Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 646: 111

Answer

The exact value of expression is $\frac{\pi }{3}$.

Work Step by Step

By using the inverse property, ${{\sin }^{-1}}\left( \sin x \right)=x$ Therefore, the above property is applicable for every $x$ in $\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$. Then, simplify ${{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)$. We see that $\frac{2\pi }{3}$ is not in the interval. So, first simplify $\left( \sin \frac{2\pi }{3} \right)$. Use the property: $\left( \sin \frac{\pi }{2}+\theta \right)=\cos \theta $ Put the $\left( \sin \frac{2\pi }{3} \right)$ in the above property. So, $\begin{align} & \left( \sin \frac{\pi }{2}+\frac{\pi }{6} \right)=\cos \frac{\pi }{6} \\ & =\frac{\sqrt{3}}{2} \end{align}$ Finally, simplify ${{\sin }^{-1}}\frac{\sqrt{3}}{2}=\frac{\pi }{3}$ Thus, ${{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)=\frac{\pi }{3}$.
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