CosA-CosB=SinC,求証三角形ABC為直角三角形(10分)

cosA-cosB
=-2 sin[(A+B)/2] sin[(A-B)/2]
=-2 sin[(180-C)/2] sin[(A-B)/2]
=-2 cos(C/2)sin[(A-B)/2]
so
-2 cos(C/2)sin[(A-B)/2]=sinC
-2 cos(C/2)sin[(A-B)/2]=2sin(C/2)cos(C/2)
-sin[(A-B)/2]=sin(C/2)
sin[(B-A)/2]=sin(C/2)
(B-A)/2=C/2
B-A=C
B=A+C
that is
A+B+C=180
A+A+C+C=180
A+C=90
angle B is an right angle
三角形ABC為直角三角形