Answer
$$\cos75^\circ=\frac{\sqrt 6-\sqrt 2}{4}$$
Work Step by Step
$$A=\cos75^\circ$$
We can rewrite $75^\circ$ into the sum of $30^\circ$ and $45^\circ$, which means $$A=\cos(30^\circ+45^\circ)$$
Now we apply the identity for cosine of a sum: $$A=\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ$$ $$A=\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}-\frac{1}{2}\frac{\sqrt 2}{2}$$ $$A=\frac{\sqrt 6}{4}-\frac{\sqrt 2}{4}$$ $$A=\frac{\sqrt 6-\sqrt 2}{4}$$
