Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 681: 77

Answer

See the explanation below.

Work Step by Step

We use the trigonometric identity: $2{{\cos }^{2}}x-1=\cos 2x$. Now, the left side can be written as: $\begin{align} & \sin x(4{{\cos }^{2}}x-1)=\sin x(2{{\cos }^{2}}x+2{{\cos }^{2}}x-1) \\ & =\sin x(2{{\cos }^{2}}x+\cos 2x) \\ & =2\sin x{{\cos }^{2}}x+\sin x\cos 2x \\ & =2\sin x\cos x\cos x+\sin x\cos 2x \end{align}$ We use the trigonometric identities, $2\sin x\cos x=\sin 2x$ And $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ Simplified further, $\begin{align} & 2\sin x\cos x\cos x+\sin x\cos 2x=\sin 2x\cos x+\sin x\cos 2x \\ & =\sin (x+2x) \\ & =\sin 3x \end{align}$ Thus, the left side of the equation is equal to $\sin 3x$. Hence, the left side is equal to $\sin 3x$.
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