Precalculus (10th Edition)

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

Chapter 7 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 505: 1

Answer

$\frac{\pi}{2}$

Work Step by Step

We know that since $\sin (\frac{\pi}{2})=1$, then by the definition of the inverse sine function: $\sin^{-1}(1)=\frac{\pi}{2}$
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