Answer
$$\sin102^\circ-\sin95^\circ=2\cos(\frac{197^\circ}{2})\sin(\frac{7^\circ}{2})$$
Work Step by Step
$$A=\sin102^\circ-\sin95^\circ$$
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 $$A=2\cos[\frac{102^\circ+95^\circ}{2}]\sin[\frac{102^\circ-95^\circ}{2}]$$ $$A=2\cos(\frac{197^\circ}{2})\sin(\frac{7^\circ}{2})$$
