Question

If A + B + C =180º, prove that cos2A + cos2B + cos2C = -1 - 4cos A cosB cosC​

Answer 1

Answer:

Step-by-step explanation:

A+B+C = 180°

=> A+B = 180 - C

=>Cos(A+B) = Cos(180 - C) = - CosC

Cos2A + Cos2B + Cos2C

//We know that CosA + CosB = 2Cos(A+B/2)Cos(A-B/2)

= 2Cos(2A+2B/2)Cos(2A-2B/2) + Cos2C

= 2Cos(A+B)Cos(A-B) + 2Cos²C - 1

= 2(-CosC)Cos(A-B) + 2Cos²C - 1

= -1 - 2CosC[Cos(A-B) - CosC]

= -1 - 2CosC[Cos(A-B) + Cos(A+B)]

//We know that CosA + CosB = 2Cos(A+B/2)Cos(A-B/2)

= -1 -2CosC[2Cos(A-B+A+B/2)Cos(A-B-A-B/2)]

= -1 -2CosC[2Cos(2A/2)Cos(-2B/2)]

= -1 -4CosCCosACosB

= R.H.S

Hence Proved.                                       =