Use the half-angle formulas to find the exact value of the trigonometric function \( \sin 22.5^{\circ} \). \( \sin 22.5^{\circ}=\square \) (Type an exact answer, using radicals as needed.)

To find \( \sin 22.5^{\circ} \), we can use the half-angle formula for sine: \[ \sin \left( \frac{\theta}{2} \right) = \sqrt{ \frac{1 - \cos \theta}{2} } \] For \( 22.5^{\circ} \), we can set \( \theta = 45^{\circ} \). Therefore, we have \[ \sin 22.5^{\circ} = \sin \left( \frac{45^{\circ}}{2} \right) = \sqrt{ \frac{1 - \cos 45^{\circ}}{2} } \] Since \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \), we can substitute this value into the equation: \[ \sin 22.5^{\circ} = \sqrt{ \frac{1 - \frac{\sqrt{2}}{2}}{2} } \] Now simplify the expression: \[ \sin 22.5^{\circ} = \sqrt{ \frac{2 - \sqrt{2}}{4} } = \frac{\sqrt{2 - \sqrt{2}}}{2} \] Thus, the exact value of \( \sin 22.5^{\circ} \) is \[ \sin 22.5^{\circ} = \frac{\sqrt{2 - \sqrt{2}}}{2} \]