Answer
$$\sin(60^\circ+x)+\sin(60^\circ-x)=\sqrt3\cos x$$
How to verify the equation is an identity is explained in detail below.
Work Step by Step
$$\sin(60^\circ+x)+\sin(60^\circ-x)=\sqrt3\cos x$$
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.
