A+B-C=3pi then find sinA+sinB-sinC

The answer:

the full question is as follow:

if A+B-C=3pi, then find sinA+sinB-sinC

first, the main formula of sine and cosine are:

sinC = 2sin (C/2) cos (C/2)

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

therefore:

sinA+sinB-sinC = 2sin[ (A+B) / 2]cos[ (A-B) / 2] - 2sin (C/2) cos (C/2)

sin[ (A+B) / 2] = cos (C/2)

2sin (C/2) cos (C/2) = cos[ (A+B) / 2

and

A+B-C=3 pi implies A+B = 3 pi + C, so

cos[ (A+B) / 2] = cos [3 pi/2 + C/2]

and with the equivalence cos (3Pi/2 + X) = sinX

sinA+sinB-sinC = cos (C/2) + sin (C/2)