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Question Number 204344 by SEKRET last updated on 13/Feb/24
ctg^6 ((π/9))−9∙ctg^4 ((π/9))+11∙ctg^2 ((π/9))=?
$$ \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{ctg}}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{9}\centerdot\boldsymbol{\mathrm{ctg}}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{11}\centerdot\boldsymbol{\mathrm{ctg}}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)=?\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by SEKRET last updated on 14/Feb/24
solution
$$\:\:\boldsymbol{\mathrm{solution}}\: \\ $$
Commented by Frix last updated on 14/Feb/24
(1/3)
$$\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by Frix last updated on 15/Feb/24
Not as hard as it looks... x=cot^2 (π/9) c=x^3 −9x^2 +11x x=t+3 z=t^3 −16t−21 t=x−3=cot^2 (π/9) −3=((2−4cos ((2π)/9))/(cos ((2π)/9) −1))=((2−4c)/(c−1)) z=t^3 −16t−21=−((21c^3 +c^2 −17c+3)/(c^3 −3c^2 +3c−1)) c^3 =cos^3 ((2π)/9) =((6cos ((2π)/9) −1)/8)=((6c−1)/8) z=−((((21(6c−1))/8)+c^2 −17c+3)/(((6c−1)/8)−3c^2 +3c−1))= =−((c^2 −((5c)/4)+(3/8))/(−3c^2 +((15c)/4)−(9/8)))=(1/3)
$$\mathrm{Not}\:\mathrm{as}\:\mathrm{hard}\:\mathrm{as}\:\mathrm{it}\:\mathrm{looks}… \\ $$$${x}=\mathrm{cot}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}} \\ $$$${c}={x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{11}{x} \\ $$$${x}={t}+\mathrm{3} \\ $$$${z}={t}^{\mathrm{3}} −\mathrm{16}{t}−\mathrm{21} \\ $$$${t}={x}−\mathrm{3}=\mathrm{cot}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}\:−\mathrm{3}=\frac{\mathrm{2}−\mathrm{4cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}{\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\:−\mathrm{1}}=\frac{\mathrm{2}−\mathrm{4}{c}}{{c}−\mathrm{1}} \\ $$$${z}={t}^{\mathrm{3}} −\mathrm{16}{t}−\mathrm{21}=−\frac{\mathrm{21}{c}^{\mathrm{3}} +{c}^{\mathrm{2}} −\mathrm{17}{c}+\mathrm{3}}{{c}^{\mathrm{3}} −\mathrm{3}{c}^{\mathrm{2}} +\mathrm{3}{c}−\mathrm{1}} \\ $$$${c}^{\mathrm{3}} =\mathrm{cos}^{\mathrm{3}} \:\frac{\mathrm{2}\pi}{\mathrm{9}}\:=\frac{\mathrm{6cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\:−\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{6}{c}−\mathrm{1}}{\mathrm{8}} \\ $$$${z}=−\frac{\frac{\mathrm{21}\left(\mathrm{6}{c}−\mathrm{1}\right)}{\mathrm{8}}+{c}^{\mathrm{2}} −\mathrm{17}{c}+\mathrm{3}}{\frac{\mathrm{6}{c}−\mathrm{1}}{\mathrm{8}}−\mathrm{3}{c}^{\mathrm{2}} +\mathrm{3}{c}−\mathrm{1}}= \\ $$$$=−\frac{{c}^{\mathrm{2}} −\frac{\mathrm{5}{c}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{8}}}{−\mathrm{3}{c}^{\mathrm{2}} +\frac{\mathrm{15}{c}}{\mathrm{4}}−\frac{\mathrm{9}}{\mathrm{8}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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