<s>[INST] Converting $\frac{\sqrt{2-\sqrt{3}}}{2}$ into $\frac{\sqrt{3}-1}{2\sqrt{2}}$ 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?
 [/INST] $$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$
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