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.