2(sin^6x+cos^6)-3(sin^4x+cos^4x)

polo52013142022-10-04 11:39:541条回答

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水天之处共回答了17个问题 | 采纳率100%: 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; 1年前