If A+B+C=π then prove that 

Cos^A+Cos^B+Cos^C=1-2CosA.CosB.CosC

Asked by vikassuradkar2015 | 8th Dec, 2019, 11:11: PM

Expert Answer:

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 | 9th Dec, 2019, 09:25: AM