Algebra and Trigonometry 10th Edition

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

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 484: 50



Work Step by Step

$-\frac{3\pi}{2}$ does not lie in the range of $arccos~x$: ($0\leq x\leq \pi$). But, we know that the period of $cos~x=2\pi$ and that $-\frac{3\pi}{2}=\frac{\pi}{2}-2\pi$. So: $cos(-\frac{3\pi}{2})=cos(\frac{\pi}{2})$. Now, $\frac{\pi}{2}$ lies in the range of $arccos~x$. $arccos[cos(-\frac{3\pi}{2})]=\frac{\pi}{2}$
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