The following question comes from Chapter 6.4, Exercise 4 on page 156 in the set of notes Topology Without Tears.
Using Exercise 2 and 3, show that while $f: \mathbb R \to \mathbb R$ given by $f(x)=\cos(x)$ does not satisfy the condition of the Contraction Mapping Theorem, it nevertheless has a unique fixed point.
The "Exercise 2 and 3" mentioned in the question (which I've already done) were
Extend the Contraction Mapping Theorem by showing that if $f$ is a mapping of a complete metric space $(X,d)$ into itself and $f^N$ is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point.
Use the Mean Value Theorem to prove the following:
Let $f: [a,b] \to [a,b]$ be differentiable. Then $f$ is a contraction if and only of there exists an $r \in (0,1)$ such that $|f'(x)|\le r$, for all $x \in [a,b]$.
Can anybody please show me how to use those to show that $f(x)=\cos(x)$ has a unique fixed point?
I know, from a Numerical Analysis course, that this fixed point can be found using a method such as Newton's Method, but I am not sure how to prove that it actually exists (and is unique).