Question

在三角形ABC中,若sinAsinB=[(cosC/2)]^2,判断三角形形状

Answer 1

C=180-A-B

C/2=90-(A+B)/2

cosC/2

=sin[(A+B)/2]

所以[(cosC/2)]^2={sin[(A+B)/2]}^2

=[1-cos(A+B)]/2

=sinAsinB

=[cos(A-B)-cos(A+B)]/2

所以cos(A-B)=1

因为0