## Precalculus: Mathematics for Calculus, 7th Edition

$\sqrt{2}-1$
Use the half-angle formula, $\tan\frac{u}{2}=\frac{1-\cos u}{\sin u}$. $\tan 22.5^\circ$ $=\tan \frac{45^\circ}{2}$ $=\frac{1-\cos 45^\circ}{\sin 45^\circ}$ $=\frac{1-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$ $=\frac{(1-\frac{\sqrt{2}}{2})\frac{2}{\sqrt{2}}}{\frac{\sqrt{2}}{2}*\frac{2}{\sqrt{2}}}$ $=\frac{\frac{2}{\sqrt{2}}-1}{1}$ $=\sqrt{2}-1$