Prove that n∑l=1n−1∑k=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=tana−tanbtan(a−b)−1
causes the inner sum to telescope. To make full use of this, let’s extend the inner sum to k=2n:
n∑l=12n∑k=1tanlkπ2n+1tanl(k+1)π2n+1=n∑l=12n∑k=1(tanl(k+1)π2n+1−tanlkπ2n+1tanlπ2n+1−1)=n∑l=1(tanl(2n+1)π2n+1−tanlπ2n+1tanlπ2n+1−2n)=n∑l=1(0−1−2n)=−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
n∑l=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
−n∑l=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
n∑l=1tan2lπ2n+1=n(2n+1).
How to find this sum is shown at Prove that n−1∑k=1tan2kπ2n=(n−1)(2n−1)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,
n∑r=0(2n+12r+1)(coskπ2n+1)2n−2r(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
