已知tan^2α=2tan^β+1 求证:sin^2β=asin^2α-1
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/09 08:51:01
已知tan^2α=2tan^β+1 求证:sin^2β=asin^2α-1
快
快
题目有误
已知tan^2α=2tan^2β+1 求证:sin^2 β=2sin^2α-1
tan^2α=2tan^2β+1
sin^2α/cos^2α=2sin^2β/cos^2β+1
sin^2α/cos^2α=(2sin^2β+cos^2β)/cos^2β
sin^2α/cos^2α=(sin^2β+1)/cos^2β
sin^2α/(1-sin^2α)=(sin^2β+1)/(1-sin^2β)
sin^2α-sin^2α*sin^2β=sin^2β-sin^2α*sin^2β-sin^2α+1
2sin^2α=sin^2β+1
sin^2β=2sin^2α-1
已知tan^2α=2tan^2β+1 求证:sin^2 β=2sin^2α-1
tan^2α=2tan^2β+1
sin^2α/cos^2α=2sin^2β/cos^2β+1
sin^2α/cos^2α=(2sin^2β+cos^2β)/cos^2β
sin^2α/cos^2α=(sin^2β+1)/cos^2β
sin^2α/(1-sin^2α)=(sin^2β+1)/(1-sin^2β)
sin^2α-sin^2α*sin^2β=sin^2β-sin^2α*sin^2β-sin^2α+1
2sin^2α=sin^2β+1
sin^2β=2sin^2α-1
已知tan²β=2tan²α+1,求证cos2β+sin²α=0
问几道高中三角函数题1、求证:tanα*sinα/(tanα-sinα)=(tanα+sinα)/tanα*sinα2、
已知:tan(α+β)=2tanα.求证:3sinβ=sin(2α+β).
已知tan(α+β)=2tanβ,求证:3sinα=sin(α+2β)
已知sin(2α+β)+2sinβ=0.求证tanα=3tan(α+β
已知2sinβ=sin(2α+β)求证tan(α+β)=3tanα
已知sinβ=2sin(2α+β),求证 tan(α+β)=-3tanα
求证 sin^2αtanα+cos^2α/tanα+2sinαcosα=tanα+1/tanα
求证:tan(α/2)=(sin α)/(1+cos α)
sinа=msin(2α+β),求证,tan(α+β)=(1+m)/(1-m)×tanα
求证:(1+sinα)/cosα=(1+tanα/2)/(1-tanα/2)
已知1/tan^2 α=2tan^2 β+1,那么sin^2β+2sin^2α=?