Answer
$$\sin\Big(\frac{\pi}{3}+\theta\Big)-\sin\Big(\frac{\pi}{3}-\theta\Big)=\sin\theta$$
The above equation is an identity.
Work Step by Step
$$\sin\Big(\frac{\pi}{3}+\theta\Big)-\sin\Big(\frac{\pi}{3}-\theta\Big)=\sin\theta$$
The left side would be tackled first.
$$A=\sin\Big(\frac{\pi}{3}+\theta\Big)-\sin\Big(\frac{\pi}{3}-\theta\Big)$$
For this problem, obviously the identities for sum and difference of sines must be applied.
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
Therefore, $$A=\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta-\Big(\sin\frac{\pi}{3}\cos\theta-\cos\frac{\pi}{3}\sin\theta\Big)$$
$$A=\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta-\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta$$
$$A=\Big(\sin\frac{\pi}{3}\cos\theta-\sin\frac{\pi}{3}\cos\theta\Big)+\Big(\cos\frac{\pi}{3}\sin\theta+\cos\frac{\pi}{3}\sin\theta\Big)$$
$$A=2\cos\frac{\pi}{3}\sin\theta$$
$$A=2\times\frac{1}{2}\times\sin\theta$$
$$A=\sin\theta$$
Thus, the left side is equal to the right. This equation is an identity.
