# Is this expression strictly positive? $n \left(\cos\frac{k\pi}{n}\right)\left(1-\cos\frac{k\pi}{n}\right)-\sin\frac{k\pi}{n}$

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

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

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?

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

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.