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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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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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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