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2(sin^6x+cos^6)-3(sin^4x+cos^4x)
人气:337 ℃ 时间:2020-06-14 10:15:43
解答
sin^6x+cos^6x=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)
=sin^4x-sin^2xcos^2x+cos^4x
=(sin^2x+cos^2x)^2-3sin^2xcos^2x
=1-3sin^2xcos^2x
sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1--2sin^2xcos^2x
2(sin^6x+cos^6)-3(sin^4x+cos^4x)=2(1-3sin^2xcos^2x)-3(1--2sin^2xcos^2x)
=-1
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