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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 $[-1,1]$

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4 Answers 4

up vote 14 down vote accepted

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.

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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.

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Can you prove this for me! I really appreciate this idea! –  anonymous Sep 16 '10 at 11:37
4  
From a more advanced perspective, it's a special case of an "irrational rotation" (since 1 is not a rational multiple of $2\pi$). See en.wikipedia.org/wiki/Irrational_rotation . Many standard books on dynamical systems prove that the orbit of any point under an irrational rotation is dense on the circle. However, there are more direct proofs of the original question that appear in real analysis books and don't use the dynamics language. –  Carl Mummert Sep 16 '10 at 11:43

While searching on google, i got this JSTOR link: http://www.jstor.org/stable/2688681 which answers the question in an intricate way.

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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 $[-1,1]$.

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