Answer
$$\cos255^\circ=\frac{\sqrt2-\sqrt6}{4}$$
Work Step by Step
$$\cos255^\circ$$
1) We can write $255^\circ$ as the sum of $180^\circ$ and $75^\circ$.
$$\cos255^\circ=\cos(180^\circ+75^\circ)$$
Now we expand $\cos(180^\circ+75^\circ)$ according to the cosine sum identity:
$$\cos255^\circ=\cos180^\circ\cos75^\circ-\sin180^\circ\sin75^\circ$$
$$\cos255^\circ=(-1)\times\cos75^\circ-0\times\sin75^\circ$$
$$\cos255^\circ=-\cos75^\circ$$
So now the job is to calculate $-\cos75^\circ$
2) We can write $75^\circ$ as the sum of $30^\circ$ and $45^\circ$
$$-\cos75^\circ=-\cos(30^\circ+45^\circ)$$
Now we expand $\cos(30^\circ+45^\circ)$ according to the cosine sum identity:
$$-\cos75^\circ=-(\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ)$$
$$-\cos75^\circ=-\Big[\Big(\frac{\sqrt3}{2}\Big)\Big(\frac{\sqrt2}{2}\Big)-\Big(\frac{1}{2}\Big)\Big(\frac{\sqrt2}{2}\Big)\Big]$$
$$-\cos75^\circ=-\Big(\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\Big)$$
$$-\cos75^\circ=-\Big(\frac{\sqrt6-\sqrt2}{4}\Big)$$
$$-\cos75^\circ=\frac{\sqrt2-\sqrt6}{4}$$
3) Therefore, $$\cos255^\circ=-\cos75^\circ=\frac{\sqrt2-\sqrt6}{4}$$
