f4_mathe_6

given SIN(A-B)+SIN(B-C)+SIN(C-A)=0
PROVE THAT TRIANGLE ABC IS A ISOSCELES TRIANGLE
sinX + sinY = 2sin[(X+Y)/2] cos[(X-Y)/2]
cosX + cosY = 2cos((X+Y)/2)cos((X-Y)/2)
sin(A-B)+sin(B-C)
=2[sin(A-C)/2][cos(A+C-2B)/2]
So sin(A-B)+sin(B-C)+sin(C-A)
=2[sin(A-C)/2][cos(A+C-2B)/2]+sin(C-A)
=-2[sin(C-A)/2][cos(A+C-2B)/2]+2[sin(C-A)/2][cos(C-A)/2]
=2[sin(C-A)/2][cosB/2+cos(C-A)/2]
=4[sin(C-A)/2][cos(B+C-A)/2cos(B-C+A)/2]
=4[sin(C-A)/2][cos(180-2A)/2cos(180-2C)/2]
=4[sin(C-A)/2]sinAsinC
if SIN(A-B)+SIN(B-C)+SIN(C-A)=0
then 4[sin(C-A)/2]sinAsinC=0
since A and C not equal to 0
sin(C-A)/2=0
C-A=0
C=A
So TRIANGLE ABC IS A ISOSCELES TRIANGLE where angle A = angle C