Question

已知COS2α=-1/3,则SIN^6α+COS^6α=

Answer 1

SIN^6α+COS^6α=(sin^2a+cos^2a)(sin^4a-sin^2acos^2a+cos^4a)
=sin^4a-sin^2acos^2a+cos^4a
=(sin^2a+cos^2a)^2-3sin^2acos^2a
=1-3/4*(sin2a)^2 COS2α=-1/3 (COS2α)^2=1/9 (sin2a)^2=8/9
=1-3/4*8/9
=11/3

Answer 2

SIN^6α+COS^6α=(SIN^2α+COS^2α)(SIN^4α+COS^4α+SIN^2α*COS^2α)
=SIN^4α+COS^4α+SIN^2α*COS^2α=(SIN^2α+COS^2α)^2-SIN^2α*COS^2α=1-(sina*cosa)^2=1-(sin2a)^2/4,
sin2a^2=1-(-1/3)^2=8/9带进去就行