In formal, does there exist $k\in\mathbb{N}$ such that $\sin n\leq\sin k$ for all $n\in\mathbb{N}$?
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This post and other posts linked there seem related. – Martin Sleziak Oct 15 '15 at 15:19
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It is known that for all irrational $x\in\mathbb R$ the set of all the fractional parts $\{nx\}$ is dense in [0,1]. The function sin $x$ has its maximum equal to $1$ at $x=\frac {\pi}{2}$ which is irrational; hence, in theory, there are integers n such that $\frac{\pi}{2}-\{\frac {n\pi}{2}\}$ is as small as one wishes so sin $n$ is very near to $1$.
However if we take one of these integers, say $m$, there will always be another integer,$m_1$, (not necessarily greater than $m$) such that the fractional part of $\space\frac {m_1\pi}{2}$ is closer to $\frac{\pi}{2}$.
Consequently, does not exist $k\in\mathbb N$ satisfying the question.
Piquito
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Thanks, but the key thing that confused me is why the first known fact hold... – xskxzr Oct 15 '15 at 14:36
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@Shenke: http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset – sdcvvc Oct 15 '15 at 15:23
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@Shenke: User sdcvvc already answered your question. It is a very nice theorem actually. – Piquito Oct 15 '15 at 20:30
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No. Since pi is not a rational number, sin(n) will never achieve its maximum of 1. However, you can get infinitely close to the value of 1, since (n mod 2*pi) will eventually have occurrences in interval (pi +/- arcsin (epsilon))
Not sufficiently formal proof, but I think enough pointers..
Pieter21
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Maybe you want to say $(\pi/2\pm\arcsin\epsilon)$, but I can't show why $(n \pmod{2\pi})$ will eventually have occurrences in this interval, can you please prove it in detail? – xskxzr Oct 15 '15 at 13:39