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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))=?

ctg6(π9)−9⋅ctg4(π9)+11⋅ctg2(π9)=?

Commented by SEKRET last updated on 14/Feb/24

solution

solution

Commented by Frix last updated on 14/Feb/24

(1/3)

13

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)

Notashardasitlooks...x=cot2π9c=x3−9x2+11xx=t+3z=t3−16t−21t=x−3=cot2π9−3=2−4cos2π9cos2π9−1=2−4cc−1z=t3−16t−21=−21c3+c2−17c+3c3−3c2+3c−1c3=cos32π9=6cos2π9−18=6c−18z=−21(6c−1)8+c2−17c+36c−18−3c2+3c−1==−c2−5c4+38−3c2+15c4−98=13

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