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Working around, I found some Tan Binomial formulas.

Let's $S$ be a set such that:

$$ S=\left\{\text{ }\tan ^2\left(\frac{1\pi }{n}\right), \tan^2\left(\frac{2\pi }{n}\right), \tan^2\left(\frac{3\pi }{n}\right)\text{ },\text{...},\tan^2\left(\frac{k \pi }{n}\right) \right\} \text{for } k \text{ in } \text{range } \left[ 1,\left\lfloor \frac{(n-1)}{2}\right\rfloor \right] $$

and let's $S_k$ be a k-subset of $S$. For example, for k=2, we have:

$$ S_k = \left\{\text{ }\left\{\tan ^2\left(\frac{1\pi }{n}\right), \tan ^2\left(\frac{2\pi }{n}\right)\right\},\left\{\tan ^2\left(\frac{1\pi }{n}\right), \tan ^2\left(\frac{3\pi }{n}\right)\right\},\text{...} ,\left\{\tan ^2\left(\frac{(k-1)\pi }{n}\right), \tan ^2\left(\frac{k \pi }{n}\right)\right\}\right\} $$

Then one formula gives:

$$ \left\{ \begin{array}{c} \text{if n Even},\text{ }n\times \left(\text{Sum of the Product of k-Subset }S_k\right)=\left( \begin{array}{c} n \\ 2k+1 \end{array} \right) \\ \text{if n Odd},\text{ }\left( \text{Sum of the Product of k-Subset } S_k \right)=\left( \begin{array}{c} n \\ 2k \end{array} \right) \end{array} \right. $$

Here are some examples with n=7 and k varying from 1 to 3. We have: $S=\left\{\tan ^2\left(\frac{\pi 1}{7}\right),\tan ^2\left(\frac{\pi 2}{7}\right),\tan ^2\left(\frac{\pi 3}{7}\right)\right\}$

  1. for k=1, k-subset is $S_1=\left\{ \left\{\tan ^2\left(\frac{\pi 1}{7}\right)\right\},\left\{\tan ^2\left(\frac{\pi 2}{7}\right)\right\},\left\{\tan ^2\left(\frac{\pi 3}{7}\right)\right\} \right\}$, so $\tan ^2\left(\frac{\pi 1}{7}\right)+\tan ^2\left(\frac{\pi 2}{7}\right)+\tan ^2\left(\frac{\pi 3}{7}\right)=\left( \begin{array}{c} 7 \\ 2 \end{array} \right)=21$
  2. for k=2, k-subset is $S_2=\left\{ \left\{\tan ^2\left(\frac{\pi 1}{7}\right),\tan ^2\left(\frac{\pi 2}{7}\right)\right\},\left\{\tan ^2\left(\frac{\pi 2}{7}\right),t\tan ^2\left(\frac{\pi 3}{7}\right)\right\},\left\{\tan ^2\left(\frac{\pi 1}{7}\right),t\tan ^2\left(\frac{\pi 3}{7}\right)\right\} \right\}$, so $\tan ^2\left(\frac{\pi 1}{7}\right) \tan ^2\left(\frac{\pi 2}{7}\right)+\tan ^2\left(\frac{\pi 3}{7}\right) \tan ^2\left(\frac{\pi 2}{7}\right)+\tan ^2\left(\frac{\pi 1}{7}\right) \tan ^2\left(\frac{\pi 3}{7}\right)=\left( \begin{array}{c} 7 \\ 4 \end{array} \right)=35$
  3. for k=3, k-subset is $S_3=\left\{ \left\{\tan ^2\left(\frac{\pi 1}{7}\right),\tan ^2\left(\frac{\pi 2}{7}\right),\tan ^2\left(\frac{\pi 3}{7}\right)\right\} \right\}$, so $\tan ^2\left(\frac{\pi 1}{7}\right) \tan ^2\left(\frac{\pi 2}{7}\right) \tan ^2\left(\frac{\pi 3}{7}\right)=\left( \begin{array}{c} 7 \\ 6 \end{array} \right)=7$

Tan binomial formulaes 1 Tan binomial formulaes 2 Tan binomial formulaes 3

Here is the Mathematica code corresponding to the images, if someone wants to play around:

