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We know that the sine function takes it values between $[-1,1]$. So is the set $$A = \{ \sin{n} \ : \ n \in \mathbb{N}\}$$ dense in $[-1,1]$. Generally, for showing the set is dense, one proceeds, by finding out what is $\overline{A}$ of this given set. And if $\overline{A} = [-1,1]$, we are through with the proof, but i having trouble here!
Similarly can one do this with cosine function also, that is proving $B= \{ \cos{n} \ : \ n \in \mathbb{N}\}$ being dense in $[-1,1]$
The hard part is to show that for any $x$ such that $0 \le x \le 2\pi$, and any $\epsilon>0$ there exists a real number $y$ and two integers $m$ and $n$ such that $|y-x|<\epsilon$ and $n=2\pi m+y$. Hint: break up $[0,2\pi]$ into small subintervals, remember that $\pi$ is irrational and apply the pigeonhole principle.
Jmoy's is a good hint, but I think an even nicer approach is to prove that the points $(\cos n,\sin n)$ are dense on the unit circle, and then deduce the results for the individual terms.
Try proving that for any irrational number $\alpha$, the set $A=\left \{ a+b\alpha \mid a\in \mathbb{N},b\in \mathbb{Z} \right \}$ is dense in $\mathbb{R}$. Let $\alpha =\pi $. Since the set $A$ is dense in $\mathbb{R}$, $\forall \; x\in \mathbb{R}$, there exists a sequence of terms $(z_{n})$ such that $\lim_{n\rightarrow \infty }\left ( z_{n} \right )=x$.
$(z_{n})$ can be written as the sum of two sequences of integers.
$z_{n} = x_{n} + \pi y_{n}$.
Use the continuity of $\sin(x)$, to show that a sequence from the set $\left \{\sin (n)\mid n\in \mathbb{N} \right \}$ converges to every element of $[-1,1]$.
While searching on google, i got this JSTOR link: http://www.jstor.org/stable/2688681 which answers the question in an intricate way.
An elementary proof of the fact that the set ${n+\pi k}$, $n,k\in \Bbb{Z}$ is dense in reals is equivalent to showing that the subgroup $\Bbb{Z}+\pi\Bbb{Z}$ is dense in the additive group of real line. See for detail of the proof in Theorem 0.2 in the following
