题目

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.

提问时间：2020-10-03

答案

sin^2Acos^2B-cos^2Asin^2B=（sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=sin(A+B)sin(A-B)(sinC)^2=[sin(π-A-B)]^2=[sin(A+B)]^2sin(A+B)[sin(A+B)-sin(A-B)]=0sin(A+B)cosAsinB=0sin(A+B)≠0 sinB≠0,因此只有cosA=0...