Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 457: 9

Answer

$\dfrac{\sqrt2}{2}$

Work Step by Step

Note that $\sin^{-1}{\left(\frac{\sqrt2}{2}\right)}=\dfrac{\pi}{4}$, because $\sin{\left(\frac{\pi}{4}\right)}=\frac{\sqrt2}{2}$ and $\dfrac{\pi}{4}$ is in the range of $\sin^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$ Thus, $\cos{\left(\sin^{-1}{\left(\dfrac{\sqrt2}{2}\right)}\right)}=\cos{\dfrac{\pi}{4}}=\dfrac{\sqrt2}{2}$
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