## Trigonometry (11th Edition) Clone

$$\cos(-105^\circ)=\frac{\sqrt2-\sqrt6}{4}$$
$$\cos(-105^{\circ})$$ 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 $\sin(-A)=-\sin A$. $$\cos(-105^\circ)=\cos60^\circ\cos45^\circ-[-\sin60^\circ(-\sin45^\circ)]$$ $$\cos(-105^\circ)=\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$$ $$\cos(-105^\circ)=\frac{1}{2}\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\frac{\sqrt2}{2}$$ $$\cos(-105^\circ)=\frac{\sqrt2}{4}-\frac{\sqrt6}{4}$$ $$\cos(-105^\circ)=\frac{\sqrt2-\sqrt6}{4}$$