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

\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?
 A: 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.
