Answer
$\frac{2x}{1+x^2}$
Work Step by Step
Let $tan^{-1}x=y$, we have $tan(y)=x, sin(y)=\frac{x}{\sqrt {1+x^2}},cos(y)=\frac{1}{\sqrt {1+x^2}}$
Thus $sin(2tan^{-1}x)=sin(2y)=2sin(y)cos(y)=\frac{2x}{1+x^2}$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 43
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