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: 51



Work Step by Step

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