Question

prove that   
a sin A/2 sin B-C/2 + b sin B/2 sin C-A/2 + c sin C/2 sin A-B/2 = 0

Answer 1

in a triangle ,  A/2 = (180 - B+C)/2
sin A/2 = sin (90 - (B+C)/2) = cos (B+C)/2
sin B/2 = cos (C+A)/2
sin C/2 = cos (A+B)/2

Now substitute these in the given L H S

a cos (B+C)/2 sin (B-C)/2 + b cos(A+C)/2 sin (C-A)/2 + c cos (A+B)/2 sin (A-B)/2
= a [ sin B - sin C]/2 + b [ sin C - sin A]/2 + c [sin A - sin B]/2
we know a sin B = b sin A        a sin C = c sin A    b sin C = c sin B

= 1/2 [  b sin A - c sin A + b sin C - b sin A + c sin A - c sin B ]

= 0 as the terms cancel out