Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$

Question

I have tried to wrap my head around this for some time now, and quite frankly I am stuck.

Given is that : $$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$

Then what is the value of $a/b$?

I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?

Answer 1

Here is a solution using the observation in Cauchy's proof of Basel problem.

Lemma. We have $$ \prod_{k=1}^{n} \left( t - \tan^2\left(\frac{k\pi}{2n+1}\right)\right) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k. \tag{*} $$

Using this, we immediately know that

$$ a = \sum_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{2} = n(2n+1) $$

and

$$ b = \prod_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{1} = 2n+1. $$

So the ratio $a/b$ is exactly $n$. In OP's case, $n = 8$ and hence the answer is $8$.

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Proof of Lemma. Write $m = 2n+1$. Then by the de Moivre's formula,

\begin{align*} \frac{\cos(mx) + i\sin(mx)}{\cos^m x} = \left( \frac{\cos x + i\sin x}{\cos x} \right)^m = (1 + i\tan x)^m = \sum_{k=0}^{m} \binom{m}{k} (i\tan x)^k. \end{align*}

Taking the imaginary parts only and plugging $m = 2n+1$ back,

\begin{align*} \frac{\sin((2n+1)x)}{\cos^{2n+1} x} = \sum_{k=0}^{n} (-1)^k \binom{m}{2k+1} \tan^{2k+1} x, \end{align*}

and so,

\begin{align*} (-1)^n \frac{\sin((2n+1)x)}{\sin x \cos^{2n} x} = \sum_{k=0}^{n} (-1)^{n-k} \binom{m}{2k+1} \tan^{2k} x. \end{align*}

However, the left-hand side becomes zero if $x_k = \frac{k\pi}{2n+1}$ for $k = 1, \dots, n$, and also note that $\tan^2(x_k)$ for $k = 1, \dots, n$ are all different. So, denoting the right-hand side of $\text{(*)}$ by $P(t)$, i.e., writing

$$ P(t) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k, $$

then this implies that $P(\tan^2(x_k)) = 0$ for $k = 1, \dots, n$, hence $\tan^2(x_k)$ for $k = 1, \dots, n$ are $n$ distinct zeros of $P(t)$. Since the leading coefficient of $P(t)$ is $1$, this proves $\text{(*)}$.

Answer 2

Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$,

$$\tan(2n+1)x=\dfrac{\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}}{\binom{2n+1}0-\binom{2n+1}2t^2+\cdots+(-1)^n\binom{2n+1}{2n}t^{2n}}$$

where $t=\tan x$

So, the roots of $$\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}=0$$ i.e., $$t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(-1)^n(2n+1)t=0$$ are $t_r=\tan\dfrac{r\pi}{2n+1}; r=0,\pm1,\pm2,\cdots,\pm n$

As $r=0\implies \tan\dfrac{r\pi}{2n+1}=0$, the roots of $$t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(-1)^n(2n+1)=0$$ are $t_r=\tan\dfrac{r\pi}{2n+1}; r=\pm1,\pm2,\cdots,\pm n$

Writing $t_r^2=p_r,$

the roots of $$p^n-\binom{2n+1}{2n-1}p^{n-1}+\cdots+(-1)^n(2n+1)=0$$ are $p_r=\tan^2\dfrac{r\pi}{2n+1}; r=1,2,\cdots, n$

Use Vieta's formula

Answer 3

$$𝑎=∑_{𝑟=1}^8 tan^2(\frac{𝑟𝜋}{17})=136$$

$$b=\prod_{𝑟=1}^8 tan^2(\frac{𝑟𝜋}{17})= 17$$

$$\frac{a}{b}=\frac{136}{17}=8$$

This results can be calculated by hand.

$${Tan[\frac{𝜋}{17}]^2, Tan[(2\frac{𝜋}{17}]^2, Tan[(3 \frac{𝜋}{17}]^2, Tan[(4 \frac{𝜋}{17}]^2, Cot[(7 \frac{𝜋}{34}]^2, Cot[(5 \frac{𝜋}{34}]^2, Cot[(3 \frac{𝜋}{34}]^2, Cot[\frac{𝜋}{34}]^2}$$

$${0.0349437, 0.15008, 0.383376, 0.831052, 1.75354, 4.03315, 12.3527, \ 116.461}$$

This confirms the formulas by @sangchul-lee: Cauchys proof of the Basel lemma but with the figures calculated.