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: 1


The required value is $\frac{1}{2}\left[ \cos 4x-\cos 8x \right]$.

Work Step by Step

One of the product-to-sum formulas is $\sin \alpha sin\beta =\frac{1}{2}\left[ \cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right) \right]$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is $6x$ and the value of $\beta $ is $2x$. Thus, the expression can be evaluated as pprovided below. $\begin{align} & \sin 6xsin2x=\frac{1}{2}\left[ \cos \left( 6x-2x \right)-\cos \left( 6x+2x \right) \right] \\ & =\frac{1}{2}\left[ \cos 4x-\cos 8x \right] \end{align}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.