Manipulate[
Module[
{
set1,S1,Sk1,psk1,pskEven1,
set2,S2,Sk2,psk2,pskEven2,
ProdSumSubset,binEven,bin1,bin2,
hf,rh,opt1
},
hf[x_]:=HoldForm@x;
rh[x_]:=ReleaseHold@x;

set1[n_]:=Table[ Tan[hf@( k)*Pi/hf@(n) ]^2,{k,1,Floor@((n-1)/2)}] ;
set2[n_]:=Table[ Tan[hf@(2 k-1)*Pi/hf@(2*n) ]^2,{k,1,Floor@(n/2)}] ;
ProdSumSubset[S_,k_]:=Plus@@(Times@@#&/@ Subsets[S,{k}]);

bin1[n_,k_]:=Binomial[n,hf@(2*k)];
bin2[n_,k_]:=Binomial[n,hf@(2*k+1)];

opt1={ Frame->All,Alignment->{{Center,Left},Center},ItemSize->{{Scaled@.25,Scaled@.75}},FrameStyle->GrayLevel[0.7`] };
TraditionalForm@
Grid[
{
{ "n",n },
{ "k",k },
{"",SpanFromLeft},

{ "Set S1",S1=set1[n] },
{ "k-Subset Sk1",Sk1=Subsets[S1,{k}]//StandardForm },
psk1=ProdSumSubset[S1,k];
pskEven1=If[ EvenQ@n,n*psk1,psk1 ];
binEven=If[ EvenQ@n,bin2[n,k],bin1[n,k] ];
{ "Sum of product of k-subset Sk1",Row[{ pskEven1,"=", pskEven1//rh//N,"=",binEven,"=",binEven//rh}] },
{"",SpanFromLeft},

{ "Set S2",S2=set2[n] },
{ "k-Subset Sk2",Sk2=Subsets[S2,{k}]//StandardForm },
psk2=ProdSumSubset[S2,k];
pskEven2=If[ EvenQ@n,psk2,n*psk2 ];
binEven=If[ EvenQ@n,bin1[n,k],bin2[n,k] ];
{ "Sum of product of k-subset Sk2",Row[{ pskEven2,"=", pskEven2//rh//N,"=",binEven,"=",binEven//rh}] }

},opt1
]
]
,{{n,7},1,20,1,Appearance->"Open"}
,{{k,1},1,10,1,Appearance->"Open"}
]

The same holds for a set $S$ such that:

$$ S=\left\{\text{ }\tan ^2\left(\frac{1\pi }{2n}\right), \tan ^2\left(\frac{3\pi }{2n}\right), \tan ^2\left(\frac{5\pi }{2n}\right)\text{ },\text{...},\tan ^2\left(\frac{(2k-1)\pi }{2n}\right) \right\} \text{for}\text{ }k\text{ }\text{in}\text{ }\text{range} \left[ 1,\left\lfloor \frac{n}{2}\right\rfloor \right] $$

then:

$$ \left\{ \begin{array}{c} \text{if } n \text{ Even}, \left( \text{Sum } \text{of } \text{the } \text{Product } \text{of } k-\text{Subset }S_k \right)=\left( \begin{array}{c} n \\ 2k \end{array} \right) \\ \text{if } n \text{ Odd},\text{ }n\times \left( \text{Sum } \text{of } \text{the } \text{Product } \text{of } k-\text{Subset } S_k \right)=\left( \begin{array}{c} n \\ 2k+1 \end{array} \right) \end{array} \right. $$

For example, for n=6 and k=2: $$ \tan ^2\left(\frac{\pi 1}{12}\right) \tan ^2\left(\frac{\pi 3}{12}\right)+\tan ^2\left(\frac{\pi 1}{12}\right) \tan ^2\left(\frac{\pi 5}{12}\right)+\tan ^2\left(\frac{\pi 3}{12}\right) \tan ^2\left(\frac{\pi 5}{12}\right)=\left( \begin{array}{c} 6 \\ 4 \end{array} \right)=15 $$

$ $
$ $ These formulas can be explained this way:
Let's
$z=1+i x=|z|e^{i \text{arcTan}(x)}$ and it's complex conjuguate
$z^*=1-i x=|z|e^{-i \text{arcTan}(x)}$
then:
$$ \text{Cos}(n \text{Arctan} x)=\frac{z^n+z^{*n}}{2 |z|^n}=\frac{(1+i x)^n+(1-i x)^n}{2\left(1+x^2\right)^{\frac{n}{2}}}=\frac{\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor } (-1)^k\text{ }\left( \begin{array}{c} n\\2 k\end{array} \right)x^{2 k}}{\left(1+x^2\right)^{\frac{n}{2}}} $$ Where the numerator is a binomial polynomial (For ex:$\text{Cos} (6 \text{Arctan} x)=\frac{-x^6+15 x^4-15 x^2+1}{\left(x^2+1\right)^3}$,$\text{Cos} (7 \text{Arctan} x)=\frac{-7 x^6+35 x^4-21 x^2+1}{\left(x^2+1\right)^{7/2}}$,etc...) admitting solutions in the form $\pm \text{Tan}\left(\frac{(2k+1) \pi }{2n}\right)$ for $k\in\mathbb{N}$.

