## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 6 - Review Questions - Page 501: 96

#### Answer

$cos^{-1}(\frac{\sqrt 3}{2})=\frac{\pi}{6}$

#### Work Step by Step

The range of $cos^{-1}~x$ is $[0,\pi]$. So, $cos^{-1}~(\frac{\sqrt 3}{2})=\frac{\pi}{6}$, because $cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$ and $0\leq\frac{\pi}{6}\leq\pi$

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