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.


1 Answer 1


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 $y \in [-1,1]$)


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