Answer
$\dfrac{\sqrt{6}+\sqrt{2}}{4}$
Work Step by Step
Addition formula
$$\cos{(A+B)}= \cos{A} \cos{B}-\sin{A}\sin{B}$$
$ \cos{\left(\dfrac{\pi}{12} \right)} = \cos{\left(\dfrac{\pi}{3} -\dfrac{\pi}{4} \right)}$
$\cos{\left(\dfrac{\pi}{3} -\dfrac{\pi}{4} \right)} = \cos{\left(\dfrac{\pi}{3} \right)} \cos{\left(-\dfrac{\pi}{4} \right)} - \sin{\left(\dfrac{\pi}{3} \right)} \sin{\left(-\dfrac{\pi}{4} \right)}$
$\cos{\left(\dfrac{\pi}{12} \right)} = \dfrac{1}{2} \times \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{2}}{2}$
$\cos{\left(\dfrac{\pi}{12} \right)} = \dfrac{\sqrt{6}+\sqrt{2}}{4}$
