Answer
$\dfrac{\sqrt6-\sqrt2}{4}$
Work Step by Step
Recall:
$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$
Hence,
$\cos(\frac{5\pi}{12})\\
=\cos{75^\circ}\\
=\cos{(45^\circ+30^\circ)}\\
=\cos{45^\circ}\cos{30^\circ}-\sin{45^\circ}\sin{30^\circ}\\
=\frac{\sqrt3}{2}\frac{\sqrt2}{2}-\frac{1}{2}\frac{\sqrt2}{2}\\
=\frac{\sqrt6-\sqrt2}{4}$
