4

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.

Thomas Andrews
  • 177,126
Steven Sun
  • 1,190

2 Answers2

5

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).

OR.
  • 5,941
  • Mhm..How to expand this in Taylor series? – Steven Sun Mar 20 '14 at 13:03
  • @Smy2012 First, take into account that there, most surely, are completely elementary solutions to this problem (using ad hoc tricks with trigonometric identities). This is just an algorithmic (mechanical) solution, that probably solves many other problems as well. To expand in Taylor series (at $z=0$) notice that the function is rational. Oh! I should say that we only need Taylor series as powers of $z^{2k}$. So, we can think the function is $\frac{1-Z}{1+Z}\cdot\frac{1-Z^3}{1+Z^3}$, and the replace $Z$ by $z^{2k}$ in the series. In general, to expand a rational function in ... – OR. Mar 20 '14 at 15:31
  • ... series, you can begin by computing its partial fraction decomposition (probably better over the complex numbers), and then expand each of the partial fractions in series using the binomial theorem. – OR. Mar 20 '14 at 15:33
0

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.?

Community
  • 1
lab bhattacharjee
  • 274,582