Answer
$-\frac{\sqrt3-1}{2\sqrt2}$
Work Step by Step
Using the given formula and the known values: $\sin(\frac{13\pi}{12})=\sin(\frac{\pi}{3})\cos(\frac{3\pi}{4})+\sin(\frac{3\pi}{4})\cos(\frac{\pi}{3})=\frac{\sqrt3}{2}(-\frac{1}{\sqrt2})+\frac{1}{\sqrt2}\frac{1}{2}=-\frac{\sqrt3-1}{2\sqrt2}$
