## Trigonometry (11th Edition) Clone

$$\sin\Big(\frac{7\pi}{6}+x\Big)-\cos\Big(\frac{2\pi}{3}+x\Big)=0$$ The equation is an identity.
$$\sin\Big(\frac{7\pi}{6}+x\Big)-\cos\Big(\frac{2\pi}{3}+x\Big)=0$$ First, it might be a good idea to calculate $\sin\frac{7\pi}{6}$, $\cos\frac{7\pi}{6}$, $\sin\frac{2\pi}{3}$ and $\cos\frac{2\pi}{3}$. $$\sin\frac{7\pi}{6}=\sin\Big(\pi+\frac{\pi}{6}\Big)=\sin\pi\cos\frac{\pi}{6}+\cos\pi\sin\frac{\pi}{6}$$ $$\sin\frac{7\pi}{6}=0\times\cos\frac{\pi}{6}+(-1)\times\frac{1}{2}=-\frac{1}{2}$$ $$\cos\frac{7\pi}{6}=\cos\Big(\pi+\frac{\pi}{6}\Big)=\cos\pi\cos\frac{\pi}{6}-\sin\pi\sin\frac{\pi}{6}$$ $$\cos\frac{7\pi}{6}=(-1)\times\frac{\sqrt3}{2}-0\times\frac{1}{2}=-\frac{\sqrt3}{2}$$ $$\sin\frac{2\pi}{3}=\sin\Big(\frac{\pi}{3}+\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\cos\frac{\pi}{3}+\cos\frac{\pi}{3}\sin\frac{\pi}{3}$$ $$\sin\frac{2\pi}{3}=2\sin\frac{\pi}{3}\cos\frac{\pi}{3}=2\times\frac{\sqrt3}{2}\times\frac{1}{2}=\frac{\sqrt3}{2}$$ $$\cos\frac{2\pi}{3}=\cos\Big(\frac{\pi}{3}+\frac{\pi}{3}\Big)=\cos\frac{\pi}{3}\cos\frac{\pi}{3}-\sin\frac{\pi}{3}\sin\frac{\pi}{3}$$ $$\cos\frac{2\pi}{3}=\cos^2\frac{\pi}{3}-\sin^2\frac{\pi}{3}=\Big(\frac{1}{2}\Big)^2-\Big(\frac{\sqrt3}{2}\Big)^2=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}$$ Now, we expand the left side of the equation, using the sum identity for sines: $$X=\sin\Big(\frac{7\pi}{6}+x\Big)-\cos\Big(\frac{2\pi}{3}+x\Big)$$ $$X=\sin\frac{7\pi}{6}\cos x+\cos \frac{7\pi}{6}\sin x-\Big(\cos\frac{2\pi}{3}\cos x-\sin\frac{2\pi}{3}\sin x\Big)$$ $$X=-\frac{1}{2}\cos x-\frac{\sqrt3}{2}\sin x-\Big(-\frac{1}{2}\cos x-\frac{\sqrt3}{2}\sin x\Big)$$ $$X=-\frac{1}{2}\cos x-\frac{\sqrt3}{2}\sin x+\frac{1}{2}\cos x+\frac{\sqrt3}{2}\sin x$$ $$X=0$$ Therefore the equation is indeed an identity.