(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 16:08:27
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=
化简(急)
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)
1-cos^4x-sin^4x
=1-(cos^4x+sin^4x)
=1-[(cos^2x+sin^2x)^2-2cos^2xsin^2x]
=1-(1-2cos^2xsin^2x)
=2cos^2xsin^2x
1-cos^6x-sin^6x
=1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)
=1-[(cos^2x+sin^2x)^2-3cos^2xsin^2x]
=1-[1-3cos^2xsin^2x]
=3cos^2xsin^2x
原式=2cos^2xsin^2x/3cos^2xsin^2x=2/3
用计算器算卅。。。
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=
=[(sin²x+cos²x)²-cos^4X-sin^4x)]/[1-(cos²x+sin²x)(cos²x-sin²xcos²x+sin²x)]
=2sin²xcos²x/[1-(1-sin²xcos²x)]
=2sin²xcos²x/sin²xcos²x
=2
化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
化简(1-cos^4x-sin^4x)/(1-cos^6x-sin^6x)
求值:(1-sin^6 x-cos^6 x)/(1-sin^4 x-cos^4 x)
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)
求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x
cos x/sin x+sin x/(1+cos x) 化简.
证明一个sin&cos的等式证明 1+sin^2(x)+sin^4(x)+sin^6(x)=[1-sin^8(x)]/[cos^2(x)]
求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x
求∫1/(sin^4x+cos^4x)dx,
∫[1/(sin^2(x)cos^4(x)]dx
求解∫1/(cos^4(x)sin^2(x))dx
求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2).
求证(cos^2x-sin^2x)(cos^4x+sin^4x)+1/4sin2xsin4x=cos2x
求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值
求1-cos^6x-sin^6x/1-cos^4x-sin^4x的值,