Question

Prove that sin²A + sin²B – sin²C = 2sinA · sin B. cosC

Answer 1

L.H.S. = sin2A + sin2B – sin2C

= \(\frac{1}{2}\)[1-(cos 2A + cos2B – cos2C)] 

= \(\frac{1}{2}\)[1-{{2cos(A + B). Cos(A – B) – 2cos2C+1}] 

= \(\frac{1}{2}\)[+2cosC . cos(A – B) + 2cos2c] 

= \(\frac{1}{2}\)[2cosC(cos(A – B) + cosC] 

= \(\frac{1}{2}\)[2cosC(cos(A – B) – cos(A + B))] 

= \(\frac{1}{2}\)[2cosC[-2sin A.sin(-B)]] 

= 2 sinA sinB sinc