Question Number 217871 by hardmath last updated on 22/Mar/25
![n ≥ 2 Prove that: Π_(k=1) ^n tg [ (π/3) (1 + (3^k /(3^n − 1)))] = Π_(k=1) ^n ctg [ (π/3) (1 − (3^k /(3^n − 1)))]](https://www.tinkutara.com/question/Q217871.png)
$$\mathrm{n}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\mathrm{tg}\:\left[\:\frac{\pi}{\mathrm{3}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{1}}\right)\right]\:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\mathrm{ctg}\:\left[\:\frac{\pi}{\mathrm{3}}\:\left(\mathrm{1}\:−\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{1}}\right)\right] \\ $$
Answered by MrGaster last updated on 05/Apr/25
![Π_(k=1) ^n tan[ (π/3) (1 + (3^k /(3^n − 1)))] =Π_(k=1) ^n (1/(cot[(π/3)(1+(3^k /(3^n −1)))])) Π_(k=1) ^n cot [ (π/3) (1 + (3^k /(3^n − 1)))] =Π_(k=1) ^n cot[(π/3)−(π/3)∙(3^k /(3^n −1))] cot((π/3)−x)=tan(x+(π/6)) Π_(k=1) ^n cot[(π/3)(1−(3^k /(3^n −1)))]=Π_(k=1) ^n tan[(π/3)∙(3^k /(3^n −1))+(π/6)] Π^n tan[(π/3)(1+(3^k /(3^n −1)))]=Π_(k=1) ^n tan[(π/3)+(π/3)∙(3^k /3^(n−1) )] tan((π/3)+x)=(((√3)+tan x)/(1−(√3)tan x)) Π_(k=1) ^n tan[(π/3)(1+(3^k /(3^n −1)))]=Π_(k=1) ^n (((√3)+tan((π/3)∙(3^k /(3^n −1))))/(1−(√3)tan((π/3)∙(3^k /(3^n −1))))) tan((π/3)∙(3^k /(3^n −1)))=tan((π/3)∙(3^k /(3^n −1))) Π_(k=1) ^n (((√3)+tan((π/3)∙(3^k /(3^n −1))))/(1−(√3)tan((π/3)∙(3^k /(3^n −1)))))=Π_(k=1) ^n cot[(π/3)(1−(3^k /(3^n −1)))] Π_(k=1) ^n tg [ (π/3) (1 + (3^k /(3^n − 1)))] = Π_(k=1) ^n ctg [ (π/3) (1 − (3^k /(3^n − 1)))]](https://www.tinkutara.com/question/Q218287.png)
$$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\mathrm{tan}\left[\:\frac{\pi}{\mathrm{3}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{1}}\right)\right]\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\mathrm{cot}\left[\frac{\pi}{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)\right]} \\ $$$$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\mathrm{cot}\:\left[\:\frac{\pi}{\mathrm{3}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{1}}\right)\right]\:=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cot}\left[\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right] \\ $$$$\mathrm{cot}\left(\frac{\pi}{\mathrm{3}}−{x}\right)=\mathrm{tan}\left({x}+\frac{\pi}{\mathrm{6}}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cot}\left[\frac{\pi}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)\right]=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{tan}\left[\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}+\frac{\pi}{\mathrm{6}}\right] \\ $$$$\overset{{n}} {\prod}\mathrm{tan}\left[\frac{\pi}{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)\right]=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{tan}\left[\frac{\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}−\mathrm{1}} }\right] \\ $$$$\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}+{x}\right)=\frac{\sqrt{\mathrm{3}}+\mathrm{tan}\:{x}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\:{x}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{tan}\left[\frac{\pi}{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)\right]=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\sqrt{\mathrm{3}}+\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)} \\ $$$$\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)=\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\sqrt{\mathrm{3}}+\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\centerdot\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cot}\left[\frac{\pi}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{3}^{{k}} }{\mathrm{3}^{{n}} −\mathrm{1}}\right)\right] \\ $$$$\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\mathrm{tg}\:\left[\:\frac{\pi}{\mathrm{3}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{1}}\right)\right]\:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\prod}}\:\mathrm{ctg}\:\left[\:\frac{\pi}{\mathrm{3}}\:\left(\mathrm{1}\:−\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{1}}\right)\right] \\ $$
Commented by hardmath last updated on 05/Apr/25
$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$