Answer
$cos(\frac{\pi}{12})= \boxed{\frac{\sqrt{2}+\sqrt{6}}{4}}$
Work Step by Step
$cos(\frac{\pi}{12}) = cos(\frac{\pi}{3}-\frac{\pi}{4})$
Using $cos(A-B) = cos(A)cos(B)+sin(A)sin(B)$,
$cos(\frac{\pi}{3}-\frac{\pi}{4}) = cos(\frac{\pi}{3})cos(\frac{\pi}{4})+sin(\frac{\pi}{3})sin(\frac{\pi}{4})$
$cos(\frac{\pi}{12}) = (\frac{1}{2})(\frac{\sqrt{2}}{2})+(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4} = \dfrac{\sqrt{2}+\sqrt{6}}{4}$
