## Calculus (3rd Edition)

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