化简[sinθ(1+sinθ)+cosθ(1+cosθ)]*[sinθ(1-sinθ)+cosθ(1-cosθ)]-si
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:2024/11/17 06:10:34
化简[sinθ(1+sinθ)+cosθ(1+cosθ)]*[sinθ(1-sinθ)+cosθ(1-cosθ)]-sin2θ
[sinθ(1+sinθ)+cosθ(1+cosθ)]*[sinθ(1-sinθ)+cosθ(1-cosθ)]-sin2θ
=[sinθ+(sinθ)^2+cosθ+(cosθ)^2]*[sinθ-(sinθ)^2+cosθ-(cosθ)^2]-sin2θ
=(sinθ+cosθ+1)*(sinθ+cosθ-1)-sin2θ
=(sinθ+cosθ)^2-1-sin2θ
=(sinθ)^2+(cosθ)^2+2sinθcosθ-1-sin2θ
=1+sin2θ-1-sin2θ
=0
=[sinθ+(sinθ)^2+cosθ+(cosθ)^2]*[sinθ-(sinθ)^2+cosθ-(cosθ)^2]-sin2θ
=(sinθ+cosθ+1)*(sinθ+cosθ-1)-sin2θ
=(sinθ+cosθ)^2-1-sin2θ
=(sinθ)^2+(cosθ)^2+2sinθcosθ-1-sin2θ
=1+sin2θ-1-sin2θ
=0
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
f(θ)=【sinθcosθ/(sinθ+cosθ+1)】+sinθcosθ化简
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
化简【1+sinθ-cosθ/1+sinθ+cosθ】+cot(θ/2)
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
化简1+sinθ-cosθ/1+sinθ+cosθ详细过程.
化简y=1+cosΘ-sinΘ/1-cosΘ+sinΘ
化简2cosθ/√1-sin^2θ+√1-cos^2θ/sinθ