#### Answer

$$\sin(60^\circ-x)-\sin(60^\circ+x)=-\sin x$$
The detailed explanation is written in the Work step by step.

#### Work Step by Step

$$\sin(60^\circ-x)-\sin(60^\circ+x)=-\sin x$$
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,
$$X=\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x-\Big(\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x\Big)$$
$$X=\Big(\frac{\sqrt3}{2}\cos x-\frac{\sqrt3}{2}\cos x\Big)-\frac{1}{2}\sin x-\frac{1}{2}\sin x$$
$$X=0-2\times\frac{1}{2}\sin x$$
$$X=-\sin x$$
Therefore, $$\sin(60^\circ-x)-\sin(60^\circ+x)=-\sin x$$
We have verified that the equation is an identity.