1-(sin^6x+cos^6x)=3sin^2xcos^2x 如何证明?
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1-(sin^6x+cos^6x)=3sin^2xcos^2x 如何证明?
1-(sin^6x+cos^6x)=3sin^2xcos^2x
只要证明
sin^6x+cos^6x=1-3sin^2xcos^2x
sin^6x +cos^6x
=(sin^2x +cos^2x)(sin^4x -sin^2xcos^2x +cos^4x)
=(1)(sin^4x +2cos^2xsin^2x +cos^4x -2sin^2xcos^2x -sin^2xcos^2x)
=(sin^2x +cos^2x)^2 -3sin^2xcos^2x
=(1)^2 -3sin^2xcos^2x
=1-3sin^2xcos^2x
只要证明
sin^6x+cos^6x=1-3sin^2xcos^2x
sin^6x +cos^6x
=(sin^2x +cos^2x)(sin^4x -sin^2xcos^2x +cos^4x)
=(1)(sin^4x +2cos^2xsin^2x +cos^4x -2sin^2xcos^2x -sin^2xcos^2x)
=(sin^2x +cos^2x)^2 -3sin^2xcos^2x
=(1)^2 -3sin^2xcos^2x
=1-3sin^2xcos^2x
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