Question

In ∆ABC, A + B + C = π show that

sin²A + sin²B + sin²C = 2 sinA sinB cosC​

Answer 1

Step-by-step explanation:

L.H.S. = sin2A + sin2B – sin2C = 1212[1-(cos 2A + cos2B – cos2C)] = 1212[1-{{2cos(A + B). Cos(A – B) – 2cos2C+1}] = 1212[+2cosC . cos(A – B) + 2cos2c] = 1212[2cosC(cos(A – B) + cosC] = 1212[2cosC(cos(A – B) – cos(A + B))] = 1212[2cosC[-2sin A.sin(-B)]] = 2 sinA sinB sincRead more on Sarthaks.com - https://www.sarthaks.com/643768/prove-that-sin-2a-sin-2b-sin-2c-2sina-sin-b-cosc