Answer
$$\cos\frac{11\pi}{12}=\frac{-\sqrt6-\sqrt2}{4}$$
Work Step by Step
$$\cos\frac{11\pi}{12}$$
1) We can write $\frac{11\pi}{12}$ as the difference of $\pi$ and $\frac{\pi}{12}$, since
$$\pi-\frac{\pi}{12}=\frac{12\pi}{12}-\frac{\pi}{12}=\frac{11\pi}{12}$$
So,
$$\cos\frac{11\pi}{12}=\cos(\pi-\frac{\pi}{12})$$
Now we expand $\cos(\pi-\frac{\pi}{12})$ according to the cosine difference identity:
$$\cos\frac{11\pi}{12}=\cos\pi\cos\frac{\pi}{12}+\sin\pi\sin\frac{\pi}{12}$$
$$\cos\frac{11\pi}{12}=(-1)\times\cos\frac{\pi}{12}+0\times\sin\frac{\pi}{12}$$
$$\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12}$$
So now the job is to calculate $-\cos\frac{\pi}{12}$
2) We can write $\frac{\pi}{12}$ as the difference of $\frac{\pi}{4}$ and $\frac{\pi}{6}$, since
$$\frac{\pi}{4}-\frac{\pi}{6}=\frac{3\pi}{12}-\frac{2\pi}{12}=\frac{\pi}{12}$$
So,
$$-\cos\frac{\pi}{12}=-\cos(\frac{\pi}{4}-\frac{\pi}{6})$$
Now we expand $\cos(\frac{\pi}{4}-\frac{\pi}{6})$ according to the cosine difference identity:
$$-\cos\frac{\pi}{12}=-(\cos\frac{\pi}{4}\cos\frac{\pi}{6}+\sin\frac{\pi}{4}\sin\frac{\pi}{6})$$
$$-\cos\frac{\pi}{12}=-\Big[\Big(\frac{\sqrt2}{2}\Big)\Big(\frac{\sqrt3}{2}\Big)+\Big(\frac{\sqrt2}{2}\Big)\Big(\frac{1}{2}\Big)\Big]$$
$$-\cos\frac{\pi}{12}=-\Big(\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\Big)$$
$$-\cos\frac{\pi}{12}=-\Big(\frac{\sqrt6+\sqrt2}{4}\Big)$$
$$-\cos\frac{\pi}{12}=\frac{-\sqrt6-\sqrt2}{4}$$
3) Therefore, $$\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12}=\frac{-\sqrt6-\sqrt2}{4}$$