sin a的六次方+cos a的六次方=16分之13,求sin 2a?

首页/题库/432℃/2024-05-17 16:27:10

优质解答：

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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