Answer
$-\dfrac{\pi}{10}$
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.
Precalculus (10th Edition)
by Sullivan, Michael
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
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