Find the limit of $a_{n+1}=\cos(a_n)$ 
Let $\beta\in[-1,1]$ and let $a_n$ defined by $ \begin{cases} a_1=\beta,  \\ a_{n+1}=\cos(a_n) \end{cases}$
Let $c\in \mathbb{R}$, s.a $\cos(c)=c$.
Prove that: $\lim\limits_{n\to\infty}a_n=c$

My work so far: I know that $c\in(0,1)$.
$|\cos(a_n)-c|=|\cos(a_n)-\cos(c)|=|-2\sin(\frac{a_n+c}{2})\sin(\frac{a_n-c}{2})|\le\frac{1}{2}|a_n+c||a_n-c|$
But I don't know how to continue from here.
 A: Define $f: [-1,1] \to [-1,1]$ by $f(x)= \cos x$. If $x,y \in [-1,1]$, then there is $t$ between $x$ and $y$ such that
$|f(x)-f(y)| = | \sin t| \cdot |x-y| \le q|x-y|$, 
where $q= | \sin 1| <1$.
Now apply Banach's fix point theorem
A: Elaborating a bit more on an answer already given, note that $\cos$ has $-\sin$ as derivative, and note that in the interval $[-1, 1]$, the derivative of $\cos$ in absolute value never exceeds $1$, given that $|(\cos x)'| = |\sin x| < 1$ for $x \in [-1, 1]$.
On the other hand, the mean value theorem gives that $\forall x,y\ \cos(x) - \cos(y) = (\cos a)'(x - y), a\in]x,y[$ (provided $y>x$) which translates into
$$\forall x, y\ |\cos(x) - \cos(y)| = |\sin(a)||x - y|$$
Now the truth is that $|\sin(a)| < 1$ when $a \in [-1, 1]$. Let $L$ be the maximum value $|\sin(a)|$ can attain in the interval $[-1, 1]$. Then,
$$|\cos(x) - \cos(y)| \leq L|x - y| < |x - y|$$
Now set $x = c$ and $y = \beta$:
$$|\cos(c) - \cos(\beta)| = |c - a_1| \leq L|c - \beta|$$
Now do the same thing, but with $x = c$ and $y = a_1$:
$$|\cos(c) - \cos(a_1)| = |c - a_2| \leq L|c - a_1| \leq L(L|c - \beta|) = L^2|c - \beta|$$
See where this is going? Setting $y = a_n$ gives
$$|\cos(c) - \cos(a_n)| = |c - a_{n+1}| \leq L|c - a_n| \leq L(L^n|c - \beta|) = L^{n+1}|c - \beta|$$
Now just take the limit $n \to \infty$. Because $|L| < 1, \lim_{n\to\infty}$ will give $|c - a_n| \to 0$.
