(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/30 00:30:26
(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
1-sin^6θ-cos^6θ
=1-[(sin^2θ)^3+(cos^2θ)^3]
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ+2sin^2θcos^2θ+cos^4θ-3sin^2θcos^2θ)
=1-[(sin^2θ+cos^2θ)^2-3sin^2θcos^2θ]
=3sin^2θcos^2θ
1-sin^4θ-cos^4θ
=1-(sin^4θ+cos^4θ)
=1-[(sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ)]
=1-[(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ]
=2sin^2θcos^2θ
所以:(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
=(3sin^2θcos^2θ)/(2sin^2θcos^2θ)
=3/2.
=1-[(sin^2θ)^3+(cos^2θ)^3]
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ+2sin^2θcos^2θ+cos^4θ-3sin^2θcos^2θ)
=1-[(sin^2θ+cos^2θ)^2-3sin^2θcos^2θ]
=3sin^2θcos^2θ
1-sin^4θ-cos^4θ
=1-(sin^4θ+cos^4θ)
=1-[(sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ)]
=1-[(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ]
=2sin^2θcos^2θ
所以:(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
=(3sin^2θcos^2θ)/(2sin^2θcos^2θ)
=3/2.
(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
已知sin^4θ+cos^4θ=1,求sinθ+cosθ的值
已知sinaθ-cosθ=-1/5,求sinθcosθ,sinθ^4+cosθ^4
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
高一三角函数 快!已知sin^3θ+cos^3θ=1,求sinθ+cosθ及sin^4θ+cos^4θ的值
数学4sinΘcosΘ-5sinΘ-5cosΘ-1=0那么sin三次方Θ+cos三次方Θ=
f(θ)=【sinθcosθ/(sinθ+cosθ+1)】+sinθcosθ化简
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