本题中请勿使用计算器。

1. Apply the double angle formula for cosine to relate $22.5^\circ$ to $45^\circ$

$\cos 45^\circ = 1 - 2\sin^2 22.5^\circ$

2. Substitute the exact value of $\cos 45^\circ$ and rearrange to make $\sin^2 22.5^\circ$ the subject

$\frac{\sqrt{2}}{2} = 1 - 2\sin^2 22.5^\circ$
$2\sin^2 22.5^\circ = 1 - \frac{\sqrt{2}}{2}$
$2\sin^2 22.5^\circ = \frac{2 - \sqrt{2}}{2}$
$\sin^2 22.5^\circ = \frac{2 - \sqrt{2}}{4}$

3. Take the square root to find $\sin 22.5^\circ$, noting that it is positive in the first quadrant

$\sin 22.5^\circ = \sqrt{\frac{2 - \sqrt{2}}{4}}$
$\sin 22.5^\circ = \frac{\sqrt{2 - \sqrt{2}}}{2}$

1. Express $15^\circ$ as a difference of special angles and apply the compound angle formula for tangent.

$\tan 15^\circ = \tan(45^\circ - 30^\circ)$
$\tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

2. Substitute the exact values of the trigonometric functions.

$\tan 45^\circ = 1$
$\tan 30^\circ = \frac{1}{\sqrt{3}}$
$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

3. Rationalise the denominator.

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$
$= \frac{3 - 2\sqrt{3} + 1}{3 - 1}$

4. Simplify to obtain the final result.

$= \frac{4 - 2\sqrt{3}}{2}$
$= 2 - \sqrt{3}$ (Shown)