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Find the ratio of

$\sum_{r=1}^n$ $\tan^2({rπ\over 17})$ and $\prod_{r=1}^n$$\tan^2({rπ\over 17})$

My attempt : I noticed that calculating them separately and then finding ratio is next to impossible. So I thought of somehow making a polynomial whose roots are $tan^2({rπ\over 17})$ , so that I can find the ratio by the ratio of sum and product of roots. But I am not able to form such a polynomial. I tried to write

$\sin(17 \theta)=0$

And then tried to convert it to a polynomial of $\cos(2\theta)$ , and then replace $\cos(2\theta)$ by $ {1-\tan^2(\theta)\over 1+\tan^2(\theta)} $ but that approach was too hard to follow. Maybe I need some better way to proceed with this method itself.

Or else I am pretty sure that it can be done in a less hectic way by complex analysis. Although I tried to find $\cos(17\theta)$ in terms of $\cos(2\theta)$ using de Moivre's theorem but I faced difficulties in that too.

Could someone please help me with this ?

Thanks.

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ARROW
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1 Answers1

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Using De Moivre's Theorem: $(\cos(\theta)+i\sin(\theta))^{2n+1}=\cos((2n+1)(\theta))+i\sin((2n+1)\theta)$.

Expanding using binomial theorem and comparing imaginary terms,

$$\sin((2n+1)\theta) = \sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k+1} \cos^{2n-2k}(\theta) \sin^{2k+1}(\theta)$$

$$=\tan(\theta) \cos^{2n+1}(\theta) \sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k+1} (\tan(\theta))^{2k}$$.

$$\implies \sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k+1} (\tan(\theta))^{2k}=0$$

As $\forall \theta = \frac{jπ}{2n+1}, 1 \leq j \leq 2n+1, \sin((2n+1)\theta) =0$ and $\tan(\theta), \cos(\theta) \neq 0$

So, $\tan^2(\frac{jπ}{2n+1}), 1 \leq j \leq 2n+1$, are the roots of the equation $\sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k+1} (x)^{2k}=0$.

Hence the sum and product required is the sum and product of the roots of the above equation respectively.

Thus the required ratio is $\frac{{2n+1 \choose 2}}{{2n+1 \choose 2n}} = n$. Here $n=8$.

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