## Precalculus (6th Edition)

$-\frac{\sqrt{2+\sqrt{3}}}{2}$
Use the half-angle formula, $\cos\frac{u}{2}=\pm\sqrt{\frac{1+\cos u}{2}}$. Note that $195^\circ$ is in Quadrant III, where cosine is negative, so we take the negative square root. $\cos 195^\circ$ $=\cos\frac{390^\circ}{2}$ $=\cos\frac{30^\circ}{2}$ $=-\sqrt{\frac{1+\cos 30^\circ}{2}}$ $=-\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$ $=-\sqrt{\frac{(1+\frac{\sqrt{3}}{2})*2}{2*2}}$ $=-\sqrt{\frac{2+\sqrt{3}}{4}}$ $=-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$ $=-\frac{\sqrt{2+\sqrt{3}}}{2}$