Fixed point of $\cos(\sin(x))$ 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?
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 $x\approx 0.76868..$
Since it's a contraction, just iterate. Pick an arbitrary starting point and keep applying the operator until you achieve the desired accuracy.