Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.1 Polar Cordinates - 8.1 Assess Your Understanding - Page 592: 62



Work Step by Step

Solve for $r$ using the formula $r=\sqrt {x^{2}+y^{2}}$ to obtain: $$r=\sqrt {(-2)^{2}+(-2\sqrt {3})^{2}}=4$$ Solve for the angle using the formula $\alpha= \tan^{-1}(\frac{y}{x})$ to obtain: $$\alpha=\tan^{-1}\left(\frac{-2\sqrt {3}}{-2}\right)=\frac{\pi}{3}$$ Since $(-2, -2\sqrt {3})$ is in the third quadrant, then the angle measurement is: $$\theta=\alpha+\pi=\frac{\pi}{3}+\pi=\frac{4\pi}{3}$$ Thus, the polar coordinates for the point $(-2, -2\sqrt {3})$ can be given by $(4,\frac{4\pi}{3})$.
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