Answer
$6~cos^2x-3=3~cos~2x$
Work Step by Step
$sin^2x+cos^2x=1$
$cos^2x-1=-sin^2x$
Use: $cos~2x=cos^2x-sin^2x=2~cos^2x-1$
$6~cos^2x-3=3(2~cos^2x-1)=3~cos~2x$
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: 17
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
