If A + B + C = π, prove that

sin 2A + sin 2B – sin 2C = 4cos A cos B sin C

= sin 2A + sin 2B – sin 2C

= 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C
using formula,
sin (A + B) = sin A cos B + cos A sin B
= sin 2A + sin 2B - sin 2C

Using formula

sin2A = 2sinAcosA
= 2sinAcosA + 2sinBcosB - 2sinCcosC

since A + B + C = π

And sin(π – A) = sinA
= 2sin(B + C)cos A + 2sin(A + C)cosB - 2sin(A + B)cosC
= 2 ( sin B cos C + cos B sin C ) cos A + 2(sinAcosC + cosAsinC)cosB - 2(sinAcosB + cosAsinB )cosC
= 2cosAsinBcosC + 2cosAcosBsinC + 2sinAcosBcosC + 2cosAcosBsinC– 2sinAcosBcosC – 2cosAsinBcosC
= 2cosAcosBsinC + 2cosAcosBsinC
= 4cosAcosBsinC

= R.H.S