Converting $\frac{\sqrt{2-\sqrt{3}}}{2}$ into $\frac{\sqrt{3}-1}{2\sqrt{2}}$

Question

A few days ago I had a quiz question on calculating the sine of $15^\circ$. Using the half-angle identity for sine, I did the following: $$\sin15^\circ=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$$

But the lecturer give this: $$\frac{\sqrt{3}-1}{2\sqrt{2}}$$

This is also the first answer given by WolframAlpha.

So my question is: how does one convert between the two forms? $$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$

Edit: and why would one prefer the second over the first, if the the first is so straightforward to reach?

Answer 1

$$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$