Using a set $S=\left\{\text{ }\tan ^2\left(\frac{1\pi }{2n}\right), \tan ^2\left(\frac{3\pi }{2n}\right), \tan ^2\left(\frac{5\pi }{2n}\right)\text{ },\text{...},\tan ^2\left(\frac{(2k-1)\pi }{2n}\right) \right\}$ with $k$ in range $\left[1,\left\lfloor\frac{n}{2}\right\rfloor\right]$, we can then rewrite the numerator as a same order polynomial: $$ \text{Cos}(n \text{Arctan} x)=\frac{\left(x^2-\text{Tan}^2\left(\frac{1\pi }{2n}\right)\right)\left(x^2-\text{Tan}^2\left(\frac{3\pi }{2n}\right)\right)\text{...}\left(x^2-\text{Tan}^2\left(\frac{(2k-1)\pi }{2n}\right)\right)}{\left(1+x^2\right)^{\frac{n}{2}}} $$ Expanding the numerator and matching the binomial coefficient for each order gives one formula, the other coming from: $$ \text{Sin}(n \text{Arctan} x)=\frac{z^n-z^{*n}}{2i |z|^n}=\frac{\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor } (-1)^k\text{ }\left( \begin{array}{c} n \\ 2 k+1 \end{array} \right)x^{2 k+1}}{\left(1+x^2\right)^{\frac{n}{2}}} $$ And using the same reasoning. $ $
$ $

I could not find these formulas on the web, only special cases. Could someone tell me if they exist?

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  • $\begingroup$ These sums are the coefficients of the (monic) polynomial with roots $\tan^2(k\pi/n)$, $1\le k\le(n-1)/2$, aren't they? I bet there's some literature on these, especially in the case where $n$ is prime, and we're talking about the minimal polynomial of $\tan^2(\pi/n)$. $\endgroup$ – Gerry Myerson Oct 8 '13 at 12:22
  • $\begingroup$ Yes, you're right @gerry, this is all about polynomials with roots $\tan \left(\frac{k \pi }{n}\right) $, but still, I don't find nothing about this... $\endgroup$ – Eddy Khemiri Oct 8 '13 at 17:08
  • $\begingroup$ Five years ago, there was a discussion on the Usenet newsgroup sci.math about minimal polynomials for tangents, and I supplied this list of possible references: S Beslin, V de Angelis, Math Mag 77 (2004) 146-149; D H Lehmer, Amer Math Monthly 40 (1933) 165-166; D Surowski, P McCombs, Missouri J Math Sci 15 (2003) 4-14; W Watkins, J Zeitlin, Amer Math Monthly 100 (1993) 471-474; K W Wegner, Amer Math Monthly 66 (1959) 52-53. Also corunduminium.com/Trigpolys.html may be worth a look. $\endgroup$ – Gerry Myerson Oct 8 '13 at 22:10
  • $\begingroup$ Also, I think you'll find J S Calcut, Rationality and the tangent function, available at oberlin.edu/faculty/jcalcut/tanpap.pdf to be of interest. $\endgroup$ – Gerry Myerson Oct 8 '13 at 22:20
  • $\begingroup$ So, have you looked at all the links I found for you? $\endgroup$ – Gerry Myerson Oct 12 '13 at 5:10
1
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It seems that the formulas on page 10 of J S Calcut, Rationality and the tangent function, are what OP is looking for.

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  • $\begingroup$ but note that it gives the polynomials with its sets of solutions, but do not correlate the set of solutions with the binomial coefficients.. $\endgroup$ – Eddy Khemiri Oct 14 '13 at 9:12
  • $\begingroup$ I don't know what you mean by that. And I don't know what you expect by way of an answer to your question. Your question was whether anyone had ever found these formulas relating sums of product of tangents to binomial coefficients. Calcut found formulas relating the polynomial whose roots are the tangents to binomial coefficients. The only thing missing is the observation that if the roots are tangents then the coeffcients are sums of products of tangents, but that's an immediate observation based on the distributive law, and has nothing to do with tangents. What more do you want? $\endgroup$ – Gerry Myerson Oct 15 '13 at 11:17

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