## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Review Exercises - Page 646: 110

#### Answer

$\dfrac{\pi}{3}$

#### Work Step by Step

We have $\theta= \sin [\sin^{-1} \dfrac{\pi}{3}]$ $\implies \theta=\dfrac{\pi}{3}$ Because the trigonometric functions sine and arcsine will "cancel out".

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