Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - Review Exercises - Page 552: 26


The identity is verified. $cos^3x~sin^2x=(sin^2x-sin^4x)~cos~x$

Work Step by Step

$sin^2x+cos^2x=1$ (Pythagorean identity) $cos^2x=1-sin^2x$ $cos^3x~sin^2x=cos~x~cos^2x~sin^2x=cos~x~(1-sin^2x)~sin^2x=cos~x~(sin^2x-sin^4x)=(sin^2x-sin^4x)~cos~x$
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