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Question-176668

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Question Number 176668 by mnjuly1970 last updated on 24/Sep/22
Answered by a.lgnaoui last updated on 25/Sep/22
((2π)/9)=(((3π)/9)−(π/9)); ((4π)/9) =(((3π)/9)+(π/9));((8π)/9)=(π−(π/9)) cos ((3π)/9)=cos (π/3)=(1/2)=cos (((2π)/9))cos( (π/9))−sin (((2π)/9))cos ((π/9))=(2cos^2 ((π/9))−1)cos (π/9)−2sin^2 ((π/9))cos( (π/9)) =2cos^3 ((π/9))−cos ((π/9))−2(1−cos^2 ((π/9)))cos (π/9) (1/2)=4cos^3 ((π/9))−3cos ((π/9)) ⇒cos (π/9)={0,939} (0< (π/9)<(π/2)) 3(√((2π)/9)) +^3 (√((4π)/9)) +^3 (√(((8π)/9) )) =^3 (√(2cos^2 (π/9)−1)) +^3 (√( cos ((π/3)+(π/9)))) +^3 (√(cos (π−(π/9)))) =^3 (√(2cos^2 (π/9)−1)) +^3 (√((cos (π/3)cos (π/9))−(sin (π/3)sin (π/9)))) −^( 3) (√(cos (π/9))) cos^2 (π/9)=0,881 sin (π/9)=(√(1−0,881)) =0,3349 ^3 (√(0,763)) +^3 (√((1/2) ×0,939−(((√3)/2))×0,3349)) −^3 (√(0,939)) Resultat: ^( ^( 3) ) (√((2π)/9)) +^( 3) (√((4π)/9)) +^( 3) (√((8π)/9)) =0,913+ 0,564 −0,979=0,498
$$\frac{\mathrm{2}\pi}{\mathrm{9}}=\left(\frac{\mathrm{3}\pi}{\mathrm{9}}−\frac{\pi}{\mathrm{9}}\right);\:\:\frac{\mathrm{4}\pi}{\mathrm{9}}\:=\left(\frac{\mathrm{3}\pi}{\mathrm{9}}+\frac{\pi}{\mathrm{9}}\right);\frac{\mathrm{8}\pi}{\mathrm{9}}=\left(\pi−\frac{\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{cos}\left(\:\frac{\pi}{\mathrm{9}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)=\left(\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{1}\right)\mathrm{cos}\:\frac{\pi}{\mathrm{9}}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)\mathrm{cos}\left(\:\frac{\pi}{\mathrm{9}}\right) \\ $$$$=\mathrm{2cos}\:^{\mathrm{3}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)\right)\mathrm{cos}\:\frac{\pi}{\mathrm{9}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4cos}\:^{\mathrm{3}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{3cos}\:\left(\frac{\pi}{\mathrm{9}}\right)\:\:\:\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{9}}=\left\{\mathrm{0},\mathrm{939}\right\}\:\:\:\:\left(\mathrm{0}<\:\frac{\pi}{\mathrm{9}}<\frac{\pi}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\mathrm{3}\sqrt{\frac{\mathrm{2}\pi}{\mathrm{9}}}\:+^{\mathrm{3}} \sqrt{\frac{\mathrm{4}\pi}{\mathrm{9}}}\:+^{\mathrm{3}} \sqrt{\frac{\mathrm{8}\pi}{\mathrm{9}}\:}\:\:\:\:=^{\mathrm{3}} \sqrt{\mathrm{2cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{9}}−\mathrm{1}}\:\:+^{\mathrm{3}} \:\:\sqrt{\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{9}}\right)}\:\:+^{\mathrm{3}} \sqrt{\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{9}}\right)} \\ $$$$=^{\mathrm{3}} \sqrt{\mathrm{2cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{9}}−\mathrm{1}}\:\:\:+^{\mathrm{3}} \sqrt{\left(\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{9}}\right)−\left(\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{9}}\right)}\:−^{\:\:\:\:\mathrm{3}} \:\sqrt{\mathrm{cos}\:\frac{\pi}{\mathrm{9}}} \\ $$$$\:\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{9}}=\mathrm{0},\mathrm{881}\:\:\:\:\:\mathrm{sin}\:\frac{\pi}{\mathrm{9}}=\sqrt{\mathrm{1}−\mathrm{0},\mathrm{881}}\:=\mathrm{0},\mathrm{3349} \\ $$$$\:^{\mathrm{3}} \sqrt{\mathrm{0},\mathrm{763}}\:\:+^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{2}}\:×\mathrm{0},\mathrm{939}−\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)×\mathrm{0},\mathrm{3349}}\:\:\:−\:^{\mathrm{3}} \sqrt{\mathrm{0},\mathrm{939}}\:\: \\ $$$${Resultat}: \\ $$$$\:\:^{\:\:^{\:\:\mathrm{3}} } \:\sqrt{\frac{\mathrm{2}\pi}{\mathrm{9}}}\:\:+\:^{\:\mathrm{3}} \sqrt{\frac{\mathrm{4}\pi}{\mathrm{9}}}\:+^{\:\mathrm{3}} \sqrt{\frac{\mathrm{8}\pi}{\mathrm{9}}}\:\:\:=\mathrm{0},\mathrm{913}+\:\mathrm{0},\mathrm{564}\:−\mathrm{0},\mathrm{979}=\mathrm{0},\mathrm{498} \\ $$$$ \\ $$
Answered by Peace last updated on 25/Sep/22
cos(((3.2π)/9))=cos(((3.4π)/9))=cos(((3.8π)/9))=cos(((2π)/3))=−(1/2) cos(3x)=−(1/2)⇔4cos^3 (x)−3cos(x)+(1/2)=0 ((2π)/9);((4π)/9);((8π)/9) Principal Solution let a=(cos(((2π)/9)))^(1/3) ,b=(cos(((4π)/9)))^(1/3) ,c=(cos(((8π)/9)))^(1/3) a^3 ,b^3 ,c^3 Roots of 4X^3 −3X+(1/2)=0 ⇒a^3 +b^3 +c^3 =0;Σa^3 b^3 =−(3/4);(abc)^3 =−(1/8) Using x^3 +y^3 +z^3 =(x+y+z)^3 −3(x+y+z)(xy+yz+zx)+3xyz ⇒0=(a+b+c)^3 −3(a+b+c)(ab+bc+ca)+3abc Σa^3 b^3 =(Σab)^3 −3(Σab)(abc(a+b+c))+3a^2 b^2 c^2 a+b+c=r ab+bc+ca=s⇒ { ((r^3 −3rs−(3/2)=0)),((−(3/4)=s^3 +(3/2)rs+(3/4))) :} ⇔ { ((s=((2r^3 −3)/(6r)))),((s^3 +(3/2)rs+(3/2)=0....(E))) :} r^3 =t..(E)⇔(2t−3)^3 +216t.(3/(12))(2t−3)+3.108t=0 8t^3 −36t^2 −27+54t+108t^2 −162t+324t=0 8t^3 +72t^2 +216t−27=0 Cardon ⇒t=(3/2)((9)^(1/3) −2) r=a+b+c=((cos(((2π)/9))))^(1/3) +((cos(((4π)/9))))^(1/3) +((cos(((8π)/9))))^(1/3) =(t)^(1/3) =(((3^(5/3) −6)/2))^(1/3) ≈0.49
$${cos}\left(\frac{\mathrm{3}.\mathrm{2}\pi}{\mathrm{9}}\right)={cos}\left(\frac{\mathrm{3}.\mathrm{4}\pi}{\mathrm{9}}\right)={cos}\left(\frac{\mathrm{3}.\mathrm{8}\pi}{\mathrm{9}}\right)={cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\mathrm{3}{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow\mathrm{4}{cos}^{\mathrm{3}} \left({x}\right)−\mathrm{3}{cos}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{\mathrm{2}\pi}{\mathrm{9}};\frac{\mathrm{4}\pi}{\mathrm{9}};\frac{\mathrm{8}\pi}{\mathrm{9}}\:\:{Principal}\:{Solution} \\ $$$${let}\:{a}=\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,{b}=\left({cos}\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,{c}=\left({cos}\left(\frac{\mathrm{8}\pi}{\mathrm{9}}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}^{\mathrm{3}} ,{b}^{\mathrm{3}} ,{c}^{\mathrm{3}} \:{Roots}\:{of}\:\:\mathrm{4}{X}^{\mathrm{3}} −\mathrm{3}{X}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{0};\Sigma{a}^{\mathrm{3}} {b}^{\mathrm{3}} =−\frac{\mathrm{3}}{\mathrm{4}};\left({abc}\right)^{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${Using} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\left({x}+{y}+{z}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right)+\mathrm{3}{xyz} \\ $$$$\Rightarrow\mathrm{0}=\left({a}+{b}+{c}\right)^{\mathrm{3}} −\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)+\mathrm{3}{abc} \\ $$$$\Sigma{a}^{\mathrm{3}} {b}^{\mathrm{3}} =\left(\Sigma{ab}\right)^{\mathrm{3}} −\mathrm{3}\left(\Sigma{ab}\right)\left({abc}\left({a}+{b}+{c}\right)\right)+\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$${a}+{b}+{c}={r} \\ $$$${ab}+{bc}+{ca}={s}\Rightarrow \\ $$$$\begin{cases}{{r}^{\mathrm{3}} −\mathrm{3}{rs}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0}}\\{−\frac{\mathrm{3}}{\mathrm{4}}={s}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{rs}+\frac{\mathrm{3}}{\mathrm{4}}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{{s}=\frac{\mathrm{2}{r}^{\mathrm{3}} −\mathrm{3}}{\mathrm{6}{r}}}\\{{s}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{rs}+\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0}….\left({E}\right)}\end{cases} \\ $$$${r}^{\mathrm{3}} ={t}..\left({E}\right)\Leftrightarrow\left(\mathrm{2}{t}−\mathrm{3}\right)^{\mathrm{3}} +\mathrm{216}{t}.\frac{\mathrm{3}}{\mathrm{12}}\left(\mathrm{2}{t}−\mathrm{3}\right)+\mathrm{3}.\mathrm{108}{t}=\mathrm{0} \\ $$$$\mathrm{8}{t}^{\mathrm{3}} −\mathrm{36}{t}^{\mathrm{2}} −\mathrm{27}+\mathrm{54}{t}+\mathrm{108}{t}^{\mathrm{2}} −\mathrm{162}{t}+\mathrm{324}{t}=\mathrm{0} \\ $$$$\mathrm{8}{t}^{\mathrm{3}} +\mathrm{72}{t}^{\mathrm{2}} +\mathrm{216}{t}−\mathrm{27}=\mathrm{0} \\ $$$${Cardon}\:\Rightarrow{t}=\frac{\mathrm{3}}{\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{2}\right) \\ $$$${r}={a}+{b}+{c}=\sqrt[{\mathrm{3}}]{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)}+\sqrt[{\mathrm{3}}]{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)}+\sqrt[{\mathrm{3}}]{{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{9}}\right)}=\sqrt[{\mathrm{3}}]{{t}} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}^{\frac{\mathrm{5}}{\mathrm{3}}} −\mathrm{6}}{\mathrm{2}}}\approx\mathrm{0}.\mathrm{49} \\ $$$$ \\ $$
Answered by Peace last updated on 25/Sep/22
hello sir happy to comback in this forum i lost my compt mindispower
$${hello}\:{sir}\:{happy}\:{to}\:{comback}\:{in}\:{this}\:{forum} \\ $$$${i}\:{lost}\:{my}\:{compt}\:{mindispower} \\ $$

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