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Question Number 61157 by Askash last updated on 29/May/19

if tan 5θ + tan 4θ =1 find 3θ

$${if} \\ $$$${tan}\:\mathrm{5}\theta\:+\:{tan}\:\mathrm{4}\theta\:=\mathrm{1} \\ $$$${find}\:\mathrm{3}\theta \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 30/May/19

there are totally 9 values for tan 3θ. one of them is θ=(π/4), and tan 3θ=−1. with θ=(π/4) tan 3θ=tan (((3π)/4))=−1 tan 5θ+tan 4θ=tan (((5π)/4))+tan (π)=1

$${there}\:{are}\:{totally}\:\mathrm{9}\:{values}\:{for}\:\mathrm{tan}\:\mathrm{3}\theta. \\ $$$$ \\ $$$${one}\:{of}\:{them}\:{is}\:\theta=\frac{\pi}{\mathrm{4}},\:{and}\:\mathrm{tan}\:\mathrm{3}\theta=−\mathrm{1}. \\ $$$${with}\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\mathrm{3}\theta=\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)=−\mathrm{1} \\ $$$$\mathrm{tan}\:\mathrm{5}\theta+\mathrm{tan}\:\mathrm{4}\theta=\mathrm{tan}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\right)+\mathrm{tan}\:\left(\pi\right)=\mathrm{1} \\ $$

Commented by Askash last updated on 29/May/19

tan 3θ

$${tan}\:\mathrm{3}\theta \\ $$

Commented by mr W last updated on 30/May/19

Answered by tanmay last updated on 30/May/19

tan(α+β+γ+...)=((S_1 −S_3 +S_5 ...)/(1−S_2 +S_4 −S_6 +..)) tanθ=a tan5θ=((5a−5C_3 a^3 +5C_5 a^5 )/(1−5C_2 a^2 +5C_4 a^4 ))=((5a−10a^3 +a^5 )/(1−10a^2 +5a^4 )) tan4θ=((4a−4C_3 a^3 )/(1−4C_2 a^2 +4C_4 a^4 ))=((4a−4a^3 )/(1−6a^2 +a^4 )) tan5θ+ta4θ=1 ((5a−10a^2 +a^5 )/(1−10a^2 +5a^4 ))+((4a−4a^3 )/(1−6a^2 +a^4 ))=1 wait... anither approach... here tan5θ+tan4θ=1 so θ=small tan5θ+tan4θ≈5θ+4θ 9θ=1 θ=(1/9) tan3θ≈3θ =3×(1/9)=0.33 so tan3θ=0.33 (approax)

$$ \\ $$$${tan}\left(\alpha+\beta+\gamma+...\right)=\frac{{S}_{\mathrm{1}} −{S}_{\mathrm{3}} +{S}_{\mathrm{5}} ...}{\mathrm{1}−{S}_{\mathrm{2}} +{S}_{\mathrm{4}} −{S}_{\mathrm{6}} +..} \\ $$$${tan}\theta={a} \\ $$$${tan}\mathrm{5}\theta=\frac{\mathrm{5}{a}−\mathrm{5}{C}_{\mathrm{3}} {a}^{\mathrm{3}} +\mathrm{5}{C}_{\mathrm{5}} {a}^{\mathrm{5}} }{\mathrm{1}−\mathrm{5}{C}_{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{5}{C}_{\mathrm{4}} {a}^{\mathrm{4}} }=\frac{\mathrm{5}{a}−\mathrm{10}{a}^{\mathrm{3}} +{a}^{\mathrm{5}} }{\mathrm{1}−\mathrm{10}{a}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{4}} } \\ $$$${tan}\mathrm{4}\theta=\frac{\mathrm{4}{a}−\mathrm{4}{C}_{\mathrm{3}} {a}^{\mathrm{3}} }{\mathrm{1}−\mathrm{4}{C}_{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{4}{C}_{\mathrm{4}} {a}^{\mathrm{4}} }=\frac{\mathrm{4}{a}−\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{1}−\mathrm{6}{a}^{\mathrm{2}} +{a}^{\mathrm{4}} } \\ $$$${tan}\mathrm{5}\theta+{ta}\mathrm{4}\theta=\mathrm{1} \\ $$$$\frac{\mathrm{5}{a}−\mathrm{10}{a}^{\mathrm{2}} +{a}^{\mathrm{5}} }{\mathrm{1}−\mathrm{10}{a}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{4}} }+\frac{\mathrm{4}{a}−\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{1}−\mathrm{6}{a}^{\mathrm{2}} +{a}^{\mathrm{4}} }=\mathrm{1} \\ $$$${wait}... \\ $$$${anither}\:{approach}... \\ $$$${here}\:{tan}\mathrm{5}\theta+{tan}\mathrm{4}\theta=\mathrm{1} \\ $$$${so}\:\theta={small} \\ $$$${tan}\mathrm{5}\theta+{tan}\mathrm{4}\theta\approx\mathrm{5}\theta+\mathrm{4}\theta \\ $$$$\mathrm{9}\theta=\mathrm{1} \\ $$$$\theta=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${tan}\mathrm{3}\theta\approx\mathrm{3}\theta \\ $$$$=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{0}.\mathrm{33} \\ $$$${so}\:{tan}\mathrm{3}\theta=\mathrm{0}.\mathrm{33}\:\left({approax}\right) \\ $$$$ \\ $$

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