## Trigonometry (11th Edition) Clone

$$\sin(60^\circ+x)+\sin(60^\circ-x)=\sqrt3\cos x$$ How to verify the equation is an identity is explained in detail below.
$$\sin(60^\circ+x)+\sin(60^\circ-x)=\sqrt3\cos x$$ As the left side is more complicated, it needs to be dealt with first. $$X=\sin(60^\circ+x)+\sin(60^\circ-x)$$ - According to the sum and difference identities for sine, which states $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ as we apply $A=60^\circ$ and $B=x$ to $X$, we would have $$\sin(60^\circ+x)=\sin60^\circ\cos x+\cos60^\circ\sin x$$ $$\sin(60^\circ+x)=\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x$$ and $$\sin(60^\circ-x)=\sin60^\circ\cos x-\cos60^\circ\sin x$$ $$\sin(60^\circ-x)=\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x$$ Therefore, $$X=\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x+\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x$$ $$X=\Big(\frac{\sqrt3}{2}\cos x+\frac{\sqrt3}{2}\cos x\Big)+\Big(\frac{1}{2}\sin x-\frac{1}{2}\sin x\Big)$$ $$X=2\times\frac{\sqrt3}{2}\cos x+0$$ $$X=\sqrt3\cos x$$ Therefore, $$\sin(60^\circ+x)+\sin(60^\circ-x)=\sqrt3\cos x$$ We have verified that the equation is an identity.