Answer
$\sin(\frac{\pi}{3}-\frac{\pi}{4})=\frac{\sqrt {6}-\sqrt {2}}{4}$
Work Step by Step
Using the addition formula, we have
$\sin(\frac{\pi}{3}-\frac{\pi}{4})=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}$
$=\frac{\sqrt 3}{2}\times\frac{1}{\sqrt 2}-\frac{1}{2}\times\frac{1}{\sqrt 2}=\frac{\sqrt {3}-1}{2\sqrt 2}=\frac{\sqrt {6}-\sqrt {2}}{4}$
