<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] If we are allowed to solve analytically then
\begin{align}2\sqrt{2}(2\sqrt{2}-1-\sqrt3)&=
4\sqrt{4}-2\sqrt{2}-2\sqrt{6}\\&=
8-2(\sqrt{2}+\sqrt{6})\\&=
8-2\left(2\sqrt{2+\sqrt{3}}\right)\\&=
8-4\sqrt{2+\sqrt{3}}\\&=
4\left(2-\sqrt{2+\sqrt{3}}\right)
\end{align}
so that
$$\frac{\sqrt{4\left(2-\sqrt{2+\sqrt{3}}\right)}}{4}=\frac{2\sqrt{2-\sqrt{2+\sqrt{3}}}}{4}=\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
</s>