## Precalculus (6th Edition)

$\sin(\frac{7\pi}{6}+x)-\cos(\frac{2\pi}{3}+x)=0$
Start with the left side: $\sin(\frac{7\pi}{6}+x)-\cos(\frac{2\pi}{3}+x)$ Use the addition formulas for sine and cosine: $=(\sin \frac{7\pi}{6}\cos x+\cos \frac{7\pi}{6}\sin x)-(\cos\frac{2\pi}{3}\cos x-\sin \frac{2\pi}{3}\sin x)$ Simplify: $=\sin \frac{7\pi}{6}\cos x+\cos \frac{7\pi}{6}\sin x-\cos\frac{2\pi}{3}\cos x+\sin \frac{2\pi}{3}\sin x$ $=-\frac{1}{2}\cos x+(-\frac{\sqrt{3}}{2})\sin x-(-\frac{1}{2})\cos x+\frac{\sqrt{3}}{2}\sin x$ $=-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x$ $=0$ Since this equals the right side, the identity has been proven.