Question

若tan(α+β)=2tanα,求证:3sinβ=sin(2α+β).

Answer 1

∵tan(α+β)=2tanα,
∴sin(α+β)/cos(α+β)=2sin(α)/cos(α)
∴sin(α+β)*cos(α)=2cos(α+β)*sin(α)
∴2sin(α+β)*cos(α)=4cos(α+β)*sin(α)
∴3sin(α+β)*cos(α)-3cos(α+β)*sin(α)=sin(α+β)*cos(α)+cos(α+β)*sin(α)
∴3sin[(α+β)-α]=sin(α+β)*cos(α)+cos(α+β)*sin(α)
∴3sin(β)=sin[(α+β)+α]
即:3sinβ=sin（2α+β)

Answer 2

tan(α+β)=sin(α+β)/cos(α+β)=(sinαcosβ+cosαsinβ)/(cosαcosβ-sinαsinβ)
=2tanα=2sinα/cosα
-2(sinα)^2sinβ+sin2αcosβ=sin2αcosβ/2+(cosα)^2sinβ
-(sinα)^2sinβ-sinβ=-sin2αcosβ/2
-(1-cos2α)sinβ-2sinβ=-sin2αcosβ
3sinβ=cos2αsinβ+sin2αcosβ
3sinβ=sin(β+2α)