## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 34

#### Answer

$$\sin4x=4\sin x\cos x-8\sin^3x\cos x$$ The equation is verified to be an identity as below.

#### Work Step by Step

$$\sin4x=4\sin x\cos x-8\sin^3x\cos x$$ We would tackle from the right side. $$X=4\sin x\cos x-8\sin^3x\cos x$$ $$X=4\sin x\cos x-(4\sin x\cos x)\times(2\sin^2x)$$ $$X=4\sin x\cos x(1-2\sin^2x)$$ $$X=2\times(2\sin x\cos x)\times(1-2\sin^2x)$$ - From Double-Angle Identities for sine and cosine: $$2\sin x\cos x=\sin2x$$ $$1-2\sin^2x=\cos2x$$ Therefore, $$X=2\sin2x\cos2x$$ - Finally, we can still apply Double-Angle Identity for sine ($2\sin\theta\cos\theta=\sin2\theta$) to $X$, but as $\theta=2x$. That means, $$X=\sin(2\times2x)$$ $$X=\sin4x$$ That concludes, $$\sin4x=4\sin x\cos x-8\sin^3x\cos x$$ 2 sides are equal, and the equation is an identity.

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