Showing that $\cos(x)$ is a contraction mapping on $[0,\pi]$ How do I show that $\cos(x)$ is a contraction mapping on $[0,\pi]$? I would normally use the mean value theorem and find $\max|-\sin(x)|$ on $(0,\pi)$ but I dont think this will work here.
So I think I need to look at $|\cos(x)-\cos(y)|$ but I can't see what to do to get this of the form $|\cos(x)-\cos(y)|\leq\alpha|x-y|$?
Thanks for any help

To show that $\cos(x)$ is a contraction mapping on $[0,1]$ you just need to show that it is Lipschitz with a Lipschitz constant less than $1$. Because $\cos(x)$ is continuously differentiable, there is a maximum absolute value of the derivative on each closed interval, and the mean value theorem can be used to show that maximum absolute value works as a Lipschitz constant. Since the derivative of $\cos(x)$ is bounded below $1$ in absolute value on $[0,1]$, that will give the desired result. 
