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Prove that for every natural number $N>M$ (where M is sufficiently large) we can find natural $P \leq N$ such that $|\sin(P)|<\frac{1}N$ or may be that some real positive C exists such that $|\sin(P)|<\frac{C}N$

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Savinov, the bullet-points at the top of this page are for you… – The Chaz 2.0 Oct 17 '11 at 19:10
Perhaps you mean $|\sin(P)|$. Otherwise it's too easy (just take $P = 4$). – Robert Israel Oct 17 '11 at 22:22
Oh yes. Thanks! I forgot to wright it. – Evgeny Savinov Oct 17 '11 at 22:54
up vote 1 down vote accepted

I proved it. Consider the unit circle at the origin and divide it into $N$ parts (every part has length $\frac{2\pi}{N}$). Then I make $N$ steps by one radian from point $(1,0)$. So I have $N$ points on the circle. If my last point belongs to interval of angles $(-\frac{\pi}{N},\frac{\pi}{N})$, then $|\sin(N)|<=|N \mod 2\pi|<=\frac{\pi}{N}$. But if it doesnt belong, then on some part we have two point (let first point corresponds to step number $N_1$,second - $N_2$). Then if we make step $P=|N_1-N_2|<=N$ radians as we made first time, we would get on interval of angles $(-\frac{\pi}{N},\frac{\pi}{N})$(that's obvious) Then $|\sin(P)|<=|P \mod 2\pi|<=\frac{\pi}{N}$.

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Idea: $|\sin(P)| \leq |P \mod 2\pi|$, where the modulo is taken with respect to the closest multiple of $2\pi$ (rather than the maximal multiple below $P$).

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Thanks, I thought it and it can be improved $|\sin(P)| \leq |P \mod \pi|$, but i havent idea how to estimate this. – Evgeny Savinov Oct 17 '11 at 23:16
Look up "diophantine approximation". The bound you need can be proved using continued fractions. – Yuval Filmus Oct 17 '11 at 23:31

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