Contraction mapping theorem in $\mathbb R^{2}$ The question is: 
Show that there exists a unique $(x_1,y_1)\in \mathbb R^{2}$ such that $\cos(\sin(x_1)) = y_1$ and $\sin(\cos(y_1)) = x_1.$
My attempt to a solution:
Let $F(x)= \begin{bmatrix}\sin(\cos(y_1)) \\ \cos(\sin(x_1)) \end{bmatrix}$, where $x=(x_1,y_1)\in \mathbb R^2$. I then tried proving that F is a contraction to use the contraction mapping theorem. I tried doing this by:
1) Calculating the Jacobean of $F$: $$F’(x)=\begin{bmatrix} 0 && -\cos(\cos(y_1))\sin(y_1) \\ -\sin(\sin(x_1))\cos(x_1) && 0 \end{bmatrix},$$ and using the matrix metric $\Vert.\Vert_{\infty}$ to prove $\Vert F’(x)\Vert_{\infty}\le c$, where $c<1$, and thus $F$ is a contraction by contraction property. But that didn’t work since I can’t bound $\vert \cos(\cos(y_1))\sin(y_1)\vert$ or $\vert\sin(\sin(x_1))\cos(x_1)\vert$ by anything smaller than $1$.
2) Proving F is a contraction straight from definition:
I tried doing this using the metric $d(x,y)=\max\{|x_1-y_1|, |x_2-y_2|\}$, where $x=(x_1,x_2)$ and $y=(y_1,y_2)$, but I ran into the same problem as above. I also tried Euclidean metric but had no luck.
Does anyone have any suggestions on how to solve this problem or spot any mistake in my reasoning above?
 A: As noted by the OP, $|\sin(t)|\le1<\pi/2$ implies that there exists $\lambda<1$ such that $|\sin(\sin(t))|\le\lambda$, hence $$|\sin(\sin(t))\cos(t)|\le\lambda.$$
Alas the most that can be said about the other entry in the Jacobian is that its modulus does not exceed $1$.
It turns out that this does imply that $F$ is a strict contraction in a certain weighted $\ell_1$ norm.
Writing $x=(x_1,x_2)$ as usual instead of the somewhat idiosyncratic notation in  the OP, for $\alpha>0$ define a norm on $\Bbb R^2$ by $$||x||_\alpha=|x_1|+\alpha|x_2|.$$If $A$ is a $2\times 2$ matrix let $||A||_\alpha$ be the corresponding operator norm $$||A||_\alpha=\sup_{||x||_\alpha=1}||Ax||_\alpha.$$


Straightforward Calculation $\left|\left|\begin{bmatrix}a&b\\c&d\end{bmatrix}\right|\right|_\alpha
\le\max(|a|+\alpha|c|,|b|/\alpha+|d|)$.


Hence $$||F'(x)||_\alpha\le\max(\lambda\alpha,1/\alpha).$$ So if $1<\alpha<1/\lambda$ it follows that $||F'(x)||_\alpha\le\beta<1$, hence $F$ is a strict contraction in the metric $d(x,y)=||x-y||_\alpha$.
A: Alternatively it is possible to boil down the problem to $\mathbb{R}^1$ by realizing that if there exists $(x_1,y_1)\in \mathbb R^{2}$ such that $\cos(\sin(x_1)) = y_1$ and $\sin(\cos(y_1)) = x_1$, then necessarily $x_1$ is the unique fixed point of $f(x) = \sin(\cos(\cos(\sin(x))))$.
But does the converse statement holds? That is, are these statements equivalent?
Computing $f'$ and analyzing its absolute value one can conclude that
$$ |f'(x)| = |\cos (\cos (\cos (\sin (x)))) \cdot \sin (\cos (\sin (x))) \cdot \sin (\sin (x)) \cdot \cos (x)| < 1$$
Therefore $f$ has an unique fixed point, say $x_1$. Now, by the definition of function, there exists an unique $y_1 \in \mathbb{R}$ such that $\cos(\sin(x_1)) = y_1$. Hence the result.
