## Precalculus (6th Edition)

$\displaystyle \frac{\sqrt{50-10\sqrt{5}}}{10}$
To apply the Half-Angle Identity, $\displaystyle \sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$, we need to determine the sign and find $\cos x$ With $0 < x < \displaystyle \frac{\pi}{2}\qquad/\div 2$ $0 < \displaystyle \frac{x}{2} < \frac{\pi}{4}$, we see that both $x$ and $\displaystyle \frac{x}{2}$ are in quadrant I. $\cos x$ is positive, $\displaystyle \sin\frac{x}{2}$ is positive. We find $\displaystyle \cos x=\frac{1}{\sec x}$ from the Pythagorean identity: $\sec^{2}x= \tan^{2}x+1=4+1=5$ $\sec x=\sqrt{5}$ $\displaystyle \cos x=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$ So, $\displaystyle \sin\frac{x}{2}=+\sqrt{\frac{1-\frac{\sqrt{5}}{5}}{2}}=\sqrt{\frac{5-\sqrt{5}}{10}\cdot\frac{10}{10}}=\frac{\sqrt{50-10\sqrt{5}}}{10}$