<s>[INST] Solving $\cos \frac{165}{2}$ What are the steps to solve $$\frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4}$$ into $$\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
Please explain. I got them from derivation of $$\cos(82.5^\circ)$$ in 2 different ways. 
$$\cos(60^\circ+22.5^\circ)$$ and $$\cos\frac{(90^\circ+75^\circ)}{2}$$
Thanks in advance
 [/INST] Note that $\sqrt{2\sqrt{2}} = \sqrt{4/\sqrt{2}}$ so you have
$$
\begin{split}
f &= \frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4} \\
  &= \frac12 \sqrt{\frac{2\sqrt{2}-1-\sqrt3}{\sqrt2}} \\
  &= \frac12 \sqrt{2 - \frac{1+\sqrt3}{\sqrt2}}
\end{split}
$$
so remains to prove that $$ \frac{1 + \sqrt{3}}{\sqrt2} = \sqrt{2+\sqrt{3}}$$
To do that,multiply by $\sqrt2$ and square both sides to get
$$
1^2 + 3 + 2\sqrt{3} = 2(2+\sqrt3)
$$
which are obviously equal to $4+2\sqrt3$.
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