Answer
$$\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}$$
1 should be matched with C.
Work Step by Step
$$\sin 15^\circ$$
$15^\circ$ can be rewritten as the difference of $45^\circ$ and $30^\circ$.
$$15^\circ=45^\circ-30^\circ$$
Therefore, $$\sin15^\circ=\sin(45^\circ-30^\circ)$$
Now we apply the sine difference identity, which states
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
That makes
$$\sin15^\circ=\sin 45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ$$
$$\sin15^\circ=\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\frac{1}{2}$$
$$\sin15^\circ=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$$
$$\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}$$
That means we match 1 with C.