## Precalculus (6th Edition)

$\displaystyle \frac{\sqrt{50-15\sqrt{10}}}{10}$
$\displaystyle \frac{\pi}{2} < x < \pi \qquad/\div 2$ $\displaystyle \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ So, we see that $x$ is in quadrant II, $\cos x$ is negative, and $\displaystyle \frac{x}{2}$ is in quadrant I, $\displaystyle \cos\frac{x}{2}$ is positive. $\displaystyle \cot x=-3\Rightarrow\tan x=-\frac{1}{3}$ $\displaystyle \sec^{2}x=\tan^{2}+1=\frac{1}{9}+1=\frac{10}{9}$ $\displaystyle \sec x=-\frac{\sqrt{10}}{3}$ $\displaystyle \cos x=-\frac{3}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}$ Half-Angle Identity: $\displaystyle \cos\frac{x}{2}=+\sqrt{\frac{1+\cos x}{2}}=\sqrt{\frac{1-\frac{3\sqrt{10}}{10}}{2}}$ $=\displaystyle \sqrt{\frac{10-3\sqrt{10}}{20}\cdot\frac{5}{5}}=\frac{\sqrt{50-15\sqrt{10}}}{10}$