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Let us define a function $f(k,n)$ by

\begin{equation} f(k,n)=n \left (\cos\frac{k\pi}{n}\right) \left(1-\cos\frac{k\pi}{n}\right) - \sin \frac{k\pi}{n} \end{equation}

where $\frac{k}{n}$ is irreducible with $k,n \in \mathbb{N}$, with $k \leq \left \lfloor{n/2}\right \rfloor$, and $k \geq 2,$ $n \geq 5$.

I suspect that $f(k,n) >0$.

Plugging in a few values of $k$ and $n$ and computing $f(k,n)$ numerically indeed shows that, but how do I prove / disprove this?

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  • $\begingroup$ This is connected to the concept of irrationality measure. $\endgroup$ – Lucian Feb 26 '16 at 7:00
  • $\begingroup$ Have you tried showing that $n\cos\theta (1-\cos\theta)-\sin\theta>0$ for $\frac{\pi}{n}\leq \theta <\frac{\pi}{2}$? $\endgroup$ – Bobby Grizzard Feb 26 '16 at 20:50
  • $\begingroup$ I mean for $\frac{\pi}{n}\leq \theta \leq \frac{\pi}{2}-\frac{\pi}{n}$ $\endgroup$ – Bobby Grizzard Feb 26 '16 at 21:07
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Proof: Since $\frac{k}{n}$ is irreducible and $k\le \lfloor \frac{n}{2}\rfloor$, we have $k \le \frac{n-1}{2}$.

Let $u = \tan \frac{k\pi}{2n}$. Since $2\le k \le \frac{n-1}{2}$ and $n\ge 5$, we have $$0 < \frac{\pi}{n} \le \tan \frac{\pi}{n} \le u \le \tan \Big(\frac{\pi}{4} - \frac{\pi}{4n}\Big) = \frac{1 - \tan \frac{\pi}{4n}}{1 + \tan \frac{\pi}{4n}} \le \frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}} < 1.$$ Using $\cos \frac{k\pi}{n} = \frac{1-u^2}{1+u^2}$ and $\sin \frac{k\pi}{n} = \frac{2u}{1+u^2}$, we have $$n \Big(\cos \frac{k\pi}{n}\Big) \Big(1 - \cos \frac{k\pi}{n}\Big) - \sin \frac{k\pi}{n} = \frac{2u(-nu^3-u^2+nu-1)}{(u^2+1)^2}.$$ It suffices to prove that $-nu^3-u^2+nu-1 > 0$. To proceed, we need the following fact.

Fact 1: For each positive integer $n$, $f_n(x) = -nx^3 - x^2 + nx - 1$ is concave on $[0, 1]$ since $f_n''(x) = -6nx - 2 < 0$ for $x > -\frac{1}{3n}$. As a result, for $0\le a \le x \le b \le 1$, we have $f_n(x) \ge \min(f_n(a), f_n(b))$.

Now, according to Fact 1, letting $a = \frac{\pi}{n}$ and $b = \frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}}$, since $0 < a \le u \le b < 1$, it suffices to prove that $f_n(\frac{\pi}{n}) > 0$ and $f_n(\frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}}) > 0$. We have $f_n(\frac{\pi}{n}) = -\frac{\pi^3}{n^2}+\pi-\frac{\pi^2}{n^2}-1 \ge -\frac{\pi^3}{5^2}+\pi-\frac{\pi^2}{5^2}-1 > 0$ and \begin{align} f_n(\frac{1-\frac{\pi}{4n}}{1+\frac{\pi}{4n}}) &= \frac{2n^3}{(4n+\pi)^3}\Big(-\frac{8\pi^2}{n}+32\pi - \frac{\pi^3}{n^3}-\frac{4\pi^2}{n^2}-\frac{16\pi}{n}-64\Big)\\ &\ge \frac{2n^3}{(4n+\pi)^3}\Big(-\frac{8\pi^2}{5}+32\pi - \frac{\pi^3}{5^3}-\frac{4\pi^2}{5^2}-\frac{16\pi}{5}-64\Big)\\ &> 0. \end{align} This completes the proof.

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