## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 960: 31

#### Answer

$\sin 3 \theta =\sin \theta (4 \cos^2 \theta-1)$

#### Work Step by Step

Re-arrange as: $\sin ( 2 \theta +\theta) = \sin 2 \theta \cos \theta +\cos 2 \theta \sin \theta$ or, $=2 \sin \theta \cos \theta \times \cos \theta +(1-2 \sin^2 \theta) \times \sin \theta$ or, $=\sin \theta ( 2 \cos^2 \theta+1-2 \sin^2 \theta)$ or, $=\sin \theta [ 2 \cos^2 \theta+1-2 (1-\cos^2 \theta)]$ Thus, it has been proven that $\sin 3 \theta =\sin \theta (4 \cos^2 \theta-1)$

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