How to find the sequences of $\sin n$ (n=natural number) sub-sequence limits? I know that it is a $[-1;1]$, but how to proof?
Edit: Is it true, that sin(n), with all natural numbers have different value? How to proof? If that is true, it is posible found bijection between N->[-1;1]
Outline: Consider the points $P(n)=(\cos n,\sin n)$ as $n$ ranges over the positive integers. It is enough to show that this set of points is dense in the unit circle.
Since $\pi$ is irrational, we have $P(m)\ne P(n)$ if $m\ne n$. It follows by the Pigeonhole Principle that given any $\epsilon \gt 0$, there exist $m_0$ and $n_0$, with $m_0\lt n_0$, such that $0\lt d(P(m_0),P(n_0))\lt \epsilon$. In particular, $0 \lt |\sin(m_0) -\sin(n_0)|\lt \epsilon$.
Then for any point $P$ on the unit circle, there are infinitely many positive integers $k$ such that the distance of $P(m_0+k(n_0-m_0))$ from $P$ is less than $\epsilon$. In particular, for any $x$ there exist infinitely many $k$ such that $|\sin x -\sin(m_0+k(n_0-m_0))|\lt \epsilon$.
Remark: The modified question asks whether $\sin m$ and $\sin n$ can be equal if $m\ne n$. Note that $\sin x=\sin y$ if and only if $y=x+2k\pi$ or $y=(2k+1)\pi -x$ for some integer $k$. Since $\pi$ is irrational, this cannot happen if $x=m$ and $y=n$, where $m$ and $n$ are distinct integers. In essence this fact was used in the proof outlined in the answer.
It's easy to prove that $\forall n,m\in\mathbb{N}, \sin n \neq \sin m$ if $n \neq m $.
If $\sin{n}=\sin{m}$, then $n=m+2\pi k (k\in\mathbb{Z})$ by periodicity. So, $\pi=\dfrac{n-m}{2k}\in\mathbb{Q}$ But since the left side is clearly irrational number, this contradiction implies the above fact.
