已知tg(a+b)=3tga,求证:2sin2b-sin2a=sin(2a+2b)
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已知tg(a+b)=3tga,求证:2sin2b-sin2a=sin(2a+2b)因为tg(a+b)=3tga所以 sin(a+b)/cos(a+b) = 3sina/cosa即 sin(a+b)*cosa = 3cos(a+b)*sina积化和差得：sin(2a+b)+sinb = 3sin(2a+b)-3sinb所以 2sinb=sin(2a+b) ,两边同乘cosb得：sin2b = (1/2)*[sin(2a+2b) +sin(2a)]所以2sin2b-sin2a=sin(2a+2b)