Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 627: 64


$\dfrac{x}{\sqrt {1+x^2}}$

Work Step by Step

Suppose $\theta =\tan^{-1} x $ This gives: $ x =\tan \theta $ Since, $ r=\sqrt {a^2+y^2}$ This implies that $ r=\sqrt {(1)^2+x^2}=\sqrt {1+x^2}$ Therefore, we have $\sin(\tan^{-1} x)=\dfrac{x}{\sqrt {1+x^2}}$
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