## Precalculus (6th Edition)

$\displaystyle \frac{\sqrt{10}}{4}$
$0 < x < \displaystyle \frac{\pi}{2}\qquad/\div 2$ $0 < \displaystyle \frac{x}{2} < \frac{\pi}{4}$, so, $\displaystyle \frac{x}{2}$ is in quadrant I, where the cosine is positive. Half-Angle Identity: $\displaystyle \cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}=+\sqrt{\frac{1+\frac{1}{4}}{2}}$ $=\displaystyle \sqrt{\frac{\frac{5}{4}}{2} }=\frac{\sqrt{5}}{\sqrt{8}}\cdot\frac{\sqrt{8}}{\sqrt{8}}=\frac{\sqrt{40}}{8}$ $=\displaystyle \frac{2\sqrt{10}}{8}$ $=\displaystyle \frac{\sqrt{10}}{4}$