Answer
$7~cos(-5β)~sin~3β=\frac{7}{2}(-sin~2β+sin~8β)$
Work Step by Step
$7~cos(-5β)~sin~3β=7(\frac{1}{2})[sin(-5β+3β)-sin(-5β-3β)]=\frac{7}{2}[sin(-2β)-sin(-8β)]=\frac{7}{2}(-sin~2β+sin~8β)$
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: 50
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