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...

Look over the Banach Fixed Point Theorem. Its practically the same deal. 
Hint: Show with MVT that:
$$ | \cos(a_{n}) - \cos(a_{n-1})| <= q^n|a_1 - a_0|$$
Where $q$ is a constant so $ 0 < q < 1 $. 
Use that in the Cauchy proof.
