sin(2a+b)+2sinb=0,求證:tan a =3 tan(a+b) ``

sin(2a+b)+2sinb=0,求證:tan a =3 tan(a+b) ``

sin(2a+b)=sin(a+b)*cosa+sina*cos(a+b);sinb=sin(a+b-a)=sin(a+b)*cosa-cos(a+b)*sina;代入sin(2a+b)+2sinb=0:3*sin(a+b)*cosa-cos(a+b)sina=0;………………(*)只要cosa與cos(a+b)不為0,由(*)式可得結論….