Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 55


$$\tan (x) =\frac{\sin 2x}{ 1+ \cos 2 x}$$

Work Step by Step

We use the know identities to obtain: \begin{align*} \tan (x)&= \frac{\sin (x)}{\cos (x)}\\ &= \frac{2\sin (x)\cos (x)}{2\cos^2 (x)}\\ &= \frac{\sin 2x}{ 2\cos^2 x}\\ &= \frac{\sin 2x}{ \cos^2 x+ \cos^2 x}\\ &= \frac{\sin 2x}{ 1-\sin^2 x+ \cos^2 x}\\ &=\frac{\sin 2x}{ 1+ \cos 2 x} \end{align*}
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