## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Quiz (Sections 5.1-5.4) - Page 230: 8

#### Answer

$$\sin\Big(\frac{\pi}{3}+\theta\Big)-\sin\Big(\frac{\pi}{3}-\theta\Big)=\sin\theta$$ The above equation is an identity.

#### Work Step by Step

$$\sin\Big(\frac{\pi}{3}+\theta\Big)-\sin\Big(\frac{\pi}{3}-\theta\Big)=\sin\theta$$ The left side would be tackled first. $$A=\sin\Big(\frac{\pi}{3}+\theta\Big)-\sin\Big(\frac{\pi}{3}-\theta\Big)$$ For this problem, obviously the identities for sum and difference of sines must be applied. $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ Therefore, $$A=\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta-\Big(\sin\frac{\pi}{3}\cos\theta-\cos\frac{\pi}{3}\sin\theta\Big)$$ $$A=\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta-\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta$$ $$A=\Big(\sin\frac{\pi}{3}\cos\theta-\sin\frac{\pi}{3}\cos\theta\Big)+\Big(\cos\frac{\pi}{3}\sin\theta+\cos\frac{\pi}{3}\sin\theta\Big)$$ $$A=2\cos\frac{\pi}{3}\sin\theta$$ $$A=2\times\frac{1}{2}\times\sin\theta$$ $$A=\sin\theta$$ Thus, the left side is equal to the right. This equation is an identity.

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