## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 236: 5

#### Answer

$$4\sin\frac{\pi}{3}\cos\frac{\pi}{3}=\sqrt3$$ 5 would be matched with F.

#### Work Step by Step

$$X=4\sin\frac{\pi}{3}\cos\frac{\pi}{3}$$ $$X=2\times(2\sin\frac{\pi}{3}\cos\frac{\pi}{3})$$ From the double-angle identities: $$2\sin A\cos A=\sin2A$$ Thus $2\sin\frac{\pi}{3}\cos\frac{\pi}{3}$ can be seen here as $2\sin A\cos A$ with $A=\frac{\pi}{3}$. Therefore, $$X=2\times\Big[\sin\Big(2\times\frac{\pi}{3}\Big)\Big]$$ $$X=2\sin\frac{2\pi}{3}$$ $|\sin\frac{2\pi}{3}|=\sin\frac{\pi}{3}$. As $\frac{2\pi}{3}$ is in quadrant II, in which $\sin\theta\gt0$, so $\sin\frac{2\pi}{3}=\sin\frac{\pi}{3}=\frac{\sqrt3}{2}$. $$X=2\times\frac{\sqrt3}{2}$$ $$X=\sqrt3$$ So, $$4\sin\frac{\pi}{3}\cos\frac{\pi}{3}=\sqrt3$$ 5 would be matched with F.

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