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If A+B+C=π then prove that  Cos^A+Cos^B+Cos^C=1-2CosA.CosB.CosC
The question which you have posted should be like this

Cos2A + Cos2B + Cos2C =  1- 2CosA.CosB.CosC

LHS = Cos2A + Cos2B + Cos2C
= (1 + cos 2A)/2 + (1 + cos 2B)/2 + (1 + cos 2C)/2
= 1/2 + 1/2 + 1/2 + 1/2 (cos 2A + cos 2B + cos 2C)
= 3/2 + 1/2 [2 cos (A + B) cos (A - B) + 2cos2 C - 1]
= 3/2 - 1/2 + 1/2 [2 cos (A + B) cos (A - B) + 2cos2 C]
= 1 + cos (A + B) cos (A - B) + cos2 C
= 1 + [-cos C cos (A - B) + cos 2 C]
= 1 - cos C [cos (A - B) - cos C]
= 1 - cos C [cos (A - B) + cos (A + B)]
= 1 - cos C 2cos A cos B
= 1 - 2 cos A cos B cos C
= RHS

Hence,
Cos2A + Cos2B + Cos2C =  1- 2CosA.CosB.CosC
Answered by Sneha shidid | 09 Dec, 2019, 09:25: AM

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