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Quiz 7
Math 1151, Precalculus 2
Name:
Summer 2012, Section 002
Problem 1.(10 points.) Match the left side trigonometric expressions with the correct
formula on the right side.
• sin(2θ) = 2 sin(θ) cos(θ)
• cos(2θ) = cos2 (θ) − sin2 (θ)
q
θ
• cos( 2 ) = ± 1+cos(θ)
2
• tan( 2θ ) =
• sin( 2θ ) =
1−cos(θ)
sin(θ)
q
± 1−cos(θ)
2
Problem 2.(10 points) If cos(θ) =
3
5
where − π2 < θ < 0, find sin(2θ).
Since we already have the value of cosine we need to find the value of sine.
√
sin θ = − 1 − cos2 θ = − 45 . SO we find that sin(2θ) = − 24
.
25
Problem 3.(10 points) Solve the equation below on the interval [0, 2π).
√
3 sin(θ) + cos(θ) = 1
We want to utilize the identity
sin(θ+φ) = cos φ sin θ+sin φ cos θ. So first we divide both
q
√
√ 2
side of the equation by
3 + 12 = 2 obtaining 23 sin(θ) + 12 cos(θ) = 21 . Equating
√
this with the identity above we must set cos φ =
that φ =
θ+
π
6
=
π
6
π
.
6
3
2
and sin φ =
Now the equation condenses into sin(θ + π6 ) =
+ 2πk or θ +
π
6
=
5π
6
+ 2πk. Then θ = 2πk or θ =
2π
3
1
.
2
1
.
2
So it must be that
+ 2πk for k an integer.
The solutions that lie in [0, 2π) are 0, 2π
.
3
Extra credit.(8 points) Let f (x) = 2x2 − x + 18 . Complete the square.
f (x) = 2(x2 − 1/2x) + 1/8
= 2(x − 1/4)2 + 1/8 − 2(1/4)2
= 2(x − 1/4)2 .
Then we find