Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 785: 116



Work Step by Step

Formula of $\sin 2x$ is $\sin 2x=2\sin x\cos x$. It is known that, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Divide both sides by ${{\sin }^{2}}x+{{\cos }^{2}}x$ in the above equation: $\begin{align} & \frac{\sin 2x}{{{\sin }^{2}}x+{{\cos }^{2}}x}=\frac{2\sin x\cos x}{{{\sin }^{2}}x+{{\cos }^{2}}x} \\ & \frac{\sin 2x}{1}=\frac{2\sin x\cos x}{{{\sin }^{2}}x+{{\cos }^{2}}x} \end{align}$ Divideby ${{\cos }^{2}}x$ in nominator as well as denominator and get, $\begin{align} & \sin 2x=\frac{\left( \frac{2\sin x\cos x}{{{\cos }^{2}}x} \right)}{\left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x} \right)} \\ & =2\frac{\left( \frac{2\sin x}{\cos x} \right)}{\left( 1+\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)} \\ & =\frac{2\tan x}{1+{{\tan }^{2}}x} \end{align}$
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