Answer
$cos~4x=4~cos^2x-3$
Work Step by Step
$cos~4x=cos~2(2x)=cos^22x-sin^22x=cos^22x+cos^22x-1=2~cos^22x-1=2(cos^2x-sin^2x)-1=2(cos^2x+cos^2x-1)-1=4~cos^2x-3$
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: 25
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