Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 689: 5

Answer

The required solution is $\frac{1}{2}\left[ \sin 3x-\sin x \right]$

Work Step by Step

One of the product-to-sum formulas is $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)+\sin \left( \alpha -\beta \right) \right]$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is $x$ and the value of $\beta $ is $2x$. Now, the expression can be evaluated as provided below: $\begin{align} & \sin x\cos 2x=\frac{1}{2}\left[ \sin \left( x+2x \right)+\sin \left( x-2x \right) \right] \\ & =\frac{1}{2}\left[ \sin 3x+\sin \left( -x \right) \right] \end{align}$ Thus, applying the even-odd identity, which is $sin(-x)=-\sin x$, the expression can be further evaluated as given below: $\frac{1}{2}\left[ \sin 3x+\sin \left( -x \right) \right]=\frac{1}{2}\left[ \sin 3x-\sin x \right]$
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