Stitz-Zeager_College_Algebra_e-book

Theorem 1113 given constants 0 e 0 and the graph of

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Unformatted text preview: y that tan(θ) = x = −22 3 = − 3. This tells us θ has a reference angle of π , and since P 3 lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π , π π π so we choose θ = 53 . Hence, our answer is 4, 53 . To check, we convert (r, θ) = 4, 53 1 5π back to rectangular coordinates and we find x = r cos(θ) = 4 cos 3 = 4 2 = 2 and √ √ π y = r sin(θ) = 4 sin 53 = 4 − 23 = −2 3, as required. 2. The point √(−3, −3) lies in Quadrant III. Using x = y = −3, we get r2 = (−3)2√ (−3)2 = 18 Q + √ so r = ± 18 = ±3 2. Since we are asked for r ≥ 0, we choose r = 3 2. We find −3 tan(θ) = −3 = 1, which means θ has a reference angle of π . Since Q lies in Quadrant III, 4 π we choose θ = 54 , which satisfies the requirement that 0 ≤ θ < 2π . Our final answer is √ √π √ √ π (r, θ) = 3 2, 54 . To check, we find x = r cos(θ) = (3 2) cos 54 = (3 2) − 22 = −3 √ √ √ π and y = r sin(θ) = (3 2) sin 54 = (3 2) − 22 = −3, so we are done. y y θ=...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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