Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 8



Work Step by Step

Since $\sin\frac{4\pi}{3}=-\frac{\sqrt{2}}{2}$, we have $$\sin^{-1}\left(\sin\frac{4\pi}{3}\right)=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)=-\frac{\pi}{3}.$$
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