Answer
$\displaystyle \sin\frac{\pi}{2}-\sin\frac{\pi}{6}$
Work Step by Step
Product-to-Sum:
$\sin A \displaystyle \cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$
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$2\displaystyle \sin\frac{\pi}{6}\cos\frac{\pi}{3}=$
$=2(\displaystyle \frac{1}{2}[\sin(\frac{\pi}{6}+\frac{\pi}{3})+\sin(\frac{\pi}{6}-\frac{\pi}{3})])$
$=\displaystyle \sin\frac{3\pi}{6}+\sin(-\frac{\pi}{6})$
$=\displaystyle \sin\frac{\pi}{2}-\sin\frac{\pi}{6}$
