differentiability of functions from $\mathbb R^2$ to $\mathbb R^2$

I have a question where i solved half of it but couldn't continue, any help would be great and thanks in advance.

We have the function $$F(x,y)=( \cos x -\sin y, \sin x -\cos y)$$ the function is defined on $\mathbb R^2$ with the usual euclidean norm.

1) Show that $F$ is $C^1$ over $\mathbb R^2$.

Answer: $F(x,y)=(F_1(x,y),F_2(x,y))$ and \begin{align} \frac{\partial F_1}{\partial x}(x,y)&=-\sin x,\\ \frac{\partial F_1}{\partial y}(x,y)&=-\cos y,\\ \frac{\partial F_2}{\partial x}(x,y)&=\cos x,\\ \frac{\partial F_2}{\partial y}(x,y)&=\sin y. \end{align}

The four partial derivatives are continuous so $F$ is of class $C^1$ over $\mathbb R^2$.

2) Show that $\|dF(x,y)\| \leq \sqrt{2}$ for every $(x,y)$. \begin{align} dF(x,y)&=\left(\frac{\partial F_1}{\partial x}(x,y) + \frac{\partial F_1}{\partial y}(x,y) , \frac{\partial F_2}{\partial x}(x,y) +\frac{\partial F_2}{\partial y}(x,y)\right)\\ &= (-\sin x -\cos y, \cos x+ \sin y). \end{align} so $$\|dF(x,y)\| = \sqrt{ (-\sin x -\cos y)^2 +(\cos x+ \sin y )^2 } =\sqrt{2 +2(\sin x\cdot \cos y + \cos x\cdot \sin y)}.$$

but then I don't find it less than $\sqrt{2}$.

3) Show that the following recurrence sequence defined by $x_0, y_0\in \mathbb R$ and $$x_{n+1}=0.5(\cos x_n -\sin y_n) \quad\text{and}\quad y_{n+1}= 0.5(\sin x_n - \cos y_n)$$ is convergent.

How to solve part 3?

Thanks to all.

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Please do use LaTeX to write mathematics in this, otherwise it is hard to understand what you mean. You can find direction in the FAQ section –  DonAntonio Feb 22 '13 at 11:59
@jim Please do the edits in one go. Don't change only tiny bits in several turns. –  Julian Kuelshammer Feb 22 '13 at 12:02
@JulianKuelshammer sorry for that –  jim Feb 22 '13 at 12:04

Your derivate should be a $2\times 2$-Matrix. For the sequence: Write $(x_n,y_n)$ in terms of $F$. Whats the derivative of that "function". Potentially, you can find a contraction.