Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 96260 by bemath last updated on 31/May/20

tan ((π/9))+tan (((4π)/9))+tan (((7π)/9)) =?

$$\mathrm{tan}\:\left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{7}\pi}{\mathrm{9}}\right)\:=? \\ $$

Answered by john santu last updated on 31/May/20

set x = (π/9), ((4π)/9), ((7π)/9) 3x = (π/3), ((4π)/3), ((7π)/3) tan 3x = tan (π/3)=tan ((4π)/3)=tan ((7π)/3) = (√3) let q = tan x ⇒tan 3x = ((3q−q^3 )/(1−3q^2 )) ⇒(√3) −3(√3) q^2 = 3q−q^3 q^3 −3(√3) q^2 −3q+(√3) = 0 have roots are tan (π/9), tan ((4π)/9), tan ((7π)/9) by Vieta′s rule tan (π/9) + tan ((4π)/9) + tan ((7π)/9) = 3(√3)

$$\mathrm{set}\:{x}\:=\:\frac{\pi}{\mathrm{9}},\:\frac{\mathrm{4}\pi}{\mathrm{9}},\:\frac{\mathrm{7}\pi}{\mathrm{9}} \\ $$$$\mathrm{3}{x}\:=\:\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{4}\pi}{\mathrm{3}},\:\frac{\mathrm{7}\pi}{\mathrm{3}}\: \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:=\:\mathrm{tan}\:\frac{\pi}{\mathrm{3}}=\mathrm{tan}\:\frac{\mathrm{4}\pi}{\mathrm{3}}=\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{3}}\:=\:\sqrt{\mathrm{3}} \\ $$$$\mathrm{let}\:{q}\:=\:\mathrm{tan}\:{x}\:\Rightarrow\mathrm{tan}\:\mathrm{3}{x}\:=\:\frac{\mathrm{3}{q}−{q}^{\mathrm{3}} }{\mathrm{1}−\mathrm{3}{q}^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\:−\mathrm{3}\sqrt{\mathrm{3}}\:{q}^{\mathrm{2}} \:=\:\mathrm{3}{q}−{q}^{\mathrm{3}} \\ $$$${q}^{\mathrm{3}} −\mathrm{3}\sqrt{\mathrm{3}}\:{q}^{\mathrm{2}} \:−\mathrm{3}{q}+\sqrt{\mathrm{3}}\:=\:\mathrm{0} \\ $$$${have}\:{roots}\:{are}\:\mathrm{tan}\:\frac{\pi}{\mathrm{9}},\:\mathrm{tan}\:\frac{\mathrm{4}\pi}{\mathrm{9}},\:\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{9}} \\ $$$$\mathrm{by}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{rule}\: \\ $$$$\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\:+\:\mathrm{tan}\:\frac{\mathrm{4}\pi}{\mathrm{9}}\:+\:\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{9}}\:=\:\mathrm{3}\sqrt{\mathrm{3}}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com