Answer
$-\dfrac{\sqrt 6+\sqrt 2}{4}$
Work Step by Step
Here, we have $\cos (\dfrac{11\pi}{12})=\cos (\dfrac{2\pi}{3}+\dfrac{\pi}{4})$
or, $\cos (\dfrac{2\pi}{3}+\dfrac{\pi}{4})=\cos (\dfrac{2\pi}{3}+\dfrac{\pi}{4})$
or, $\cos \dfrac{2\pi}{3} \cos \dfrac{\pi}{4}-\sin \dfrac{2\pi}{3} \sin \dfrac{\pi}{4})=-\dfrac{1}{2} \cdot \dfrac{\sqrt 2}{2}- \dfrac{\sqrt 3}{2} \cdot \dfrac{\sqrt 2}{2}$
Thus, $\cos (\dfrac{11\pi}{12})=-\dfrac{\sqrt 6+\sqrt 2}{4}$
