A+B+C =180. The question is to prove sin2A-sin2B+sin2C=-4sinAcosBcosC. I get sin2A-sin2B+sin2C=-4sinBcosAcosC, am I right?

It is given that A + B + C = 180. We have to prove that
sin 2A - sin 2B + sin 2C = -4*sin A*cos B*cos C

sin 2A -
sin 2B + sin 2C

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

=> 2*sin (A - B)*cos(A + B) + sin
2C

=> -2*sin (A - B)*cos C + 2*sin C*cos
C

=> -2*cos C[sin(A - B) - sin
C]

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

=> -4*cos C*sin[(A + A - 180)/2]*cos[(180 - B
- B)/2]

=> -4*cos C*sin(A - 90)*cos(90 -
B)

=> -4*cos C*cos A*sin
B

The result that you have got is
right.