How to prove the set $\{\cos(n) \mid n \in \mathbb{N}\}$ is dense in $[-1,1]$ Prove that the set  $\{\cos(n) \mid n \in \mathbb{N}\}$ is dense in $[-1,1]$.
 A: It suffices to prove that the points $\{(\cos(n),\sin(n)): n \in \Bbb N\}$ (that is, the "rotations" by $1$ radian) are dense in the unit circle. Or, equivalently, the fractional parts of the multiples of $1/(2 \pi)$ form a dense set in $[0,1]$.
This question (if not the one you've originally posed)
has been answered several times on this site. For example, see the proof outlined here.  Or, for a brief argument that I prefer, this one.
A: Actually I am using the result...

Theorem :- Suppose, $r$ is an irrational number. Then, $Z+rZ$ is dense in $\mathbb{R}$.

Taking $r=2\pi$, it follows $Z+2\pi Z$ is dense in $\mathbb{R}$. For any $c \in [-1,1]$ there exists a $d \in \mathbb{R}$ such that, $\cos(d)=c$.
Now, there exists a sequence $x(n)+2\pi y(n)$ in $Z+2\pi Z$ such that, $\lim (x(n)+2\pi y(n))=d$. Note that, cosine function is continuous on $\mathbb{R}$. Hence, 
$$\lim ~\cos(x(n)+2\pi y(n))=\cos(d) = c \Longrightarrow \lim ~\cos (x(n))=c$$
since $x(n)$ and $y(n)$ are sequences in $Z$. Now, $\cos(x) = \cos (-x)$ holds for all $x \in \mathbb{R}$. Hence, Define, $p(n)$ by, 
$$p(n)=\begin{cases}
 x(n) &\text{ if } x(n) \ge 1 \\
 -x(n) &\text{ if } x(n) \le 0 \end{cases}$$
Hence, $p(n)$ is a sequence in $\mathbb N \cup \{0\}$. Therefore, $\{\cos(n) ~\vert~ n \in \mathbb N \cup \{0\}\}$ is dense in $[-1,~1]$. Hence, we can drop one single point which is in this case, $\cos 0=1$ out. We have, $\{\cos(n) ~\vert~ n\in \mathbb N\}$ is dense in $[-1,1]$. 
