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Prove that for each $\epsilon > 0$ it's possible to find such n, so $| \sin(n)- \frac 1 2| < \epsilon$.

Let's represent $n$ as $n = 2\pi k +x,\space\space x \in(-\pi,\pi),\space\space k \ge0,\space k \in \mathbb{N}$
Then we could say, rewrite original statement like that:
$\exists x: x \in (\arcsin(\frac 1 2 -\epsilon),\arcsin(\frac 1 2 +\epsilon))$
But I'm not sure where to go from here.
I'm really not experienced with such kind of proofs, so I would appreciate a suggestion.
Purpose of solving this is pure self-educational for me, so I would like to receive a suggestion only, not the whole ready answer. Thank you.

Michael Hardy
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Dmitri K
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  • What about $|n-2k\pi-\pi/6|$ ? See equidistribution theorem – Jean-Claude Arbaut May 07 '14 at 20:08
  • Following your way might work, but I think using the derivative of the sine function might also work. You can give an upper bound for the difference between two points by taking the the (absolute value of the) maximum of the derivative on the closed interval between the points and multiply that by by the delta-x between the points. I hope this makes sense! – Ruben May 07 '14 at 20:11

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