I can show that $\cos(\sin(x))$ is a contraction on $\mathbb{R}$ and hence by the Contraction Mapping Theorem it will have a unique fixed point. But what is the process for finding this fixed point? This is in the context of metric spaces, I know in numerical analysis it can be done trivially with fixed point iteration. Is there a method of finding it analytically?
[Math] Fixed point of $\cos(\sin(x))$
general-topologymetric-spacesreal-analysis
Related Solutions
This isn't the suggested proof at all, but it's a lot simpler, and I hope it's of some value. @Jonas Meyer's comment seems to give a good route to solving the problem as suggested, so there's no reason to duplicate.
A fixed point of $x \mapsto \cos x$ must be between $-1$ and $1$, because at a fixed point $\cos x = x$, and the codomain of $\cos$ is $[-1, 1]$.
For $-1 < x < 0$, cosine is positive but $x$ is negative, so we need only consider $0 \le x \le 1$.
At $0$, $g(x) = \cos x - x$ is positive. At $1$, $g(1)$ is negative (by direct evaluation). Throughout the interval, $g'(x) < 0$, so $g$ has at most one zero, say $b$, in the interval, and at that point $\cos b - b = 0$, so $\cos b = b$.
"Completeness can't be omitted" is just saying that if $X$ is not complete, you can't be certain that a fixed point will exist for any given function. It doesn't mean you will never find one for specific functions.
It's like saying that "if $X$ is not complete, not all cauchy sequences converge". But you can definitely find some cauchy squences that do converge.
EDIT It's now interesting to investigate how the theorem fails without completeness:
Firstly, here's an example of a contraction defined on a non-complete metric space with no fixed point:
$f:(0,1)\rightarrow(0,1)$ such that $f(x) = \dfrac{x}{2}$.
Clearly this is a contraction because $|f(x)-f(y)|=\Big|\dfrac{x}{2}-\dfrac{y}{2}\Big| = \dfrac{1}{2}|x-y|$ for any $x, y \in (0, 1)$. However, for any $x\in(0,1)$, $f(x)\neq x$, so there is no fixed point.
(On the natural domain of the function, the fixed point would be at $0$, but you can see that this is on the boundary of our non-complete set.)
Secondly, are there any cases where the metric space being incomplete results in more than one fixed point? The answer is no. Suppose $f:X\rightarrow X$ is a contraction and $X$ is any metric space (complete or not complete). Suppose $f$ has two fixed points at $x$ and $y$. Then $|f(x)-f(y)| = |x - y|$. This contradicts the definition of a contraction which requires $|f(a)-f(b)|<|a-b|$ for all $a,b\in X$.
Best Answer
The Jacobi-Anger expansion gives an expression for your formula as:
$\cos(\sin(x)) = J_0(1)+2 \sum_{n=1}^{\infty} J_{2n}(1) \cos(2nx)$.
Since the "harmonics" in the sum rapidly damp to zero, to second order the equation for the fixed point can be represented as:
$x= J_0(1) + 2[J_2(1)(\cos(2x)) + J_4(1)(\cos(4x))]$.
Using Wolfram Alpha to solve this I get $x\approx 0.76868..$