sina+cosa=1/3 求tan^3+cot^3

sina+cosa=1/3
两边平方
sin²a+cos²a+2sinacosa=1/9
1+2sinacosa=1/9
sinacosa=-4/9

tan³a+cot³a
=sin³a/cos³a+cos³a/sin³a
=(sin^6a+cos^6a)/sin³a*cos³a

分子=(sin²a+cos²a)³-3sin^4acos²a-3sin²acos^4a
=1-3sin²cos²(sin²a+cos²)
=1-3*(sinacosa)²*1
=11/27
分母=(sinacosa)³=-64/729

所以原式=-297/64

sina+cosa=1/3
1+2sinacosa=1/9
sinacosa=-4/9
tan^3a+cot^3a=sin^3a/cos^3a+cos^3a/sin^3a
=[sin^6a+cos^6a]/[sin^3acos^3a]
=[(sin^2a+cos^2a)(sin^4a-sin^2acos^2a+cos^4a)]/[(sinacosa)^3]
=[(sin^2a+cos^2a)^2-2sinacosa-sin^2acos^2a]/(-4^3/9^3)
=[1+2*4/9-16/81]/(-64/729)
=-59049/8768

答案是-297/64

(sina)^2+(cosa)^2=1,
(sina)^2+(cosa)^2+2sinacosa=1/9,
sinacosa=-4/9,
tana+cota=sina/cosa+cosa/sina=((sina)^2+(cosa)^2))/sinacosa
=1/(-4/9)=-9/4,
tan^2a+cot^2