#### Answer

$$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$
5 comes with C.

#### Work Step by Step

$$\sin15^\circ$$
$15^\circ$ can be written as half of the angle $30^\circ$.
$$15^\circ=\frac{30^\circ}{2}$$
Thus, $$\sin15^\circ=\sin\Big(\frac{30^\circ}{2}\Big)$$
We can apply here the Half-Angle Identity for sine, which states
$$\sin\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1-\cos A}{2}}$$
with $A=30^\circ$:
$$\sin15^\circ=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$
$15^\circ$ is in quadrant I. As $\sin\theta\gt0$ as $\theta$ is in quadrant I, $\sin15^\circ\gt0$.
So we would pick the positive square root.
$$\sin15^\circ=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}$$
$$\sin15^\circ=\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}$$
$$\sin15^\circ=\sqrt{\frac{2-\sqrt3}{4}}$$
$$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$
Therefore, 5 needs to be matched with C.