Answer
$\dfrac{\sqrt{{2-\sqrt2}}}{2}$
Work Step by Step
The half-angle Identity for sine can be expressed as:
$\sin {(\dfrac{\theta}{2})}=\pm\sqrt{\dfrac{1-\cos{\theta}}{2}}$
Therefore,
$\sin{(22.5^{\circ})}=\sqrt{\dfrac{1-\cos{(45^{\circ})}}{2}}\\=\sqrt{\dfrac{1-\dfrac{\sqrt2}{2}}{2}}\\=\sqrt{\dfrac{2-\sqrt2}{4}}\\ =\dfrac{\sqrt{{2-\sqrt2}}}{2}$