−∞, 0) and decreasing on (0, ∞).
105. f (x) = 4x5
SOLUTION: Evaluate the function for several x-values in its domain.
x
f(x)
−3
−972
−2
−128
−1
−4
0
0
1
4
2 - Powered
128
eSolutions Manual
by Cognero
3
972
Page 60
There are no breaks, holes, or gaps in the graph, so it is continuous for all real numbers.
As you read the graph
fromProduct-to-Sum
left to right, it is goingIdentities
up from negative infinity to 0, and then going down from 0 to
5-5 Multiple-Angle
and
positive infinity, so the graph is increasing on (−∞, 0) and decreasing on (0, ∞).
105. f (x) = 4x5
SOLUTION: Evaluate the function for several x-values in its domain.
x
−3
−2
−1
0
1
2
3
f(x)
−972
−128
−4
0
4
128
972
Use these points to construct a graph.
The function is a monomial with an odd degree and a positive value for a.
All values of x are included in the graph, so the function exists for all values of x, and D = (−∞, ∞).
All values of y are included in the graph, so the function exists for all values of x, and R = (−∞, ∞).
The only time the graph intersects the axes is when it goes through the origin, so the x- and y-intercepts are both 0.
The y-value approaches negative infinity as x approaches negative infinity, and positive infinity as x approaches
positive infinity, so
and
.
There are no breaks, holes, or gaps in the graph, so it is continuous for all real numbers.
As you read the graph from left to right, it is going up from negative infinity to positive infinity, so the graph is
increasing on (−∞, ∞).
106. REVIEW Identify the equation for the graph.
A y = 3 cos 2
By =
cos 2
C y = 3 cos
Dy =
cos eSolutions Manual - Powered by Cognero
Page 61
5-5
There are no breaks, holes, or gaps in the graph, so it is continuous for all real numbers.
As you read the graph
fromProduct-to-Sum
left to right, it is goingIdentities
up from negative infinity to positive infinity, so the graph is
Multiple-Angle
and
increasing on (−∞, ∞).
106. REVIEW Identify the equation for the graph.
A y = 3 cos 2
By =
cos 2
C y = 3 cos
Dy =
cos SOLUTION: Half of the distance between the maximum and minimum values of the function appears to be about
a=
. The graph appears to complete one cycle on the interval [0,
], so the period is
of a unit, so
. Use the period to find
b.
So, one sinusoidal equation that corresponds to the graph is y =
cos 2 . Therefore, the correct answer is B.
107. REVIEW From a lookout point on a cliff above a lake, the angle of depression to a boat on the water is 12 . The
boat is 3 kilometers from the shore just below the cliff. What is the height of the cliff from the surface of the water
to the lookout point?
F
G
H
J 3 tan 12°
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SOLUTION: Page 62
The angle of elevation from the boat to the top of the cliff is congruent to the angle of depression from the top of
the cliff to the boat because they are alternate interior angles of parallel lines.
5-5 Multiple-Angle
and Product-to-Sum
So, one sinusoidal equation
that corresponds to theIdentities
graph is y =
cos 2 . Therefore, the correct answer is B.
107. REVIEW From a lookout point on a cliff above a lake, the angle of depression to a boat on the water is 12 . The
boat is 3 kilometers from the shore just below the cliff. What is the height of the cliff from the surface of the water
to the lookout point?
F
G
H
J 3 tan 12°
SOLUTION: The angle of elevation from the boat to the top of the cliff is congruent to the angle of depression from the top of
the cliff to the boat because they are alternate interior angles of parallel lines.
Because the side adjacent to the 12º angle of elevation is given, the tangent function can be used to find the height
of the cliff.
Therefore, the correct answer is J.
108. FREE RESPONSE Use the graph to answer each of the following.
a. Write a function of the form f (x) = a cos (bx + c) + d that corresponds to the graph.
b. Rewrite f (x) as a sine function.
c. Rewrite f (x) as a cosine function of a single angle.
d. Find all solutions of f (x) = 0.
e . How do the solutions that you found in part d relate to the graph of f (x)?
SOLUTION: a. Sample answer: Half of the distance between the maximum and minimum values of the function appears to be 2
units, so a = 2. The graph appears to complete one cycle on the interval [0, ], so the period is . Use the period
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Page 63
to find b.
c. Rewrite f (x) as a cosine function of a single angle.
d. Find all solutions of f (x) = 0.
e . How do the solutions that you found in part d relate to the graph of f (x)?
5-5 Multiple-Angle
and Product-to-Sum Identities
SOLUTION: a. Sample answer: Half of the distance between the maximum and minimum values of the function appears to be 2
units, so a = 2. The graph appears to complete one cycle on the interval [0, ], so the period is . Use the period
to find b.
So, f (x) = 2 cos 2x.
b. Sample answer: Because sine and cosine are confunctions, cos
= 2 cos 2x can be rewritten in terms of sine as f (x) = 2 · sin = sin So, substituting 2x for
, f (x)
or f (x) = 2 sin
c. Sample answer: Use the double-angle identity for cosine to write 2 cos 2x in terms of a single angle.
d. Sample answer: Let f (x) = 0 and solve for x.
The period of f (x) = 2 cos 2x is
, so the solutions are x =
+ nπ, where n is an integer.
e . Sample answer: The solutions or zeros of a function are the x-intercepts of the graph of that function. So, the
2
solutions of f (x) = 4 cos x – 2 are the x-intercepts of the graph of f (x).
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