3

Find value of:

$$\displaystyle \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg)$$

The solution I tried:

Assume $$P = \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg),$$ with the help of $\tan(\pi-\theta)=-\tan \theta$, then $$P=\tan\bigg(\frac{13\pi}{25}\bigg)\cdot \tan\bigg(\frac{14\pi}{25}\bigg)\cdot \tan\bigg(\frac{15\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{24\pi}{25}\bigg)$$ which gives $$P^2=\prod^{24}_{r=1}\tan\bigg(\frac{r\pi}{25}\bigg).$$

How do I proceed from here?

Leucippus
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jacky
  • 5,194

2 Answers2

4

Consider the polynomial

$$f(t) = \frac{1}{2i}\left((1+it)^{25} - (1-it)^{25}\right) = t^{25} + \cdots + 25 t$$ When $t = \tan\theta$, we have

$$f(\tan\theta) = \frac{1}{2i}\left[\left(\frac{e^{i\theta}}{\cos\theta}\right)^{25} - \left(\frac{e^{-i\theta}}{\cos\theta}\right)^{25}\right] = \frac{1}{\cos^{25}\theta}\sin(25\theta)$$ which vanishes at $\theta = 0, \pm \frac{\pi}{25},\ldots,\pm\frac{12\pi}{25}$.

This implies the $25$ roots of $f(t)$ are $0, \pm \tan\frac{\pi}{25},\ldots,\tan\frac{12\pi}{25}$. Apply Vieta's formula to the coefficient of $t$ in $f(t)$ and notice $\tan\frac{\pi}{25}, \ldots, \tan\frac{12\pi}{25}$ are positive, we obtain:

$$\prod_{k=1}^{12}\left(-\tan^2\frac{k\pi}{25}\right) = (-1)^{24}25 \quad\implies\quad \prod_{k=1}^{12} \tan\frac{k\pi}{25} = \sqrt{25} = 5$$

achille hui
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2

Like Sum of tangent functions where arguments are in specific arithmetic series,

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

If $\tan(2n+1)x=0,(2n+1)x=m\pi$ where $m$ is any integer

So, the roots of $t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(2n+1)(-1)^nt=0$ are $\tan\dfrac{m\pi}{(2n+1)}$ where $0\le m\le 2n$

So, the roots of $t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(2n+1)(-1)^n=0$ are $\tan\dfrac{m\pi}{(2n+1)}$ where $1\le m\le 2n$

$$\implies\prod_{m=1}^{2n}\tan\dfrac{m\pi}{(2n+1)}=(-1)^n(2n+1)$$

$$\implies\prod_{m=1}^n\tan\dfrac{m\pi}{(2n+1)}=+\sqrt{2n+1}$$

as $0<\dfrac{m\pi}{(2n+1)}<\dfrac\pi2$ as $0<m\le n$

and as $\tan(\pi-y)=-\tan y,$ $\implies\tan\dfrac{m\pi}{(2n+1)}=-\tan\left(\pi-\dfrac{m\pi}{(2n+1)}\right)=-\tan\dfrac{(2n+1-m)\pi}{2n+1}$

lab bhattacharjee
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