What is the formula to find the sum of $\cot^2$ for a given range?

[Illmatic1]

Find the value of $ \displaystyle \sum_{k=1}^{n} \cot^2\bigg(\frac{\pi k}{2n+1} \bigg) $

 

[kaliprasad]

Spoiler

We have
$(\cos\, mt + i \sin\, mt) = (\cos\, t + i \sin\, t)^m$
= $\sum\limits_{k=0}^m {m \choose k } (\sin\,t)^k (i \cos\,t) ^{m-k} $
now compare imaginary parts on both sides to get
$\sin\, mt = {m \choose 1 }\cos^{m-1}a \sin\, a - {m \choose 3 }\cos^{m-3}a \sin ^3 a + ... $
putting m = 2n + 1 we get
$\sin(2n+1) t = {2n+ 1 \choose 1 }\cos^{2n}a \sin\, a - {2n+1 \choose 3 }\cos^{2n-2}a \sin ^3 a + ... $
= $\sin^{2n+1}a ({2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a + ...) $
The LHS hand hence RHS becomes zero for $t = \frac{k\pi}{2n+1}$
so for above t
${2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a +... = 0 $
so $\cot^2\frac{\pi}{2n+1}a$, $\cot^2\frac{2\pi}{2n+1}a$ ... $\cot^2\frac{n\pi}{2n+1}a$
are roots of the equation
$({2n+ 1 \choose 1 }x^n - {2n+1 \choose 3 }x^(n-2) + ...=0$
so sum of roots
= $\sum\limits_{k=1}^n \cot^2\frac{k}{2n+1} =\frac{{2n+1 \choose 3 }}{{2n+1 \choose 1 }} $
= $\frac{n(2n-1)}{3}$

 

[lfdahl]

Thankyou very much, kaliprasad, for a very clever solution! I have looked at it and have some questions.
Please excuse my lack of understanding. Thankyou in advance!

Spoiler

Questions:

(1). The argument, $a$, is it a shift of the variable $t$: $a = \frac{\pi}{2}-t$?

(2). Are the roots: $cot^2(\frac{\pi}{2n+1}a), cot^2(\frac{2\pi}{2n+1}a)$, ...?
I´d expect the roots to be: $cot^2(\frac{\pi}{2n+1}), cot^2(\frac{2\pi}{2n+1})$, ...

(3). The polynomial of degree $n$: shouldn´t it be written as:
${2n+1 \choose 1}x^n-{2n+1 \choose 3}x^{n-1}...$?

 

[kaliprasad]

Thankyou very much, kaliprasad, for a very clever solution! I have looked at it and have some questions.
Please excuse my lack of understanding. Thankyou in advance!

Spoiler

Questions:

(1). The argument, $a$, is it a shift of the variable $t$: $a = \frac{\pi}{2}-t$?

(2). Are the roots: $cot^2(\frac{\pi}{2n+1}a), cot^2(\frac{2\pi}{2n+1}a)$, ...?
I´d expect the roots to be: $cot^2(\frac{\pi}{2n+1}), cot^2(\frac{2\pi}{2n+1})$, ...

(3). The polynomial of degree $n$: shouldn´t it be written as:
${2n+1 \choose 1}x^n-{2n+1 \choose 3}x^{n-1}...$?

Thanks for the good observation

Spoiler

t and a are same. Sorry for the mistake the 1st line in expansion was wrong.
(2) and (3) you are right

I correct my solution post it below

Spoiler

We have
$(\cos\, ma + i \sin\, ma) = (\cos\, a + i \sin\, a)^m$
= $\sum\limits_{k=0}^m {m \choose k } (i\sin\,a)^k (\cos\,a) ^{m-k} $
now compare imaginary parts on both sides to get
$\sin\, ma = {m \choose 1 }\cos^{m-1}a \sin\, a - {m \choose 3 }\cos^{m-3}a \sin ^3 a + ... $
putting m = 2n + 1 we get
$\sin(2n+1) a = {2n+ 1 \choose 1 }\cos^{2n}a \sin\, a - {2n+1 \choose 3 }\cos^{2n-2}a \sin ^3 a + ... $
= $\sin^{2n+1}a ({2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a + ...) $
The LHS hand hence RHS becomes zero for $a = \frac{k\pi}{2n+1}$
so for above t
${2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a +... = 0 $
so $\cot^2\frac{\pi}{2n+1}$, $\cot^2\frac{2\pi}{2n+1}$ ... $\cot^2\frac{n\pi}{2n+1}$
are roots of the equation
$({2n+ 1 \choose 1 }x^n - {2n+1 \choose 3 }x^{n-1} + ...=0$
so sum of roots
= $\sum\limits_{k=1}^n \cot^2\frac{k}{2n+1} =\frac{{2n+1 \choose 3 }}{{2n+1 \choose 1 }} $
= $\frac{n(2n-1)}{3}$