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: 102


The exact value of the provided expression is $-\frac{\sqrt{3}}{3}$.

Work Step by Step

Let us assume $\theta ={{\sin }^{-1}}\left( -\frac{1}{2} \right)$. Then $\sin \theta =-\frac{1}{2}$ We know that for the sine function, the interval for the angle is $\left[ -\frac{\pi }{2},\frac{\pi }{2} \right]$. Therefore, the only angle that satisfies $\sin \theta =-\frac{1}{2}$ is $\left( -\frac{\pi }{6} \right)$. Hence, $\theta =\left( -\frac{\pi }{6} \right)$ and, The exact value of the given expression is $\begin{align} & \tan \left[ {{\sin }^{-1}}\left( -\frac{1}{2} \right) \right]=\tan \left( \theta \right) \\ & =\tan \left( -\frac{\pi }{6} \right) \\ & =-\frac{\sqrt{3}}{3} \end{align}$
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