本题中不得使用计算器。

1. Express $15^\circ$ as the difference of two special angles

$\sin 15^\circ = \sin(45^\circ - 30^\circ)$

2. Apply the compound angle formula for sine

$\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

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

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

4. Simplify the expression to obtain the required form

$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

1. Substitute the value of $\sin 15^\circ$ from part (a) and square the expression

$\sin^2 15^\circ = \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right)^2$

2. Expand the numerator and simplify the surds

$= \frac{(\sqrt{6})^2 - 2(\sqrt{6})(\sqrt{2}) + (\sqrt{2})^2}{16}$
$= \frac{6 - 2\sqrt{12} + 2}{16}$
$= \frac{8 - 2(2\sqrt{3})}{16}$
$= \frac{8 - 4\sqrt{3}}{16}$

3. Reduce the fraction to the required form

$= \frac{4(2 - \sqrt{3})}{16}$
$= \frac{2 - \sqrt{3}}{4}$

1. State the Pythagorean identity relating sine and cosine

$\cos^2 15^\circ = 1 - \sin^2 15^\circ$

2. Substitute the value of $\sin^2 15^\circ$ from part (b) and simplify

From part (b), $\sin^2 15^\circ = \frac{2 - \sqrt{3}}{4}$.
$\cos^2 15^\circ = 1 - \left(\frac{2 - \sqrt{3}}{4}\right)$
$\cos^2 15^\circ = \frac{4 - (2 - \sqrt{3})}{4}$
$\cos^2 15^\circ = \frac{2 + \sqrt{3}}{4}$