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orl wrote: Let $n$ and $k$ be positive integers such that $1 \leq n \leq N+1$, $1 \leq k \leq N+1$. Show that: \[ \min\limits_{n \neq k} |\sin n - \sin k| < \dfrac{2}{N}. \] Assume the contrary: Assume that $\min\limits_{n\neq k}\left|\sin n-\sin k\right|\geq\dfrac{2}{N}$. Then, for any two n and k with $n\neq k$, we have $\left|\sin n-\sin k\right|\geq\dfrac{2}{N}$. Now let p be a permutation of the set {1; 2; ...; N + 1} such that $\sin p\left(1\right)\leq\sin p\left(2\right)\leq ...\leq\sin p\left(N+1\right)$ (in other words, p is the permutation which arranges the numbers sin 1, sin 2, ..., sin (N + 1) in increasing order). Then, it follows that for any integer i with $1\leq i\leq N$, we have $\left|\sin p\left(i+1\right)-\sin p\left(i\right)\right|\geq\dfrac{2}{N}$. Since $\sin p\left(i\right)\leq\sin p\left(i+1\right)$, we have $\left|\sin p\left(i+1\right)-\sin p\left(i\right)\right|=\sin p\left(i+1\right)-\sin p\left(i\right)$, and thus $\sin p\left(i+1\right)-\sin p\left(i\right)\geq\dfrac{2}{N}$. Hence, $\sin p\left(N+1\right)-\sin p\left(1\right)=\sum\limits_{i=1}^N\left(\sin p\left(i+1\right)-\sin p\left(i\right)\right)\geq\sum\limits_{i=1}^N\dfrac{2}{N}=N\cdot\dfrac{2}{N}=2$. But this is impossible, since $\sin p\left(N+1\right)<1$ (in fact, $\sin p\left(N+1\right)\leq 1$, and equality can occur only if $p\left(N+1\right)$ is a multiple of $\dfrac{\pi}{2}$, what is impossible since $p\left(N+1\right)$ is a natural number and $\pi$ is irrational) and similarly $\sin p\left(1\right)>-1$. Thus, we get a contradiction, i. e. our assumption that $\min\limits_{n\neq k}\left|\sin n-\sin k\right|\geq\dfrac{2}{N}$ was invalid, and thus we have $\min\limits_{n\neq k}\left|\sin n-\sin k\right|<\dfrac{2}{N}$. darij