Home » Forum » Trigonometry » ProofProve that Cos(b-c)+Cos(c-a)+Cos(a-b)=-3/2 if Cosa+Cosb+Cosc=0 and Sina+Sinb+Sinc=06 years agoAnswers : (1)Dear vinay, squaring the equations Cosa+Cosb+Cosc=0 and Sina+Sinb+Sinc=0 we get cos2 a + cos2 b+ cos2 c + 2cosacosb+2cosbcosc+2cosacosc = 0and similarly sin2 a +sin2 b +sin2 c+ 2 sinasinb+2 sinasinb+2 sinasinb=0adding the above two equations we get 3 + 2(sinasinb+cosacosb+sinasinb+cosacosb+sinasinb+cosacosb)=0and by formula sinasinb + cosacosb = cos(a-b) and replacingwe get Cos(b-c)+Cos(c-a)+Cos(a-b)= -3/2 hence proved
