已知tg(a+b)=3tga,求證:2sin2b-sin2a=sin(2a+2b)

熱心網友

已知tg(a+b)=3tga,求證:2sin2b-sin2a=sin(2a+2b)因為tg(a+b)=3tga所以 sin(a+b)/cos(a+b) = 3sina/cosa即 sin(a+b)*cosa = 3cos(a+b)*sina積化和差得：sin(2a+b)+sinb = 3sin(2a+b)-3sinb所以 2sinb=sin(2a+b) ,兩邊同乘cosb得：sin2b = (1/2)*[sin(2a+2b) +sin(2a)]所以2sin2b-sin2a=sin(2a+2b)