Answer
$$\sin9x-\sin3x=2\cos 6x\sin3x$$
Work Step by Step
$$A=\sin9x-\sin3x$$
The sum-to-product identity that will be applied here is $$\sin X-\sin Y=2\cos(\frac{X+Y}{2})\sin(\frac{X-Y}{2})$$
Therefore, A would be $$A=2\cos[\frac{9x+3x}{2}]\sin[\frac{9x-3x}{2}]$$ $$A=2\cos(\frac{12x}{2})\sin(\frac{6x}{2})$$ $$A=2\cos 6x\sin3x$$