A+b-c=3pi then find sina+sinb-sinc

The
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) 

Answer: the correct answer is 10step-by-step explanation: let the number be n, then the expression becomes,then“twelve more than the quotient of a number and two, decreased by four” iswhen n=4,we substitute 4 whetever we see n to get,we simplify to obtain,we further simplify to get,therefore the correct answer is c.