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: 33


$2 \sin^2 x \tan (\dfrac{x}{2})=2\sin x - \sin 2x$

Work Step by Step

The half -angle formula can be defined as: $\tan (\dfrac{x}{2})=\dfrac{1- \cos x}{\sin x}$ Now, we have $ 2 \sin^2 x \tan (\dfrac{x}{2})=2 \sin^2 x \times \dfrac{1- \cos x}{\sin x}$ or, $=\dfrac{2 \sin^2 x-2 \sin^2 x \cos x}{\sin x}$ or, $=2 \sin x - 2\sin x \cos x$ or, $= 2\sin x - \sin 2x$ Thus, it has been proven that $2 \sin^2 x \tan (\dfrac{x}{2})=2\sin x - \sin 2x$
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