Stitz-Zeager_College_Algebra_e-book

6 hooked on conics again 841 2 2 a r 1cos is a

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Unformatted text preview: ometrically. 11.4 Polar Coordinates 789 y x 4 = −3 = − 4 , and since this isn’t the tangent of one the common angles, we resort to using 3 the arctangent function. Using a reference angle approach,5 we find α = arctan 4 is the 3 reference angle for θ. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π , we choose θ = π − arctan 4 radians. Hence, our answer is (r, θ) = 5, π − arctan 4 ≈ (5, 2.21). To 3 3 check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos π − arctan 4 and sin π − arctan 4 . These are good review exercises and are hence 3 3 4 4 left to the reader. We find cos π − arctan 4 = − 3 and sin π − arctan 3 = 5 , so that 3 5 4 x = r cos(θ) = (5) − 3 = −3 and y = r sin(θ) = (5) 5 = 4 which confirms our answer. 5 y y S θ= 3π 2 θ = π − arctan x 4 3 x R R has rectangular coordinates (0, −3) π R has polar coordinates 3, 32 S has rectangular coordinates (−3, 4) 4 S has polar coordinates 5, π − arctan 3 Now th...
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