Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 55

Answer

$\frac{1}{2}(\sin 5x-\sin x)$

Work Step by Step

Using the product-to-sum formulas on page 559, $\sin u\cos v=\frac{1}{2}[\sin(u+v)+\sin(u-v)]$: $\sin 2x\cos 3x$ =$\frac{1}{2}[\sin(2x+3x)+\sin(2x-3x)]$ =$\frac{1}{2}[\sin 5x+\sin (-x)]$ =$\frac{1}{2}(\sin 5x-\sin x)$
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