5-5 Multiple-Angle and Product-to-Sum Identities

5-5 Multiple-Angle and Product-to-Sum Identities
Find the values of sin 2 , cos 2 , and tan 2
2. tan
= for the given value and interval.
, (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
ANSWER: ;
4. sin
=
;
,
SOLUTION: Since
on the interval , one point on the terminal side of
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
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has y-coordinate −7 and a distance
or .
Now use the double-angle identities for
and cos 2 .
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5-5 Multiple-Angle and Product-to-Sum Identities
Use the definition of tangent to find tan 2 .
ANSWER: 6. tan
= ,
SOLUTION: If tan
= , then tan
= x-coordinate 1 and y-coordinate
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. Since
on the interval , one point on the terminal side of θ has
as shown. The distance from the point to the origin is or 2.
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SOLUTION: If tan
= , then tan
= on the interval . Since
, one point on the terminal side of θ has
5-5 Multiple-Angle and Product-to-Sum Identities
x-coordinate 1 and y-coordinate
Using this point, we find that
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 .
ANSWER: Solve each equation on the interval [0, 2 ].
= cos
10. cos 2
SOLUTION: eSolutions Manual - Powered by Cognero
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ANSWER: 5-5 Multiple-Angle and Product-to-Sum Identities
Solve each equation on the interval [0, 2 ].
10. cos 2 = cos
SOLUTION: or =
and θ =
when =
and = when
On the interval [0, 2π),
and cos
= 1 when
= 0.
ANSWER: 0,
,
13. sin 2
csc
= 1
SOLUTION: On the interval [0, 2π), cos
= .
ANSWER: ,
14. cos 2
+ 4 cos
= –3
SOLUTION: On the interval [0, 2π), cos
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ANSWER: = −1 when
= .
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On the interval [0, 2π), cos
= when =
and = .
ANSWER: 5-5 Multiple-Angle
and Product-to-Sum Identities
,
14. cos 2
+ 4 cos
= –3
SOLUTION: On the interval [0, 2π), cos
= −1 when
= .
ANSWER: 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.
ANSWER: 30. cos
SOLUTION: Notice that
is half of . Therefore, apply the half-angle identity for cosine, noting that since
I, its cosine is positive.
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lies in Quadrant Page 6
ANSWER: 5-5 Multiple-Angle and Product-to-Sum Identities
30. cos
SOLUTION: Notice that
is half of . Therefore, apply the half-angle identity for cosine, noting that since
I, its cosine is positive.
lies in Quadrant ANSWER: 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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ANSWER: Page 7
ANSWER: 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.
ANSWER: 32. sin
SOLUTION: Notice that
. Therefore, apply the half-angle identity for sine, noting that since
is half of Quadrant II, its sine is positive.
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ANSWER: lies in Page 8
ANSWER: 5-5 Multiple-Angle and Product-to-Sum Identities
32. sin
SOLUTION: Notice that
. Therefore, apply the half-angle identity for sine, noting that since
is half of Quadrant II, its sine is positive.
lies in ANSWER: Solve each equation on the interval [0, 2 ].
34. tan
= sin SOLUTION: On the interval [0, 2 ),
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when
. Check this solution in the original equation.
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ANSWER: 5-5 Multiple-Angle and Product-to-Sum Identities
Solve each equation on the interval [0, 2 ].
34. tan
= 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.
ANSWER: 0
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