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1-sin6x-cos6x/1-sin4x-cos4x
题目详情
1-sin6x-cos6x/1-sin4x-cos4x
▼优质解答
答案和解析
(1-sin6x-cos6x)/(1-sin4x-cos4x)
=[1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)]/(1-sin^4x-cos^4x)
=(1-sin^4x-cos^4x+sin^2xcos^2x)/(1-sin^4x-cos^4x)
=(1-1+sin^2xcos^2x+sin^2xcos^2x)/(1-1+2sin^2xcos^2x)
=3sin^2xcos^2x/2sin^2xcos^2x
=3/2
=[1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)]/(1-sin^4x-cos^4x)
=(1-sin^4x-cos^4x+sin^2xcos^2x)/(1-sin^4x-cos^4x)
=(1-1+sin^2xcos^2x+sin^2xcos^2x)/(1-1+2sin^2xcos^2x)
=3sin^2xcos^2x/2sin^2xcos^2x
=3/2
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