A function inequality Is it true that the following function 
$$\frac{\pi ^2 \left(t^2-4 (-1+t) \text{cos}\left[\frac{\pi }{m}\right]^2\right) \text{csc}\left[\frac{\pi  (-2+t)}{m}\right]^2}{m^2}, t\in[0,1]$$  attains its maximum in 0 and 1. Here $m>3$.
 A: The function $\sin((2-t)\pi/m)$ can be linearized by a function $at +b$ attaining the same values in endpoints $0,1$. Then $at +b\le \sin((2-t)\pi/m)$ implying that our function $f(t)$ is smaller than a rational function $h(t)=(At^2+B t+C)/(at + b)^2$. Now we can easily see that $h$ has a single extreme point $x\in[0,1]$ which is a local minimum. Q.E.D.
A: Let us denote $$f(t) :=  \left(t^2-4(-1+t)\cos^2\left(\frac \pi m \right)\right)\csc^2 \left(\frac {\pi(-2+t)} m \right) .$$ We start from the verification $$f(0)=f(1)= \csc^2\left( \frac \pi m \right) .   $$ Next, in order to prove the statement under consideration, it is enough to prove the convexity of $f(t)$ on $(0,1)$. Because $f(t) \in C^2(0,1)$, the convexity is equivalent to $f''(t) \ge 0$ there. After calculating
$$f''(t)=-2\, \left( \csc \left( {\frac {\pi \, \left( t-2 \right) }{m}}
 \right)  \right) ^{2}$$ $$ \left(12\pi^{2} \left( \cos \left( 
\frac {\pi} {m} \right)  \right)^{2} \left( \cot \left( \frac {\pi
 \left( t-2 \right) }{m} \right)  \right) ^{2}t-\right.$$  $$12\,{\pi }^{2}
 \left( \cos \left( {\frac {\pi }{m}} \right)  \right) ^{2} \left(
\cot \left( {\frac {\pi \, \left( t-2 \right) }{m}} \right)  \right) ^
{2}-$$ $$3\,{\pi }^{2} \left( \cot \left( {\frac {\pi \, \left( t-2
 \right) }{m}} \right)  \right) ^{2}{t}^{2}+4\,{\pi }^{2} \left( \cos
 \left( {\frac {\pi }{m}} \right)  \right) ^{2}t-$$ $$8\,\pi \, \left( \cos
 \left( {\frac {\pi }{m}} \right)  \right) ^{2}\cot \left( {\frac {
\pi \, \left( t-2 \right) }{m}} \right) m-4\,{\pi }^{2} \left( \cos
 \left( {\frac {\pi }{m}} \right)  \right) ^{2}-$$ $$\left.{\pi }^{2}{t}^{2}+4\,
\pi \,\cot \left( {\frac {\pi \, \left( t-2 \right) }{m}} \right) mt-{
m}^{2} \right) {m}^{-2}
$$
 we find its asymtotics in $m$ with help of Maple:
 $$
 f''(t)=\frac 2 3 {\frac {{\pi }^{2} \left( {t}^{4}-8\,{t}^{3}+24\,{t}^{2}-20\,t+28
 \right) }{ \left( t-2 \right) ^{4}}}
 +O\left(\frac 1 {m^2}\right), m \to \infty.
 $$ 
  This implies the convexity of $f(x)$ on $(0,1)$ for big values of $m$.
