Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 960: 32


$\dfrac{1}{2} \sin \dfrac{2x}{3}=\sin \dfrac{x}{3} \cos \dfrac{x}{3}$

Work Step by Step

We need to verify that $\dfrac{1}{2} \sin \dfrac{2x}{3}=\sin \dfrac{x}{3} \cos \dfrac{x}{3}$ We need to recall that: $\sin 2 \theta =2 \sin \theta \cos \theta $ $\dfrac{1}{2} ( 2 \sin \dfrac{x}{3}\cos \dfrac{x}{3}) =\sin \dfrac{x}{3} \cos \dfrac{x}{3}$ Thus, it has been proven that $\sin \dfrac{x}{3} \cos \dfrac{x}{3}=\sin \dfrac{x}{3} \cos \dfrac{x}{3}$
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