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化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
人气:301 ℃ 时间:2020-05-05 14:20:09
解答
原式=[1-(sin^2x+con^2x)^2-2sin^2x*con^2x]/[1-(sin^2x+con^2x)*(sin^4x+con^4x-sin^2x*con^2x)]=(2sin^2x*con^2x)/{1-[(sin^2x+con^2x)^2-2sin^2x*con^2x-sin^2x*con^2x]}=(2sin^2x*con^2x)/(3sin^2x*con^2x)=2/3...
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