Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 52



Work Step by Step

Let $u=arcsin\frac{4}{5}$. Then: $sin~u=\frac{4}{5}$ The range $arcsin~x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $arcsin\frac{4}{5}\gt0$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant) $cos^2u+sin^2u=1$ $cos^2u=1-sin^2u$ $cos^2u=1-(\frac{4}{5})^2$ $cos^2u=1-\frac{16}{25}=\frac{9}{25}$ $cos~u=\frac{3}{5}$ (First Quadrant) $cos(arcsin\frac{4}{5})=\frac{3}{5}$
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