作业帮化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/02 23:59:11
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化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
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
化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
化简(1-cos^4x-sin^4x)/(1-cos^6x-sin^6x)
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
cos x/sin x+sin x/(1+cos x) 化简.
化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x
求值：(1-sin^6 x-cos^6 x)/(1-sin^4 x-cos^4 x)
求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)
化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x
化简1+sin x/cos x·sin2x/2cos²（π/4-x/2）
化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
化简sin^4x-sin^2x+cos^2x
化简:sin^4 x-sin^2 x+cos^2 x
化简cos^4x+sin^4x
化简:cos^4(X)-sin^4(X)
化简sin^4x+cos^2x
cos x/( 1-sin x)化简