在△ABC中 若sin^2Acos^2B-cos^2Asin^2B=sin^2c,判断△ABC的形状.但不知错哪了,
在△ABC中 若sin^2Acos^2B-cos^2Asin^2B=sin^2c,判断△ABC的形状.但不知错哪了,
sin^2Acos^2B-cos^2Asin^2B=sin^2c=sin^2(π-(A+B))=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+2sinAcosBsinBcosA+sin^2Bcos^2A
约分得,2cos^2Asin^2B+2sinAcosBsinBcosA=0,
sinAcosB+cosAsinB=0,sin(A+B)=0 ,根本不是三角形嘛.