本题中不得使用计算器。

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

Using $\cos 2A = 1 - 2\sin^2 A$ with $A = 22.5^\circ$:
$\cos 45^\circ = 1 - 2\sin^2 22.5^\circ$

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

We know $\cos 45^\circ = \frac{\sqrt{2}}{2}$.
$\frac{\sqrt{2}}{2} = 1 - 2\sin^2 22.5^\circ$
$2\sin^2 22.5^\circ = 1 - \frac{\sqrt{2}}{2}$
$2\sin^2 22.5^\circ = \frac{2 - \sqrt{2}}{2}$
$\sin^2 22.5^\circ = \frac{2 - \sqrt{2}}{4}$

3. Take the square root to find $\sin 22.5^\circ$ in the required form.

Since $22.5^\circ$ is in the first quadrant, $\sin 22.5^\circ > 0$.
$\sin 22.5^\circ = \sqrt{\frac{2 - \sqrt{2}}{4}}$
$\sin 22.5^\circ = \frac{\sqrt{2 - \sqrt{2}}}{2}$

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

$\tan 15^\circ = \tan(60^\circ - 45^\circ)$
$= \frac{\tan 60^\circ - \tan 45^\circ}{1 + \tan 60^\circ \tan 45^\circ}$

2. Substitute the exact values of the trigonometric functions for the special angles

$\tan 60^\circ = \sqrt{3}, \quad \tan 45^\circ = 1$
$\tan 15^\circ = \frac{\sqrt{3} - 1}{1 + (\sqrt{3})(1)} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

3. Rationalise the denominator to simplify the expression

$\tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$
$= \frac{3 - 2\sqrt{3} + 1}{3 - 1}$
$= \frac{4 - 2\sqrt{3}}{2}$

4. Simplify to obtain the final shown result

$\tan 15^\circ = 2 - \sqrt{3} \quad \text{(shown)}$