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: 59


$\left(\sqrt 2, -\frac{\pi}{4}\right)$

Work Step by Step

Let the polar coordinates be $(r,\theta)$. Then, $r=\sqrt {x^{2}+y^{2}}=\sqrt {(1)^{2}+(-1)^{2}}=\sqrt {2}$ $(1,-1)$ lies in quadrant IV, so to find the angle, use the formula $\theta=\tan^{-1}\frac{y}{x}$ to obtain: $$\theta=\tan^{-1}{\left(\frac{-1}{1}\right)}=-\frac{\pi}{4}$$ A set of polar coordinates for the point $(1,-1)$ is $(\sqrt 2, -\frac{\pi}{4})$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.