Answer
$4-8~sin^2x=4~cos~2x$
Work Step by Step
$sin^2x+cos^2x=1$
$cos^2x=1-sin^2x$
Use: $cos~2x=cos^2x-sin^2x=1-2~sin^2x$
$4-8~sin^2x=4(1-2~sin^2x)=4~cos~2x$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 7 - 7.5 - Multiple-Angle and Product-to-Sum Formulas - 7.5 Exercises - Page 548: 19
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