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

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

已知sinβ=msin(2α+β)且α+β≠π/2+kπ(k∈Z),α≠kπ/2(k∈Z),m≠1.求证tan(α+β)=(1+m)/(1-m)tanα
sinβ=msin(2α+β）
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=(m+1)cos(a+b)sina
又α+β≠π/2+kπ(k∈Z),α≠kπ/2(k∈Z),m≠1
所以
tan(α+β）=(1+m)/（1-m）tanα