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: 64


$sin(arctan~x)=±\frac{x}{\sqrt {x^2+1}}$

Work Step by Step

Let $u=arctan~x~~$ (Range: $-\frac{\pi}{2}\lt u\lt\frac{\pi}{2}$). Then: $x=tan~u$ $cot~u=\frac{1}{tan~u}=\frac{1}{x}$ $csc^2u=1+cot^2u$ $csc^2u=1+(\frac{1}{x})^2$ $csc^2u=1+\frac{1}{x^2}=\frac{x^2+1}{x^2}$ $sin^2u=\frac{1}{csc^2u}=\frac{x^2}{x^2+1}$ $sin~u=±\frac{x}{\sqrt {x^2+1}}$
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