(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)

来源：学生作业帮助网编辑：作业帮时间：2024/11/28 11:20:09

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(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)

(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^6θ-cos^6θ)/(1-sin^4θ-cos^4θ) 若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ 已知cosθ+cosθ^2=1,则sinθ^2+sinθ^6+sinθ^8= 已知cosθ+cosθ=1,则sinθ+sin∧6θ+sin∧8θ= 化简：1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ 设cosθ+cos^2θ=1,则sin^2θ+sin^6θ+sin^8θ的值为求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ 2sinθ-cosθ=1 (sinθ+cosθ+1)/(sinθ-cosθ+1)如题已知2sinθ-cosθ=1 求(sinθ+cosθ+1)/(sinθ-cosθ+1) cos θ/(1-tan θ)怎样化成cos θ/[(cosθ-sinθ)/cosθ] f(θ)=【sinθcosθ/(sinθ+cosθ+1)】+sinθcosθ化简求证：(1+sinθ-cosθ)/(1+sinθ+cosθ)=sinθ/（cosθ+1）求证sinθ-cosθ+1/sinθ+cosθ-1＝1＋sinθ/cosθ 求证:（1+cosθ+cosθ/2） /（sinθ+sinθ/2）=sinθ/1-cosθ 化简[sinθ(1+sinθ)+cosθ(1+cosθ)]*[sinθ(1-sinθ)+cosθ(1-cosθ)]-sin2θ 求证【sinθ(1+sinθ)+cosθ(1+cosθ)】×【sinθ(1-sinθ)+cosθ(1-cosθ)】=sin2θ 1-cosθ/(2sinθ/2) 化简1+cosθ-sinθ化简(1+cosθ-sinθ)/(1-cosθ-sinθ)+(1-cosθ-sinθ)/(1+cosθ-sinθ)(1+cosθ-sinθ)/(1-cosθ-sinθ)+(1-cosθ-sinθ)/(1+cosθ-sinθ) 证明(1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)/(1-2sinθcosθ)