已知3sinB=sin（2A+B）,求证：tan（A+B）=2tan A.

3sinB=sin（2A+B）
3sinB=sinAcos(A+B)+cosAsin(A+B)
3sinB=sinA(cosAcosB-sinAsinB)+cosA(sinAcosB+cosAsinB)
3sinB=sinAcosAcosB-sinAsinAsinB+cosAsinAcosB+cosAcosAsinB
3sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
3(sinAsinA+cosAcosA)sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
4sinAsinAsinB+2cosAcosAsinB=2cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=2cosAsinAcosB-cosAsinAcosB
2sinAcosAcosB-2sinAsinAsinB=cosAsinAcosB+cosAcosAsinB
2sinA(cosAcosB-sinAsinB)=cosA(sinAcosB+cosAsinB)
2sinAcos(A+B)=cosAsin(A+B)
tan（A+B）=2tan A

3sinb=sin(2a+b)可得sin(2a+b)-sinb=2sinb
由两角正弦差的公式：
sin(2a+b)-sinb=2cos[(2a+b+b)/2]sin[(2a+b-b)/2]=2cos(a+b)sina
因此：cos(a+b)sina=sinb，即：cos(a+b)=sinb/sina
则：sin(2a+b)=sin[a+(a+b)]=sin...

解: 因为 3sinB=sin(2A+B)
所以 3sin(A+B-A)=sin(A+A+B)
3[sin(A+B)cosA-cos(A+B)sinA]=sinAcos(A+B)+cosAsin(A+B)
2sin(A+B)cosA=4cos(A+B)sinA
即:tan(A+B)=2tanA