Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$

Question

Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$.

I tried some elementary ways while all failed.

Answer 1

Let $z:=e^{i\pi/13}$. Then $\sin(k\pi/13)=(z^k-z^{-k})/2i$ and $\cos(k\pi/13)=(z^k+z^{-k})/2$.

We get $$\tan(\frac{k\pi}{13})\cdot\tan(\frac{3k\pi}{13})=-\frac{z^k-z^{-k}}{z^k+z^{-k}}\cdot\frac{z^{3k}-z^{-3k}}{z^{3k}+z^{-3k}}=-\frac{1-z^{2k}}{1+z^{2k}}\cdot\frac{1-z^{6k}}{1+z^{6k}}$$

Expand this in Taylor series (the coefficients are going to be sums of four or so geometric progressions) and sum these series for $k=1,...,12$. Notice that $1+z^k+z^{2k}+...+z^{12k}=0$, for $k$ not multiple of $13$ and equal to $0$ otherwise. This is because $z^{13}+1=0$.

Now the sum becomes the sum of four or so geometric series (geometric series we know how to add).

Answer 2

From Sum of tangent functions where arguments are in specific arithmetic series and Prove the trigonometric identity $(35)$,

$$\tan13x=\frac{\binom{13}1t-\binom{13}3t^3+\binom{13}5t^5-\binom{13}7t^7+\binom{13}9t^9-\binom{13}{11}t^{11}+t^{13}}{\cdots}$$

where $\displaystyle t=\tan x$

Now if $\displaystyle\tan13x=0,13x=n\pi$ where $n$ is any integer

$\displaystyle\implies x=\frac{n\pi}{13}$ where $0\le n\le13-1$

As $\displaystyle\tan0=0,\tan\frac{n\pi}{13},1\le n\le12$ are the roots of

$\displaystyle\binom{13}1-\binom{13}3t^2+\binom{13}5t^4-\binom{13}7t^6+\binom{13}9t^8-\binom{13}{11}t^{10}+t^{12}=0\ \ \ \ (1)$

As $\displaystyle\tan(r\pi-y)=-\tan y,$ for any integer $r$

the given relation $\displaystyle(i)\sum\limits_{k=1}^{12}\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}=2\sum\limits_{k=1}^6\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}$

$\displaystyle(ii)u_k=\tan^2\frac{k\pi}{13},1\le k\le6$

shall be the roots of

$\displaystyle\binom{13}1-\binom{13}3u+\binom{13}5u^2-\binom{13}7u^3+\binom{13}9u^4-\binom{13}{11}u^5+u^6=0\ \ \ \ (2)$

Now, $\displaystyle y_k=\tan\frac{k\pi}{13}\cdot \tan\frac{3k\pi}{13}=\frac{3u_k-u_k^2}{1-3u_k}$ (using $\tan3A$ formula )

$\displaystyle\implies u_k^2=3u_k-y_k(1-3u_k)=3u_k(1+y_k)-y_k$

So, we need to transform Equation $\#(2)$ in terms of $y$

Then apply Vieta's Formulas.

But my Question is why $\displaystyle\tan\frac{k\pi}{13}$ is paired with $\displaystyle\tan\frac{3k\pi}{13},$ but not with $\displaystyle\tan\frac{2k\pi}{13}$ or $\displaystyle\tan\frac{5k\pi}{13}$ etc.?