Question

SIN 2A + SIN 2B + SIN 2C = 1 - 2 SIN A SIN B SIN C​

Answer 1

Answer:

L.H.S. = sin2A + sin2B – sin2C = 1 2

12[1-(cos 2A + cos2B – cos2C)] = 1 2 12[1-{{2cos(A + B). Cos(A – B) – 2cos2C+1}] = 1 2 12[+2cosC . cos(A – B) + 2cos2c] = 1 2 12[2cosC(cos(A – B) + cosC] = 1 2 12[2cosC(cos(A – B) – cos(A + B))] = 1 2 12[2cosC[-2sin A.sin(-B)]] = 2 sinA sinB /prove-that-sin-2a-sin-2b-sin-2c-2sina-sin-b-cosc

Answer 2

Answer:

I don't know sorry sorry