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: 91



Work Step by Step

The range of $arcsin~x$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$. So, $arcsin(-\frac{1}{2})=-\frac{\pi}{6}$ because $sin(-\frac{\pi}{6})=-\frac{1}{2}$ and $-\frac{\pi}{2}\leq-\frac{\pi}{6}\leq\frac{\pi}{2}$
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