sin a的六次方+cos a的六次方=16分之13,求sin 2a?

sin^6A+cos^6A
=(sin^2A+cos^2A)(sin^4A-sin^2A*cos^2A+cos^4A)
=sin^4A-sin^2A*cos^2A+cos^4A
=sin^4A+2sin^2A*cos^2A+cos^4A-3sin^2A*cos^2A
=(sin^2A+cos^2A)^2-3sin^2A*cos^2A
=1-3sin^2A*cos^2A
=1-3(1/2*sin2A)^2
=1-3/4*(sin2A)^2
1-3/4*(sin2A)^2=13/16
3/4*(sin2A)^2=3/16
(sin2A)^2=1/4
sin2a=±1/2
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