Answer
$$\cos(-105^\circ)=\frac{\sqrt2-\sqrt6}{4}$$
Work Step by Step
$$\cos(-105^{\circ})$$
As in the hint: $$-105^\circ=-60^\circ+(-45^\circ)$$
That means $$\cos(-105^\circ)=\cos[-60^\circ+(-45^\circ)]$$
Now we apply cosine of a sum:
$$\cos(-105^\circ)=\cos(-60^\circ)\cos(-45^\circ)-\sin(-60^\circ)\sin(-45^\circ)$$
Remember that $\cos(-A)=\cos A$, while $\sin(-A)=-\sin A$.
$$\cos(-105^\circ)=\cos60^\circ\cos45^\circ-[-\sin60^\circ(-\sin45^\circ)]$$
$$\cos(-105^\circ)=\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$$
$$\cos(-105^\circ)=\frac{1}{2}\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\frac{\sqrt2}{2}$$
$$\cos(-105^\circ)=\frac{\sqrt2}{4}-\frac{\sqrt6}{4}$$
$$\cos(-105^\circ)=\frac{\sqrt2-\sqrt6}{4}$$
