Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.7 Product-to-Sum and Sum-to-Product Formulas - 6.7 Assess Your Understanding - Page 524: 7

Answer

$\dfrac{1}{2} [\cos (2 \theta) -\cos (6 \theta) ] $

Work Step by Step

Recall the Product to Sum Identities: $a) \sin x \cos y =\dfrac{1}{2} [\sin (x-y) +\sin (x+y)] \\ b) \sin x \sin y =\dfrac{1}{2} [\cos (x-y) -\cos (x+y)] \\ c) \cos x \cos y =\dfrac{1}{2} [\cos (x-y) +\cos (x+y)]$ By identity $(b)$, we have: $\sin 4 \theta \ \sin 2 \theta=\dfrac{1}{2} [\cos (4 \theta-2 \theta) -\cos (4 \theta+2 \theta) ] \\=\dfrac{1}{2} [\cos (2 \theta) -\cos (6 \theta) ] $
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.