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: 25

Answer

$(\sqrt 2,\sqrt 2)$
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Work Step by Step

Step 1. See plot; move $\theta=\frac{5\pi}{4}$, then $r=-2$ to get the point. Step 2. Using the formulas $x=r\ cos\theta$ and $y=r\ sin\theta$, we have $x=-2cos\frac{5\pi}{4}=\sqrt 2$ and $y=-2sin\frac{5\pi}{4}=\sqrt 2 $, giving a point $(\sqrt 2,\sqrt 2)$
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