Answer
$$2\sin\frac{\pi}{6}\cos\frac{\pi}{3}=\sin\frac{\pi}{2}-\sin\frac{\pi}{6}$$
Work Step by Step
$$A=2\sin\frac{\pi}{6}\cos\frac{\pi}{3}$$
The product-to-sum identity that will be applied here is $$\sin X\cos Y=\frac{1}{2}[\sin(X+Y)+\sin(X-Y)]$$
Therefore, A would be $$A=2\times\frac{1}{2}[\sin(\frac{\pi}{6}+\frac{\pi}{3})+\sin(\frac{\pi}{6}-\frac{\pi}{3})]$$ $$A=\sin\frac{\pi}{2}+\sin(-\frac{\pi}{6})$$
We know that $\sin(-X)=-\sin X$. That means $$A=\sin\frac{\pi}{2}-\sin\frac{\pi}{6}$$
(Remember that the question only asks for the sum or difference of trigonometric functions. You might go as far as reach the final number, but I believe that is not necessary.)
