在△ABC中,若sinA=2sinBcosC,cos^2C-cos^2A=sin^2B,试判断△ABC的形状

问题描述:

在△ABC中,若sinA=2sinBcosC,cos^2C-cos^2A=sin^2B,试判断△ABC的形状

sinA=2sinBcosCsin(180-B-C)=2sinBcosCsin(B+C)=2sinBcosCsinBcosC+cosBsinC=2sinBcosCsinBcosC-cosBsinC=0sin(B-C)=0B=C代入cos²C-cos²A=sin²Bcos²C-sin²C=cos²(180-B-C)cos2C=cos&s...