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: 55



Work Step by Step

Let $u=arcsin\frac{5}{13}$. Then: $sin~u=\frac{5}{13}$ The range $arcsin~x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $arcsin\frac{5}{13}\gt0$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant) $cos^2u+sin^2u=1$ $cos^2u=1-sin^2u$ $cos^2u=1-(\frac{5}{13})^2=1-\frac{25}{169}=\frac{144}{169}$ $cos~u=\frac{12}{13}~~$ (First Quadrant) $sec(arcsin\frac{5}{13})=sec~u=\frac{1}{cos~u}=\frac{13}{12}$
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