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: 100


The exact value of the provided expression is $\frac{\sqrt{2}}{2}$.

Work Step by Step

Let us assume $\theta ={{\sin }^{-1}}\left( \frac{\sqrt{2}}{2} \right)$. Then, we have $\sin \theta =\left( \frac{\sqrt{2}}{2} \right)$ We know that for the sine function, the interval for the angle is $\left[ -\frac{\pi }{2},\frac{\pi }{2} \right]$. So, the only angle that satisfies $\sin \theta =\left( \frac{\sqrt{2}}{2} \right)$ is $\frac{\pi }{4}$. Hence, $\theta =\left( \frac{\pi }{4} \right)$ and the exact value of the given expression is $\begin{align} & \cos \left( {{\sin }^{-1}}\frac{\sqrt{2}}{2} \right)=\cos \left( \theta \right) \\ & =\cos \left( \frac{\pi }{4} \right) \\ & =\frac{\sqrt{2}}{2} \end{align}$
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