Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - Review Questions - Page 501: 94


$arccos(\frac{\sqrt 2}{2})=\frac{\pi}{4}$

Work Step by Step

The range of $arccos~x$ is $[0,\pi]$. So, $arccos(\frac{\sqrt 2}{2})=\frac{\pi}{4}$, because $cos(\frac{\pi}{4})=\frac{\sqrt 2}{2}$ and $0\leq\frac{\pi}{4}\leq\pi$
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