How to find the equations whose roots are equal to the following numbers? How to find the equations whose roots are equal to the following numbers ?
(a) $\sin^2\frac{\pi}{2n+1}$,$\sin^2\frac{2\pi}{2n+1}$,$\sin^2\frac{3\pi}{2n+1}$,...,$\sin^2\frac{n\pi}{2n+1}$
(b)$\cot^2\frac{\pi}{2n+1}$,$\cot^2\frac{2\pi}{2n+1}$,$\cot^2\frac{3\pi}{2n+1}$,...,$\cot^2\frac{n\pi}{2n+1}$
I got stuck while solving this problem.This was from a complex number chapter practice problem exercise. I tried to fit in Demoivre's theorem considering roots of unity.But no idea how to bring about the squared sine terms.  Any suggestions?

For $(b),$ see Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$
For $(a),$  
Method $\#1:$ $$\cot^2y=\dfrac{1-\sin^2y}{\sin^2y}$$
Method $\#2:$ 
Using De Moivre's formula,
$$\cos(2m+1)x+i\sin(2m+1)x=(\cos y+i\sin y)^{2m+1}=\cdots$$
Equating the imaginary parts (See this),
$$\sin(2m+1)x=(2n+1)\sin x-\cdots+(-1)^{m-1}2^{2m-2}(2m+1)\sin^{2m-1}x+(-1)^m2^{2m}\sin^{2m+1}x$$
Now if $\sin(2m+1)x=0,(2m+1)x=r\pi$ where $r$ is any integer
$x=\dfrac{r\pi}{2m+1}$ where $r\equiv0,\pm1,\pm2,\cdots,\pm m\pmod{2m+1}$
Can you take it from here?
