Question

Prove that sin4A + sin4B + sin4C = -4sin2A . sin2B · sin2C

Answer 1

L.H.S = sin4A + sin4B + sin4C [A+B+C = 180] 

= 2 sin(2A + 2B) · cos(2A – 2B) + 2 sinc.cos2C [2A + 2B + 2C = 360] 

= (-sin2C) · cos(2A – 2B) + (2 sin2C · cos2C) [2A + 2B = 360 – 20] 

= -2 sin2C [cos(2A – 2B) – cos2C] [sin(2A + 2B) = – sin2C] 

= – 2 sin2C [cos(2A – 2B) – cos(2A + 2B)] [cos(2A + 2B) = cos 2C] 

= -2 sin2C[-2 sin2A · sin(-2B)] 

= – 4 sin2A · sin2B · sin2C 

= R.H.S.