Proving n∑l=1n−1∑k=1tanlkπ2n+1tanl(k+1)π2n+1=0\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0

Prove that nl=1n1k=1tanlkπ2n+1tanl(k+1)π2n+1=0


It is very easy to prove this identity for each fixed n . For example let n=6; writing out all terms in a 5×6 matrix, we obtain:

tanπ13tan2π13tan2π13tan3π13tan3π13tan4π13tan4π13tan5π13tan5π13tan6π13tan2π13tan4π13tan4π13tan6π13tan6π13tan8π13tan8π13tan10π13tan10π13tan12π13tan3π13tan6π13tan6π13tan9π13tan9π13tan12π13tan12π13tan15π13tan15π13tan18π13tan4π13tan8π13tan8π13tan12π13tan12π13tan16π13tan16π13tan20π13tan20π13tan24π13tan5π13tan10π13tan10π13tan15π13tan15π13tan20π13tan20π13tan25π13tan25π13tan30π13tan6π13tan12π13tan12π13tan18π13tan18π13tan24π13tan24π13tan30π13tan30π13tan36π13

one can notice then, that the first column vanish the fourth one :

tanπ13tan2π13=tan12π13tan15π13

tan2π13tan4π13=tan24π13tan30π13

tan3π13tan6π13=tan16π13tan20π13

tan4π13tan8π13=tan4π13tan5π13

tan5π13tan10π13=tan8π13tan10π13

tan6π13tan12π13=tan20π13tan25π13

and the third column vanish the fifth one :

tan3π13tan4π13=tan30π13tan36π13

tan6π13tan8π13=tan5π13tan6π13

tan9π13tan12π13=tan25π13tan30π13

tan12π13tan16π13=tan10π13tan12π13

tan15π13tan20π13=tan20π13tan24π13

tan18π13tan24π13=tan15π13tan18π13

while the second column is self-vanishing:

tan2π13tan3π13=tan10π13tan15π13

tan4π13tan6π13=tan6π13tan9π13

tan8π13tan12π13=tan12π13tan18π13 .

So the equality occurs.
But how to generalize the proof?

Answer

Micah already pointed the way in a comment: The identity

tanatanb=tanatanbtan(ab)1

causes the inner sum to telescope. To make full use of this, let’s extend the inner sum to k=2n:

nl=12nk=1tanlkπ2n+1tanl(k+1)π2n+1=nl=12nk=1(tanl(k+1)π2n+1tanlkπ2n+1tanlπ2n+11)=nl=1(tanl(2n+1)π2n+1tanlπ2n+1tanlπ2n+12n)=nl=1(012n)=n(2n+1).

This sum contains each of the terms we want to sum twice, with mirror symmetry, and it contains one additional term in the middle for k=n. Thus, for our sum to vanish, we need

nl=1tanlnπ2n+1tanl(n+1)π2n+1=n(2n+1).

The arguments of the two factors add to lπ, so they’re negatives of each other, so we’re looking for

nl=1tan2lnπ2n+1.

We can again extend the sum to 2n to double it, since the arguments form pairs that add up to nπ; then, since n and 2n+1 are coprime, we can replace ln by l while traversing the same arguments; and then we can set the upper limit back to n, since the arguments still add up to π in pairs. Thus, what we need is

nl=1tan2lπ2n+1=n(2n+1).

How to find this sum is shown at Prove that n1k=1tan2kπ2n=(n1)(2n1)3. Adapting the argument in the accepted answer there for our odd denominator, we obtain

(coskπ2n+1+isinkπ2n+1)2n+1=(1)k,

taking the imaginary part,

nr=0(2n+12r+1)(coskπ2n+1)2n2r(isinkπ2n+1)2r+1=0,

dividing by \left(\cos\frac{k\pi}{2n+1}\right)^{2n+1},


\sum_{r=0}^n\binom{2n+1}{2r+1}\left(\mathrm i\tan\frac{k\pi}{2n+1}\right)^{2r+1}=0\;,

and dividing by \tan\frac{k\pi}{2n+1} and letting x=\tan^2\frac{k\pi}{2n+1},


\sum_{r=0}^n\binom{2n+1}{2r+1}(-x)^r=0\;.

Then Vieta’s formula shows that the sum of the roots of this equation is

\frac{\binom{2n+1}2}{\binom{2n+1}0}=n(2n+1)\;,

as required.

[Update:]

This answer suggests an alternative, more elementary way to calculate the sum of the squares of the tangents: On the grid \frac{l\pi}{2n+1}, the tangent decomposes into n mutually orthogonal sines, each of whose dot product with itself is 2n+1, so the dot product of the tangent with itself is n(2n+1).

Attribution
Source : Link , Question Author : Maria Mikolayevskaya , Answer Author : Community

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