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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