Question

10. If A+B+C = π, prove that sin² A + sin²B + sin²C= 2 + 2 cos A cos B cos C
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Answer 1

Answer:

Step-by-step explanation:

Given, A+B+C = π

L.H.S

sin² A + sin²B + sin²C

= [1 - Cos2A]/2 + [1 - Cos2B]/2 + [1 - Cos2C]/2 (∵ Cos2θ =  1 - Sin²θ)

= 1/2[3 - (Cos2A+Cos2B+Cos2C)] (∵CosA + CosB=2Cos(A+B/2)Cos (A-B/2)

= 1/2 ( 3 - [2Cos(A+B)Cos(A-B)+ Cos2C] )

= 1/2(3 - [2 * (-CosC).Cos(A-B)+ Cos2C]) (∵Cos (A+B) = Cos ( π - C) = - CosC)

= 1/2 (3 - [ -2CosC.Cos(A-B) + 2Cos²C -1]) (∵ Cos2C = 2Cos²C - 1)

= 1/2(3 + 1 + 2CosCCos(A-B) - 2Cos²C)

= 1/2(4 + 2CosC [Cos(A-B) - CosC]) (∵CosC=Cos(π - (A+B))= - Cos(A+B))

= 1/2(4 + 2CosC[Cos(A-B) + Cos(A+B)])

= 1/2(4 + 2CosC(2CosACosB))  (∵CosA + CosB=2Cos(A+B/2)Cos (A-B/2)

= 1/2(4 + 4 CosACosBCosc)

= 2 + 2CosACosBCosC

= R.H.S

Hence proved.