证明如下:
由3sinb=sin(2a+b)知
3sinb=sin2acosb+sinbcos2a移项得
sinb(3-cos2a)=sin2acosb
sinb(2+2sina^2)=2sinacosacosb再次移项化简得
sinb=sina(cosacosb-sinbsina)=sinacos(a+b)
又tan(a+b)-2tana=(sin(a+b)cosa-2sinacos(a+b))/(coa(a+b)cosa)=
(sinb-sinacos(a+b))/(coa(a+b)cosa)由知tan(a+b)-2tana=0