Menu Close

2cos-pi-9-1-3-2cos-2pi-9-1-3-2cos-4pi-9-1-3-

Tinku Tara 0 Comments
Pin




Question Number 90876 by john santu last updated on 26/Apr/20
((2cos ((π/9))))^(1/(3 )) −((2cos (((2π)/9))))^(1/(3 )) −((2cos (((4π)/9))))^(1/(3 )) = ?
$$\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{9}}\right)}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)}−\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)}\:=\:? \\ $$
Commented by john santu last updated on 26/Apr/20
Ramanujan theorem let α, β ,γ be a roots x^3 −ax^2 +bx−1=0 then ((α ))^(1/(3 )) + (β)^(1/(3 )) + (γ)^(1/(3 )) = ((a+6+3t))^(1/(3 )) with t^3 −3(a+b+3)t−(ab+6(a+b)+9)=0 ⇒ (α)^(1/(3 )) +(β)^(1/(3 )) +(γ)^(1/(3 )) = ((a+6+3t))^(1/(3 )) = ϕ ((αβ))^(1/(3 )) + ((βγ))^(1/(3 )) + ((αγ))^(1/(3 )) = ((b+6+3t))^(1/(3 )) = θ (t+3)^3 = (ϕθ)^3 consider x^3 −3x−1=0 x = t+(1/t) ⇒t^9 +1 = 0 t^9 = −1 = e^(iπ+2πki) = e^(iπ(1+2k)) ,k∈Z t = e^(iπ(((1+2k)/9))) , k = 0,1,2,...,8 t = e^((iπ)/9) , e^((i5π)/9) , e^((i7π)/9) t = e^(iθ) = cos θ+isin θ x^3 −3x−1 = 0 roots α = 2cos ((π/9)) β= −2cos (((2π)/9)) γ = −2cos (((4π)/9)) ∴ ((2cos ((π/9))))^(1/(3 )) −((2cos (((2π)/9))))^(1/(3 )) −((2cos (((4π)/9))))^(1/(3 )) = ((6−3(9)^(1/(3 )) ))^(1/(3 ))
$${Ramanujan}\:{theorem} \\ $$$${let}\:\alpha,\:\beta\:,\gamma\:{be}\:{a}\:{roots}\: \\ $$$${x}^{\mathrm{3}} −{ax}^{\mathrm{2}} +{bx}−\mathrm{1}=\mathrm{0} \\ $$$${then}\:\sqrt[{\mathrm{3}\:}]{\alpha\:}\:+\:\sqrt[{\mathrm{3}\:}]{\beta}\:+\:\sqrt[{\mathrm{3}\:}]{\gamma}\:=\:\sqrt[{\mathrm{3}\:}]{{a}+\mathrm{6}+\mathrm{3}{t}} \\ $$$${with}\:{t}^{\mathrm{3}} −\mathrm{3}\left({a}+{b}+\mathrm{3}\right){t}−\left({ab}+\mathrm{6}\left({a}+{b}\right)+\mathrm{9}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\sqrt[{\mathrm{3}\:}]{\alpha}+\sqrt[{\mathrm{3}\:}]{\beta}\:+\sqrt[{\mathrm{3}\:}]{\gamma}\:=\:\sqrt[{\mathrm{3}\:}]{{a}+\mathrm{6}+\mathrm{3}{t}}\:=\:\varphi \\ $$$$\sqrt[{\mathrm{3}\:}]{\alpha\beta}\:+\:\sqrt[{\mathrm{3}\:\:}]{\beta\gamma}\:+\:\sqrt[{\mathrm{3}\:\:}]{\alpha\gamma}\:=\:\sqrt[{\mathrm{3}\:}]{{b}+\mathrm{6}+\mathrm{3}{t}}\:=\:\theta \\ $$$$\left({t}+\mathrm{3}\right)^{\mathrm{3}} \:=\:\left(\varphi\theta\right)^{\mathrm{3}} \\ $$$${consider}\:{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}\:=\:{t}+\frac{\mathrm{1}}{{t}}\:\Rightarrow{t}^{\mathrm{9}} \:+\mathrm{1}\:=\:\mathrm{0} \\ $$$${t}^{\mathrm{9}} \:=\:−\mathrm{1}\:=\:{e}^{{i}\pi+\mathrm{2}\pi{ki}} \:=\:{e}^{{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)} \:,{k}\in\mathbb{Z} \\ $$$${t}\:=\:{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \:,\:{k}\:=\:\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{8} \\ $$$${t}\:=\:{e}^{\frac{{i}\pi}{\mathrm{9}}} \:,\:{e}^{\frac{{i}\mathrm{5}\pi}{\mathrm{9}}} ,\:{e}^{\frac{{i}\mathrm{7}\pi}{\mathrm{9}}} \\ $$$${t}\:=\:{e}^{{i}\theta} \:=\:\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{1}\:=\:\mathrm{0}\:{roots}\: \\ $$$$\alpha\:=\:\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{9}}\right) \\ $$$$\beta=\:−\mathrm{2cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\gamma\:=\:−\mathrm{2cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right) \\ $$$$\therefore\:\sqrt[{\mathrm{3}\:}]{\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{9}}\right)}−\sqrt[{\mathrm{3}\:}]{\mathrm{2cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)}\: \\ $$$$−\sqrt[{\mathrm{3}\:\:}]{\mathrm{2cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)}\:=\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{6}−\mathrm{3}\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}}}\: \\ $$$$ \\ $$
Commented by jagoll last updated on 26/Apr/20

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *