1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程

来源：学生作业帮编辑：搜搜做题作业网作业帮分类：数学作业时间：2024/08/18 09:05:15

1-sin^2θ-cos^2θ/1-sinθ-cosθ
=1-1/1-sinθ-cosθ
=0
1-sin^6θ-cos^6θ/1-sin^4θ-cos^4θ
=1-(sin^6θ+cos^6θ)/1-sin^4θ-cos^4θ
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)/1-sin^4θ-cos^4θ
=1-1×(sin^4θ-sin^2θcos^2θ+cos^4θ)/1-sin^4θ-cos^4θ
=1-sin^4θ+sin^2θcos^2θ-cos^4θ]/1-sin^4θ-cos^4θ
=1 - (sin^2θcos^2θ/1-sin^4θ-cos^4θ)
=1 - (sin^2θcos^2θ/sin^2θ+cos^2θ-sin^4θ-cos^4θ)
=1 - (sin^2θcos^2θ/sin^2θ(1-sin^2θ)+cos^2θ(1-cos^2θ))
=1 - (sin^2θcos^2θ/sin^2θcos^2θ+cos^2θsin^2θ)
=1 - (sin^2θcos^2θ/2sin^2θcos^2θ)
=1-1/2
=1/2

1-sin^2θ-cos^2θ/1-sinθ-cosθ 请给一个具体一点的过程化简：1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ 求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ 求证:（1+cosθ+cosθ/2） /（sinθ+sinθ/2）=sinθ/1-cosθ sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-( 求证（1+sinθ+cosθ）/（1+sinθ-cosθ）+（1-cosθ+sinθ）/（1+cosθ+sinθ）=2/ 求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以（1+2sinθcos 为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1) θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小化简1+sinθ-cosθ/1+sinθ+cosθ详细过程. 化简【1+sinθ-cosθ/1+sinθ+cosθ】+cot（θ/2） sinΘ-2cosΘ=0 sin平方Θ+(2cosΘ)平方=1 求sinΘ和cosΘ的值