1.

Prove that sin2A + sin2B + sin2C = 4sinA · sinB · sin C

Answer»

L.H.S. = sin2A + sin2B + sin2C 

= 2sin(A + B) · cos(A – B) + 2sinCcosC 

= 2sinC.cos(A – B) + 2 sinC cosC 

= 2sinC [cos(A – B) – cos(A + B)] 

= 2 sinC[-2 sinA · sin(-B)] 

= 4sinA . sinB . sinC 

∵ sin(-B) = R.H.S. = SinB.



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