$x_{n+1}=x_n\cos(y_n)-y_n\sin(y_n), y_{n+1}= x_n\sin(y_n)+y_n\cos(y_n) \implies \lim_{x \to \infty}x_n, \lim_{y \to \infty}y_n$ Question_

The sequences $\{x_n\}$ and $\{y_n\}$ are defined as $(x_1, y_1)=(0.8, 0.6)$, 
$x_{n+1}=x_n\cos(y_n)-y_n\sin(y_n)$ and 
$y_{n+1}= x_n\sin(y_n)+y_n\cos(y_n)$. 
Then, if $$\lim_{x \to \infty}x_n, \space\space\space \lim_{y \to \infty}y_n$$ exist, find the value of them. If they does not exist, prove it. 

I'm not sure whether they have values or not. Here's the procedure I've made:
My Own Procedure_

Let $\{z_n\}$ be a sequence given by $\tan\left(z_n\right)=y_n/x_n$.
We can drive: 
  $$x_{n+1} = \sqrt{x_n^2+y_n^2}\cos(y_n+z_n)$$
$$y_{n+1} = \sqrt{x_n^2+y_n^2}\sin(y_n+z_n)$$
Using two equations, we can make a new equation about $y_n$ and $z_n$:
  $$\tan(z_{n+1}) = \tan(y_n+z_n)$$
  Subsequently,
  $$z_{n+1} = y_n+z_n+k\pi$$

And I'm stuck here. Do you think that the procedure is going in the right way? Could you give me other key ideas about the problem? Thanks very much.
 A: If $x_0=y_0=0$, then $x_n=y_n=0$. Assume that $x_n^2+y_n^2>0$. Set
$$
x_n+iy_n=r_n\mathrm{e}^{i\vartheta_n},
$$
then the recursion provides
$$
r_{n+1}\mathrm{e}^{i\vartheta_{n+1}}=x_{n+1}+iy_{n+1}=\mathrm{e}^{iy_n}(x_{n}+iy_{n})=
r_n\mathrm{e}^{iy_n}\mathrm{e}^{i\vartheta_n}
$$
Hence, $r_{n+1}=r_n=\cdots=r_0$, and
$$
\vartheta_{n+1}=\vartheta_n+y_n=\vartheta_n+r_0\sin \vartheta_n,
$$
modulo $2\pi$.
I have run a program and it does not always converge. 
Convergence can be proved if $r_0<1/6$ and $\vartheta_0$ near $2k\pi$, in which case
$$
x_n\to r_0\quad\text{and}\quad y_n\to 0.
$$
A: If the sequence converges, it will converge to a fixed point of 
$$G(x,y)=(x \cos y-y \sin y, x \sin y+y \cos y)$$
From the second equation, you can obtain $x = \frac{y(1-\cos y)}{\sin y}$, and substituting in the first equation you get
$$
\frac{y(1-\cos y)}{\sin y} (\cos y -1 )= y \sin y \Leftrightarrow -1 + 2 \cos y-\cos^2 y = 1- \cos ^2y \Leftrightarrow \cos y = 1
$$
So, assuming that $y\ne 0$ and $\sin y \ne 0$, we must have that $\cos y=1$, i.e. $y = 2 k \pi, k \in \mathbb{Z}$, which would mean that $\sin y = 0$... so, we can only have solutions if $y=0$ or $\sin y=0$.
If $y = 0$, $x$ can take any value. If $\sin y=0$, we must have $\cos y = 1$ and no restriction on $x$.
Bottom line: The points of the form $(x^*,2 k \pi), x^* \in \mathbb{R}, k \in \mathbb{Z}$ are fixed points of $G$ and are therefore possible values for $\lim (x_n , y_n)$.
Depending on the initial "guess", the sequence may or may not converge, but when it converges, it will converge to one of these points. In the specific case of $(x_1,y_1)=(0.8, 0.6)$ the sequence converges to $(-1,0)$.
A: As shown by @Yiorgos, using the polar form the iterations are equivalent to
$$\theta_{n+1}=\theta_n+r\sin\theta_n,$$ with $r$ constant. With the given initial values, $r=1$, $\theta_1=\arctan\dfrac34$. Convergence is possible to a multiple of $\pi$, and by computing the first iterates, we do observe convergence to $\pi$.
Indeed, the function $\theta+\sin\theta$ is contractant around $\pi$ in the whole range $(0,2\pi)$, and the iterations will converge from $\theta_1$.
In fact, the function is asymptotically cubic around the fixed-point, and convergence is extremely fast.

$$0.6435011087933,\\1.2435011087933,\\2.1904164564551,\\3.0045155724143,\\3.1411637741144,\\3.1415926535766,\\\cdots$$
A: It's easy to notice that the sum of the squares of the $2$ sequences is constant (i.e. $x_1^2+y_1^2$) here $x_n^2+y_n^2 = x_1^2+y_1^2 = 1$
Now,
$|x_{n+1}| = |x_n\cdot\cos(y_n) - y_n\cdot\sin(y_n)|$ , which is bounded above at $(x_n^2+y_n^2)^{1/2} = 1$
Also,
$x_{n+1} = x_n\cdot\cos((1-x_n^2)^{1/2}) - ((1-x_n^2)^{1/2})\cdot\sin(y_n) \le x_n - (1-x_n^2)^{1/2} < x_n$.
So by monotonic convergence theorem we can assert that $x_n$ is convergent
Similar arguments provide that , $|y_n| \le1$. Due to the nature of sine function when $y_n$ is less than $1$ we also conclude that $y_n$ is non-negative .
We also get that $y_n$ is increasing owing to the fact that $x_n$ is increasing and sum of the $2$ sequences is $1$ .
So we can again assert by monotone convergence theorem that $y_n$ converges .
Now passing over the lim gives $\lim y_n = 0$ and $\lim x_n^2=1$
Now notice that after a certain cutoff say $N$ belonging to set of Natural numbers $x_n$ is negative ($n>N$) so $\lim x_n < 0 \to \lim x_n = -1$.
