Dense set on unit circle [duplicate]

I have a task. Prove that $(\cos(n\alpha), \sin(n\alpha))$, with $n\in\mathbb{N}$ is dense on unit circle.($\alpha$ chosen such that our set is infinite ) We want to show, that for every element's neighbourhood there is an element of that that is in the neighborhood. We know that there all points are different (there is no cycle on the circle). But why can we use pigeonhole principle to prove that if we divide the circle in $k$ arcs then there exists two elements which a closer than $\frac{2\pi}{k}$?

EDIT: Thanks everyone. Now everything is clear. This question is [SOLVED]

marked as duplicate by rtybase, Claude Leibovici, Guy Fsone, Paul Sinclair, user223391 Jan 18 '18 at 19:03

• Can you rewrite the second sentence? It's wrong the way it is stated. – zhw. Jan 18 '18 at 3:17
• As written, the statement is wrong. $\alpha=\pi$ is irrational and $(\cos(\pi n),\sin(\pi n))$ is not dense in the unit circle. – Jack D'Aurizio Jan 18 '18 at 10:56
• It's a lot easier to use the theorem saying that if $f$-continuous function and $\left(x_n\right)_{n\in\mathbb{N}}$ is dense, then $\left(f(x_n)\right)_{n\in\mathbb{N}}$ is also dense in $f$'s range. In this case $f(x)=e^{i\cdot \alpha \cdot x}$ and $f$'s range is the unit circle. – rtybase Jan 18 '18 at 12:45
• Nope, it will be $\left\{n + m \cdot \frac{2\pi}{\alpha} \mid n,m\in\mathbb{Z} \right\}$ assuming $\frac{2\pi}{\alpha}$ is irrational, as per Kronecker's approximation theorem. And $f(x)=e^{i\cdot \alpha \cdot x}$ is a periodic function. – rtybase Jan 18 '18 at 14:36
• @rtybase thanks a lot – Kirill Losev Jan 18 '18 at 15:46

The basic idea is that $(\{n\alpha\})$ is dense in $(0, 1)$. This is Weyl's equidistribution theorem: https://en.wikipedia.org/wiki/Equidistribution_theorem