## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 49

#### Answer

$$\sin4x=4\sin x\cos^3x-4\sin^3 x\cos x$$

#### Work Step by Step

$$\sin4x$$ The question asks to write $\sin4x$ in terms of a trigonometric function of $x$. We will need both Sum and Difference Identities and Double-Angle Identities to help here. $$\sin4x=\sin(2\times2x)$$ - Double-Angle Identity for sine: $\sin2A=2\sin A\cos A$ with here $A=2x$ $$\sin4x=2\sin2x\cos2x$$ - For $\sin2x$, we apply again $\sin2A=2\sin A\cos A$ with $A=x$. - For $\cos2x$, we apply Double-Angle Identity for cosine: $\cos2A=\cos^2A-\sin^2A$. $$\sin4x=2(2\sin x\cos x)(\cos^2x-\sin^2x)$$ $$\sin4x=(4\sin x\cos x)(\cos^2x-\sin^2x)$$ $$\sin4x=4\sin x\cos^3x-4\sin^3 x\cos x$$ We can stop here as all trigonometric functions are functions of $x$ now.

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