Question

在△ABC中,若sinB*sinC=COS^2(A/2),判断并证明△ABC的形状 感激!

Answer 1

cos^2(A/2)=(1+cosA)/2,因此

cosA=2sinB×sinC -1,又因A+B+C=180,故

cos(180-B-C)=2sinB×sinC -1,将cos(180-B-C)展开,得

cos(180-B-C)=cos180×cos(B+C)+sin180×sin(B+C)

=-cos(B+C)+0

=-cosB×cosC+sinB×sinC

带入原式得,

cosB×cosC+sinB×sinC=1,即

cos(B-C)=1,说明B=C

因此该三角形是等腰三角形.