Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.4 - Inverse Trigonometric Functions and Right Triangles - 6.4 Exercises - Page 507: 29

Answer

$\frac{3}{5}$

Work Step by Step

Given expression is- $\cos(\sin^{-1}\frac{4}{5})$ Assuming $\sin^{-1} \frac{4}{5}$ = $u$ , we get- $\sin u$ = $\frac{4}{5}$, [$-\frac{\pi}{2}\leq u\leq \frac{\pi}{2}$] Therefore $\cos u$ will be positive. From first Pythagorean identity- $\cos u$ = + $\sqrt {1 - \sin^{2} u}$ = $\sqrt {1 - (\frac{4}{5})^{2}} $ = $\sqrt {1 - \frac{16}{25}} $ = $\sqrt { \frac{25 - 16}{25}} $ = $\sqrt { \frac{9}{25}} $ = $\frac{3}{5}$ i.e. $\cos u$ = $\frac{3}{5}$ i.e. $\cos(\sin^{-1}\frac{4}{5})$ = $\frac{3}{5}$
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