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How does one prove the following inequality?

$$\left|\frac{\sin(\pi \alpha N)}{\sin(\pi \alpha)}\right| \leq \frac{1}{2 \| \alpha \|}$$

Here $\| \alpha \|$ denotes the distance to the nearest integer.

I'm not even sure where to get started.

Thanks.

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note that for irrational $\alpha$ this is a sharp inequality, since $\{\alpha N \mod 1\} $ is dense in $[0,1]$. –  Michalis Apr 18 '12 at 21:38
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1 Answer 1

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By looking at the graph of the $\sin(x)$ function around $0$, we see that $$|\sin( x)|\geq \frac{2}{\pi} x$$ when $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$ and also $$\sin(x)\leq 1.$$ Since $|\sin(\pi\alpha)|= |\sin(\pi \|\alpha\|)|$, and $\|\alpha\|\leq \frac{1}{2}$, we may apply the above inequalities to find $$\left| \frac{\sin(N\pi\alpha)}{\sin(\pi\alpha)}\right|\leq \left| \frac{1}{\sin(\pi\|\alpha\|)}\right|\leq \frac{1}{2\|\alpha\|}.$$

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