Question

If A + B + C = 180°. Prove that: sin2A – sin2B + sin2C = 4cosA · sinB · cosC

Answer 1

Given A + B + C = 180°

2A + 2B + 2C = 360° 

2A + 2B = 360° – 2C 

sin(2A + 2B) = sin(360° – 20) = – sin2C 

cos(2A + 2B) = cos(360° – 2C) = cos 2C 

L.H.S = sin2A – sin2B + sin2C 

= 2cos(A + B) · sin(A – B) + 2sinC. cosC 

= – 2cosC.sin(A – B) + 2 sinc.cosC 

= 2 cosC [sinC – sin (A – B)] 

= 2 cosC [sin(A + B) – sin(A – B)] 

= 2 cos C [2cosA . sinB] 

= 4 cosA sinB.cosC = R.H.S.