Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 47


$2\cos 6x\cdot\cos 2x$

Work Step by Step

Sum-to-Product: $\displaystyle \cos A+\cos B=2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$ $\cos(-A)=\cos A$ ---------- $\displaystyle \frac{A+B}{2}=\frac{4x+8x}{2}=6x$ $\displaystyle \frac{A-B}{2}=\frac{4x-8x}{2}=-2x$ $\cos A+\cos B=2\cos 6x\cos(-2x)$ $= 2\cos 6x\cdot\cos 2x$
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.