Proving $\displaystyle\lim_{n\to\infty}a_n=\alpha$ $a_1=\frac\pi4, a_n = \cos(a_{n-1})$ 
Let $a_1=\frac\pi4, a_n = \cos(a_{n-1})$
Prove $\displaystyle\lim_{n\to\infty}a_n=\alpha$.
Where $\alpha$ is the solution for $\cos x=x$.
Hint: check that $(a_n)$ is a cauchy sequence and use Lagrange's theorem.

Well I tried to show that it's a cauchy sequence: $|a_m-a_n|<\epsilon , \ \forall m,n$ but I just don't see how it's done with a trig recursion sequence...

Note that
$$
\cos(x):\left[\frac1{\sqrt2},\frac\pi4\right]\mapsto\left[\frac1{\sqrt2},\frac\pi4\right]
$$
and on $\left[\frac1{\sqrt2},\frac\pi4\right]$,
$$
\left|\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)\right|=|\sin(x)|\le\frac1{\sqrt2}\lt1
$$
Thus, the Mean Value Theorem ensures that $\cos(x)$ is a contraction mapping on $\left[\frac1{\sqrt2},\cos\left(\frac1{\sqrt2}\right)\right]$. The Contraction Mapping Theorem says that $\cos(x)$ has a unique fixed point in $\left[\frac1{\sqrt2},\cos\left(\frac1{\sqrt2}\right)\right]$. Furthermore, for $n\ge3$,
$$
\left|a_n-a_{n-1}\right|\le\left(\frac1{\sqrt2}\right)^{n-3}\left|a_3-a_2\right|
$$
Thus, $a_n$ is a Cauchy sequence. The point to which it converges must be the unique fixed point where $\cos(x)=x$.
