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It is given that $\sin(A - B) = \sin A \cos B - \cos A \sin B$. Use it to evaluate $\sin 15^\circ$.
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$\sin 15^\circ = \sin (45^\circ - 30^\circ)$ (1/2 Mark)
$= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$ (1/2 Mark)
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}$ (1 Mark)
$= \frac{\sqrt{3}-1}{2\sqrt{2}}$ or $\frac{\sqrt{6}-\sqrt{2}}{4}$ (1/2 Mark)
$= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$ (1/2 Mark)
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}$ (1 Mark)
$= \frac{\sqrt{3}-1}{2\sqrt{2}}$ or $\frac{\sqrt{6}-\sqrt{2}}{4}$ (1/2 Mark)