Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Review Exercises - Page 798: 32


$(3 \sqrt 2, \dfrac{7 \pi}{4})$

Work Step by Step

Here, $r=\sqrt {x^2+y^2}=\sqrt{(3)^2+(-3)^2}=3 \sqrt 2$ $\tan \theta =\dfrac{y}{x}$ and $\theta=arctan[\dfrac{3}{-3}]=arctan {-1}$ Thus, $\theta =\dfrac{7 \pi}{4}$ Hence, $(3 \sqrt 2, \dfrac{7 \pi}{4})$
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.