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: 25


The identity is verified. $sin^5x~cos^2x=(cos^2x-2~cos^4x+cos^6x)sin~x$

Work Step by Step

$sin^2x+cos^2x=1$ (Pythagorean identity) $sin^2x=1-cos^2x$ $sin^5x~cos^2x=sin~x~sin^4x~cos^2x=sin~x~(sin^2x)^2~cos^2x=sin~x~(1-cos^2x)^2~cos^2x=sin~x~(1-2~cos^2x+cos^4x)~cos^2x=sin~x~(cos^2x-2~cos^4x+cos^6x)=(cos^2x-2~cos^4x+cos^6x)sin~x$
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