The convergence of an infinite radical involving $\cos(\alpha/3)$ By using the triple angle formula for the cosine, $\cos 3\alpha$, we get the cubic equation $ 4x^3-3x = \cos \alpha $. Now, by expressing $ x $ as
$ x = \frac{1}{2}\sqrt{3+\frac{\cos \alpha}{x}}$
And iterate for $x$
$ (1) \cos (\frac{\alpha}{3})= \frac{1}{2}\sqrt{3+\frac{\cos \alpha}{\frac{1}{2}\sqrt{3+\frac{\cos \alpha}{\frac{1}{2}\sqrt{3+\frac{\cos \alpha}{\frac{1}{2}\sqrt{3+...}}}}}}} $
I have verified that this radical converges to the left hand side for $\alpha \in [0,\pi]$
The question is: how to prove the convergence by the conditions stated above? Also, I want to see an analysis of the speed of convergence, that's to say, the precision of the approximations that gives this infinite radical.
 A: This is a classical fixed-point iteration. Assume that $\cos\alpha >0$ and the sequence $\{a_n\}_{n\geq 0}$ is given by $a_0\in(0,1)$ and $a_{n+1}=f(a_n)$ with 
$$ f(x)=\frac{1}{2}\sqrt{3+\frac{\cos\alpha}{x}}.\tag{1}$$
$f(x)$ is a convex and decreasing function over $\mathbb{R}^+$, hence the subsequences $a_1,a_3,a_5,\ldots$ and $a_0,a_2,a_4,\ldots$ are both monotonic (one increasing, the other one decreasing) and convergent.  By Banach fixed-point theorem, by setting $a_{\infty}=\lim_{n\to +\infty}a_n=\cos(\alpha/3)$ we have:
$$ \left|a_n-a_{\infty}\right|\sim C\cdot M^n, \tag{2}$$
(linear convergence) with $M=\left|\,f'(a_\infty)\right|$, i.e.
$$ M = \frac{\cos(\alpha)}{8 \cos^3(\alpha/3)}\leq\frac{1}{8}.\tag{3} $$
The expected accuracy increase of this method is so $\approx 3$ bits at iteration.
Newton's method applied to $T_3$, $a_{n+1}=\frac{\cos(\alpha)+8a_n^3}{12a_n^2-3}$, converges much faster (since in this case $f'(a_\infty)=0$), but it is more sensible to the initial choice of $a_0$, since $f$ has a singularity at $\frac{1}{2}$.
