Solving infinite nested square roots of 2 converging to finite nested radical Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating infinitely)
Converging to finite nested radical of $2\cos12° = \frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$
The finite nested radical, I was able to derive $\cos12° = \cos(30-18)°$ as follows
$$\cos30°\cdot\cos18° + \sin30°\cdot\sin18°$$
$$= \frac{√3}{2}\cdot\frac{\sqrt{2+2\cos36°}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2-2\cos36°}}{2}$$
Where $\cos18° = \frac{\sqrt{2+2\cos36°}}{2}$ (by Half angle cosine formula) and
$\sin18° = \frac{\sqrt{2-2\cos36°}}{2}$ (solving again by half angle cosine formula)
$2\cos36° =\frac{ \sqrt5 +1}{2}$ which is golden ratio
$\frac{\sqrt3}{2}\cdot\frac{\sqrt{10+2\sqrt5}}{4}+ \frac{1}{2}\cdot\frac{\sqrt{5}-1}{4} = \frac{\sqrt{30+6\sqrt5}}{8}+ \frac{\sqrt5-1}{8}$
Further steps finally lead to the finite nested radical
Method actually I tried to solve infinite nested square roots of 2 is as follows.
$2\cos\theta = \sqrt{2+2\cos2\theta}$ and
$2\sin\theta = \sqrt{2-2\cos2\theta}$
Now simplifying infinite nested square roots of 2, we will get the following as simplified nested radical
$$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos12°}}}}$$
Simplifying step by step as follows
$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\sin6°}}}$ then
$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\cos84°}}}$ (by $\sin\theta = \cos(90-\theta)$
$2\cos12° = \sqrt{2+\sqrt{2+2\sin42°}}$
$2\cos12° = \sqrt{2+\sqrt{2+2\cos48°}}$
$2\cos12° = \sqrt{2+2\cos24°}$
$2\cos12° = 2\cos12°$
We are back to $\sqrt1$
Actually this is how I got stuck!
But for infinite nested square roots of 2(as depicted), if I run program in python I am able to get good approximation ( Perhaps if we run large number of nested square roots in python we get more number of digits matching the finite nested radical), because I'm not able get anywhere solving such a kind of infinite cyclic nested square roots of 2.
Dear friends, is there anyway to find the solution by any other means like solving infinite nested square roots
Thanks in advance.

Somehow I got the answer from my subsequent post (after a long homework for cyclic infinite nested square roots of 2)
Sivakumar Krishnamoorthi (https://math.stackexchange.com/users/686991/sivakumar-krishnamoorthi), Solving cyclic infinite nested square roots of 2 as cosine functions, URL (version: 2020-09-26): https://math.stackexchange.com/q/3841605
$2\cos48°$ or $2\cos\frac{4}{15}π$ is cyclic infinite nested square roots of 2 of form $cin\sqrt2[2-2+]$ i.e. $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+...}}}}$
According to half angle cosine formula within next 2 steps we get $2\cos12°$ or $2\cos\frac{π}{9}$ as $cin\sqrt2[2+2-]$ i.e. cyclic infinite nested square roots of 2 as $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-...}}}}$

Therefore $2\cos12°$ as a multiple of 3 it can be solved as nested radicals as in LHS of my question and on RHS it is cyclic infinite nested square roots of 2 which is also equivalent. As both the result belong to same $\cos \frac{π}{15}$ cyclic infinite nested square roots of 2 on RHS and finite nested radical are the same in terms of results.

