# Inverse at a single point using Banach Fixed Point

The Problem:

Show that there exists a unique $$(x,y)\in \mathbb{R}^2$$ so that $$\cos(\sin(x))=y$$ and $$\sin(\cos(y))=x$$.

I believe you can use the Banach Fixed Point theorem, although I'm sure there is a way without it.

My attempt so far:

Let $$\phi :\mathbb{R}^2\rightarrow \mathbb{R}^2$$ defined as $$\phi((x,y))=(x + c_1[\cos(\sin(x))-y], y + c_2[\sin(\cos(y)) - x])$$ with some choice of $$c_1, c_2\in \mathbb{R}-\{0\}$$. The Jacobian matrix for this map is as follows: \begin{align} D\phi_{|_{(x,y)}}&= \begin{bmatrix} 1+c_1(-\sin(sin(x))\cos(x) & -c_2 \\ -c_1 & 1+c_2(\cos(\cos(y))(-\sin(y)) \\ \end{bmatrix} \end{align} My idea was that if I can show that $$||D\phi_{|_{(x,y)}}||_{op}=\underset{x\neq 0}{\sup}\frac{||D\phi_{|_{(x,y)}}(x,y)||_{\mathbb{R}}}{||(x,y)||_{\mathbb{R}}}<1$$ , then the Banach Fixed Point Theorem would apply via the Mean Value Theorem. But the calculation is quite messy and I think I'm at a deadend. If anyone could supply a solution or a good hint It'd be very grateful.

No need for 2-dimensional calculations. You can use the standard banach fixed point theorem in $$\Bbb R$$.
If you have such a pair $$(x,y)\in\Bbb R^2$$ then it follows directly that for $$x$$ it has to hold
$$\sin(\cos(\cos(\sin(x)))) = x$$
But this is $$x \in \Bbb R$$ is unique.
Obviously $$x \in [-1,1]$$ has to hold. By defining $$f(x) = \sin(\cos(\cos(\sin(x))))$$ we get $$f:[-1,1] \to [-1,1]$$ as well as $$|f'(x)| \le |\sin(\cos(x))| \le \sin(1) < 1$$ hence $$L$$ is a self-mapping contradiction and the banach fixed point theorem gives the wanted unique $$x \in [-1,1]$$.
Defining $$y:=\cos(\sin(x))$$ finishes the proof (although you could use the same argument as above for getting the unique $$y \in [-1,1]$$)