Show that the following system: $x= \frac{\sin(x+y)}{2}, y= \frac{\cos(x-y)}{2} $ admits at most one solution

Question

Question:
Show that the following system: $x= \frac{\sin(x+y)}{2}, y= \frac{\cos(x-y)}{2} $ admits at most one solution.

I have tried to solve this question by bounding the norm of the differential of $$f(x;y)=\frac{1}{2} (\sin(x+y);\cos(x-y)),$$ but I am stuck. I will be really happy if someone can help me please.

Thank you.

Answer 1

First of all we know that $(\mathbb{R}^N; ||.||) $ is complete.

$Df(x;y)=Jf(x;y)=\frac{1}{2} \begin{pmatrix} cos(x+y) & cos(x+y) \\ -sin(x-y)& sin(x-y) \end{pmatrix}$
Hence $||Df(x;y)||_{\infty} \leq \frac {1}{2} Max (|cos(x+y)|;|sin(x-y)|) \leq \frac{1}{2}$
As it is true $\forall (x;y), (w;z) \in \mathbb{R^2}$ by the Mean value theorem we get that $|| f(x;y) -f(w;z)||_{\infty} \leq || (x;y) -(w;z)||_{\infty} \frac {1}{2} $. Hence $f$ is a contraction mapping.

Hence by the fixed point theorem this finish the question.

Is it correct?