sin^2β+cos^4β+sin^2βcos^2β
来源：学生作业帮助网 编辑：六六作业网 时间：2024/05/17 04:00:52
sin^2β+cos^4β+sin^2βcos^2βsin^2β+cos^4β+sin^2βcos^2βsin^2β+cos^4β+sin^2βcos^2βb=β,sin^2(b)+cos^2(b)=
sin^2β+cos^4β+sin^2βcos^2β
sin^2β+cos^4β+sin^2βcos^2β
sin^2β+cos^4β+sin^2βcos^2β
b=β,sin^2(b)+cos^2(b)=1
sin^2(b)+cos^4(b)+sin^2(b)*cos^2(b)
=sin^2(b)+[cos^4(b)+sin^2(b)*cos^2(b)]
=sin^2(b)+cos^2(b)*[cos^2(b)+sin^2(b)]
=sin^2(b)+cos^2(b)*1
=sin^2(b)+cos^2(b)
=1
sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β)
已知sinΘ+cosΘ=2sinα,sinΘ*cosΘ=sin²β,求证：4cos²2α=cos²2β
已知sinθ+cosθ=2sinα,sinθ*cosθ=sin²β,求证：4cos²2α=cos²2β
化简sin^2α+sin^2β+2sinαsinβcos(α+β)
求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
证明：sin(2α+β)/sinα - 2cos(α+β)=sinβ/sinα
sin(α+β)-2cosαsinβ/2cosαcosβ-cos(α+β)
已知(sinβ-cosβ)/(sinβ+cosβ)=1/3,求sin^β-4sinβcosβ+2=
sin^2β+cos^4β+sin^2βcos^2β
(sinα-sinβ)^2+(cosα-cosβ)^2=4sin^2(α+β)/2
利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ
证明cosα(cosα-cosβ)+sinα(sinα-sinβ)=2sin^2(α-β/2)第二个 证明sin(α+β)cosα-1/2[sin(2α+β)-sinβ]=sinβ
化简sin(α+β)-2sinαsinβ/2sinαsinβ+cos(α+β)
求证：sin²α+sin²β+2sinαsinβcos(α+β)=sin²(α+β)
sin（α+β）-2sinαcosβ/2sinαsinβ+cos（α+β）化简