已知3sinB=sin(2A+B),求证tan(A+B)=2tanA

1个回答

  • 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)]=sinacos(a+b)+cosasin(a+b)……(*)

    将sin(2a+b)=3sinb,cos(a+b)sina=sinb代入等式(*):

    3sinb=sinb+cosasin(a+b),因此sin(a+b)=2sinb/cosa

    则:tan(a+b)=sin(a+b)/cos(a+b)=(2sinb/cosa)/(sinb/sina)

    =2(sinb/cosa)*(sina/sinb)=2sina/cosa=2tana