Convergent subsequences of $x_n = \sin n$ and $y_n = \cos n$... As in title, $x_n = \sin n$ and $y_n = \cos n$.  Can we find some index sequence $\{n_k\}$ such that both $\{x_{nk}\}$ and $\{y_{nk}\}$ converge?  (Not necessarily to the same limit) 
I'm fairly lost; I was thinking of using the density of $\sin n$ in $[-1, 1]$ but I'm not sure how to approach the problem really.  Just hints please, this is for homework.  
 A: It is well-known that, if $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, the sequence $\{n\alpha\pmod{1}\}_{n\in\mathbb{N}}$ is dense in $[0,1]$.$^{(*)}$ 
Since $\pi$ is an irrational number, the sequence $\{e^{i n}\}_{n\in\mathbb{N}}$ is dense in $S^1$, so for any couple of real numbers $(x,y)$ such that $x^2+y^2=1$ there exists a sequence $n_1,n_2,\ldots$ such that $\sin n_k\to y$, $\cos n_k \to x$.
$^{(*)}$ Proof: let $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$ two consecutive convergents of the continued fraction of $\alpha$. Since $\alpha$ is between them and $\left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|=\frac{1}{q_n q_{n+1}}<\frac{1}{q_n^2}$, we have $\left|p_n-\alpha q_n\right|<\frac{1}{q_n}$, so zero is an accumulation point for the sequence $\{n\alpha\pmod{1}\}_{n\in\mathbb{N}}$. Since such a sequence is closed under sum and difference $\pmod{1}$, it follows that such a sequence is dense in $[0,1]$.
A: Try to build a monotone sequence, using the remainder of the division of n for $2 \pi$.
Plus think of $1 = sin(n)^2+cos(n)^2$: you can see that once you find a convergent $x_n$ you get the $y_n$ for free. (at least, I hope I got it right)
