4

Prove that $$\sum_{1\leq k < j\leq n}\tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4}$$

Mithlesh Upadhyay
  • 4,787
hxthanh
  • 1,522
  • 1
    Where does this problem come from ? What have you attempted ? – Jean Marie Aug 02 '16 at 06:32
  • It's nice identity, I have just detected it – hxthanh Aug 02 '16 at 06:51
  • Related http://math.stackexchange.com/questions/217240/reference-for-a-tangent-squared-sum-identity . This identity can be proved by combining the sums for squares and fourth powers of tangents from this link http://functions.wolfram.com/ElementaryFunctions/Tan/23/01/ – Martin Nicholson Aug 02 '16 at 07:10
  • See http://math.stackexchange.com/questions/1562037/question-regarding-fn-cot2-left-frac-pi-n-right-cot2-left-frac2-pin – lab bhattacharjee Aug 06 '16 at 05:21

1 Answers1

5

It is enough to recall Cauchy's proof of the Basel problem, relying on the identity:

$$ \binom{2n+1}{1}t^n-\binom{2n+1}{3}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{2n+1}=\prod_{k=1}^{n}\left(t-\cot^2\frac{k\pi}{2n+1}\right)\tag{1} $$ If we consider the "reciprocal polynomial" $$ \binom{2n+1}{2n+1}t^n-\binom{2n+1}{2n-1}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{1}=\prod_{k=1}^{n}\left(t-\tan^2\frac{k\pi}{2n+1}\right)\tag{2} $$ the original sum is just the second elementary symmetric polynomial of the roots, $e_2$, that by Vieta's theorem is given by the coefficient of $t^{n-2}$ in the LHS of $(2)$, so: $$ \sum_{1\leq j<k\leq n}\tan^2\left(\frac{j\pi}{2n+1}\right)\tan^2\left(\frac{k\pi}{2n+1}\right)=\binom{2n+1}{2n-3}=\color{red}{\binom{2n+1}{4}}\tag{3}$$ as wanted.

Jack D'Aurizio
  • 353,855