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: 7

Answer

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

Work Step by Step

One of the product-to-sum formulas is $\cos \alpha \sin \beta =\frac{1}{2}\left[ \sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right) \right]$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is $\frac{3x}{2}$ and the value of $\beta $ is $\frac{x}{2}$. Thus, the expression can be evaluated as provided below: $\begin{align} & \cos \frac{3x}{2}\sin \frac{x}{2}=\frac{1}{2}\left[ \sin \left( \frac{3x}{2}+\frac{x}{2} \right)-\sin \left( \frac{3x}{2}-\frac{x}{2} \right) \right] \\ & =\frac{1}{2}\left[ \sin \left( \frac{4x}{2} \right)-\sin \left( \frac{2x}{2} \right) \right] \\ & =\frac{1}{2}\left[ \sin 2x-\sin x \right] \end{align}$
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