Algebra and Trigonometry 10th Edition

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

Chapter 7 - 7.2 - Verifying Trigonometric Identities - 7.2 Exercises - Page 520: 37


The identity is verified. $sin^{\frac{1}{2}}x~cos~x-sin^{\frac{5}{2}}x~cos~x=cos^3x\sqrt {sin~x}$

Work Step by Step

$cos^2x+sin^2x=1$ $cos^2x=1-sin^2x$ $sin^{\frac{1}{2}}x~cos~x-sin^{\frac{5}{2}}x~cos~x=sin^{\frac{1}{2}}x~cos~x-sin^{\frac{1}{2}}x~sin^2x~cos~x=sin^{\frac{1}{2}}x~cos~x(1-sin^2x)=sin^{\frac{1}{2}}x~cos~x~cos^2x=sin^{\frac{1}{2}}x~cos^3x=cos^3x\sqrt {sin~x}$
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