求救~sinβ=msin(2α+β）sinβ=msin(2α+β）,且m≠1,α≠kπ/2,α+β≠π/2+kπ（k∈Z）,求证tan(α+β)=（1+m）/(1-m)·tanα

求救~sinβ=msin(2α+β）sinβ=msin(2α+β）,且m≠1,α≠kπ/2,α+β≠π/2+kπ（k∈Z）,求证tan(α+β)=（1+m）/(1-m)·tanα
求救~sinβ=msin(2α+β）
sinβ=msin(2α+β）,且m≠1,α≠kπ/2,α+β≠π/2+kπ（k∈Z）,求证
tan(α+β)=（1+m）/(1-m)·tanα

求救~sinβ=msin(2α+β）sinβ=msin(2α+β）,且m≠1,α≠kπ/2,α+β≠π/2+kπ（k∈Z）,求证tan(α+β)=（1+m）/(1-m)·tanα
用a和b代替
sin[(a+b)-a]=msin[(a+b)+a]
sin(a+b)cosa-cos(a+b)sina=m[sin(a+b)cosa+cos(a+b)sina]
sin(a+b)cosa-cos(a+b)sina=msin(a+b)cosa+mcos(a+b)sina
(1-m)sin(a+b)cosa=(1+m)cos(a+b)sina
sin(a+b)cosa/cos(a+b)sina=(1+m)/(1-m)
sin(a+b)/cos(a+b)=(1+m)/(1-m)*(sina/cosa)
tan(a+b)
=(1+m)/(1-m)*tana