Answer
$$\sin25^\circ+\cos(-48^\circ)=-2\sin\frac{23^\circ}{2}\cos\frac{73^\circ}{2}$$
Work Step by Step
$$A=\sin25^\circ+\cos(-48^\circ)$$
Here you can choose whether to deal with the negative $-48^\circ$ right away or to leave it alone. In this case, I will leave it alone.
The sum-to-product identity that will be applied here is $$\sin X+\sin Y=2\sin(\frac{X+Y}{2})\cos(\frac{X-Y}{2})$$
Therefore, A would be $$A=2\sin[\frac{25^\circ+(-48^\circ)}{2}]\cos[\frac{25^\circ-(-48^\circ)}{2}]$$ $$A=2\sin[\frac{25^\circ-48^\circ}{2}]\cos[\frac{25^\circ+48^\circ}{2}]$$ $$A=2\sin(\frac{-23^\circ}{2})\cos(\frac{73^\circ}{2})$$
As we know $\sin(-X)=-\sin X$, therefore, $$A=-2\sin\frac{23^\circ}{2}\cos\frac{73^\circ}{2}$$