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