Precalculus (10th Edition)

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

Chapter 7 - Trigonometry and Periodic Functions - 7.1 The Inverse Sine, Cosine, and Tangent Functions - 7.1 Assess Your Understanding - Page 451: 40



Work Step by Step

We know that $f^{-1}(f(x))=x$. Hence, $\sin^{-1}\left(\sin(\frac{-\pi}{10})\right)=-\dfrac{\pi}{10}.$ We also need that $ \frac{-\pi}{2} \leq x\leq \frac{\pi}{2}$, which is true here.
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