Answer
$cos~4θ~sin~6θ=\frac{1}{2}(sin~10θ+sin~2θ)$
Work Step by Step
$cos~4θ~sin~6θ=\frac{1}{2}[sin(4θ+6θ)-sin(4θ-6θ)]=\frac{1}{2}[sin(10θ)-sin(-2θ)]=\frac{1}{2}(sin~10θ+sin~2θ)$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 7 - Review Exercises - Page 553: 75
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