Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 683: 112


$ \frac{1}{16}(5-7cos(2x)+3cos(4x)-cos(4x)cos(2x))$

Work Step by Step

Step 1. Using the formula $sin^2x=\frac{1-cos(2x)}{2}$, we have $sin^6x=(sin^2x)^3=(\frac{1-cos(2x)}{2})^3=\frac{1}{8}(1-3cos(2x)+3cos^2(2x)-cos^3(2x))$ Step 2. Using the formula $cos^2(2x)=\frac{1+cos(4x)}{2}$, we have $sin^6x=\frac{1}{8}(1-3cos(2x)+3(\frac{1+cos(4x)}{2})-(\frac{1+cos(4x)}{2})cos(2x))=\frac{1}{16}(2-6cos(2x)+3(1+cos(4x))-(1+cos(4x))cos(2x))=\frac{1}{16}(5-7cos(2x)+3cos(4x)-cos(4x)cos(2x))$
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.