Finding value of $k$ in trigonometric equation.

Question

$$\prod^{50}_{r=1}\tan\bigg[\frac{\pi}{3}\bigg(1+\frac{3^r}{3^{50}-1}\bigg)\bigg]=k\prod^{50}_{r=1}\cot\bigg[\frac{\pi}{3}\bigg(1-\frac{3^r}{3^{50}-1}\bigg)\bigg]$$

Try:$$\tan (60^\circ+x)=\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}$$

and $$\cot(60^\circ-x)=\frac{\sqrt{3}-\tan x}{\sqrt{3}+\tan x}$$

I did not understand how to solve it other then formule above mention. Help me

Reminds me of https://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt — lab bhattacharjee, Feb 01 '18 at 06:11
There is kind of a bump in the road at $r=50$. I get something like $k\approx-1/7647619650288224399536.08654533918$. — user5713492, Feb 01 '18 at 07:27
Extremelly sir it was product instead of sigma and varies from $1$ to $50$ — DXT, Feb 01 '18 at 10:53

DXT · Accepted Answer · 2018-02-01T10:54:17.023

$$X= \prod^{50}_{r=1}\tan \left(\frac{\pi}{3}\left(1+\frac{3^r}{3^{50}-1}\right)\right)$$ and $$Y = \prod^{50}_{r=1}\cot \left(\frac{\pi}{3}\left(1-\frac{3^r}{3^{50}-1}\right)\right)$$

Assuming $$a_r=\frac{\pi}{3}\cdot \frac{3^r}{3^{50}-1}$$.Then

$$\tan(\frac{\pi}3+a_r)=\frac{\sqrt 3+\tan a_r}{1-\sqrt 3\tan a_r}$$

and $$\cot(\frac{\pi}3-a_r)=\frac{1+\sqrt 3\tan a_r}{\sqrt 3-\tan a_r}$$

So $$\frac{\tan(\frac{\pi}3+a_r)}{\cot(\frac{\pi}3-a_r)}=\frac{3-\tan^2a_r}{1-3\tan^2a_r}=\frac{\tan 3a_r}{\tan a_r}=\frac{\tan a_{r+1}}{\tan a_r}$$

So $$XY=\prod_{r=1}^{50}\frac{\tan a_{r+1}}{\tan a_r}=\frac{\tan a_{51}}{\tan a_1}$$

Now $a_{51}-a_1=\pi$ and so $\displaystyle\frac XY=1$ and so $X=Y$

Finding value of $k$ in trigonometric equation.

1 Answers1