$x_{n+1}=x_n\cos(y_n)-y_n\sin(y_n), y_{n+1}= x_n\sin(y_n)+y_n\cos(y_n) \implies \lim_{x \to \infty}x_n, \lim_{y \to \infty}y_n$ 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.
If $x_0=y_0=0$, then $x_n=y_n=0$. Assume that $x_n^2+y_n^2&gt;0$. 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 $r_0&lt;1/6$ and $\vartheta_0$ near $2k\pi$, in which case
$$
x_n\to r_0\quad\text{and}\quad y_n\to 0.
$$
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,
$x_{n+1} = x_n\cdot\cos((1-x_n^2)^{1/2}) - ((1-x_n^2)^{1/2})\cdot\sin(y_n) \le x_n - (1-x_n^2)^{1/2} &lt; x_n$.
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 ($n&gt;N$) so $\lim x_n &lt; 0 \to \lim x_n = -1$.