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: 21



Work Step by Step

Assume that $\sin\theta =\frac{2}{3}$, then by solving the triangle, we have: $$\cos \theta=\frac{\sqrt{5}}{3}.$$ Now, since $\theta =\sin^{-1}\frac{2}{3}$, we have $$\cos(\sin^{-1}\frac{2}{3})=\cos\theta=\frac{\sqrt{5}}{3}.$$
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