Precalculus (6th Edition) Blitzer

$\frac{\sqrt 5}{5}$
Step 1. Let $u=tan^{-1}\frac{1}{2}, 0\lt u\lt \frac{\pi}{2}$; we have $tan(u)=\frac{1}{2}$ Step 2. Construct a right triangle with acute angle $u$. Opposite side $=1$ and adjacent side $=2$. We have the hypotenuse $=\sqrt {1^2+2^2}=\sqrt 5$. Thus $tan(u)=\frac{1}{2}$ and $sin(u)=\frac{\sqrt 5}{5}$ Step 3. Thus we have $sin(tan^{-1}\frac{1}{2})=sin(u)=\frac{\sqrt 5}{5}$