# Prove Cos A + Cos B + CosC + Cos(A+B+C) = 4cos(A+B)/2*Cos(B+C)/2*Cos(C+A)/2

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### 1 Answer

We have to show that cos A + cos B + cos C + cos(A+B+C) = 4*cos(A+B)/2*cos(B+C)/2*cos(C+A)/2

4*[cos(A+B)/2]*[cos(B+C)/2]*[cos(C+A)/2]

use cos x*cos y = (cos(x+y)+cos(x-y))/2

=> 4*[cos(A+B+B+C)/2 + cos(A+B-B-C)/2]/2*cos(C+A)/2

=> 2*[cos(A+2B+C)/2 + cos(A-C)/2]cos(C+A)/2

=> 2*[cos(A+2B+C)/2][cos(C+A)/2]+2*[cos(A-C)/2][cos(C+A)/2]

=> 2*[cos(A+2B+C+C+A)/2 + cos(A+2B+C-C-A)/2]/2 + 2*[cos(A -C+C+A)/2 + cos(A-C-C-A)/2]/2

=> 2*[cos(2A+2B+2C)/2 + cos(2B)/2]/2 + 2*[cos(2A)/2 + cos(-2C)/2]/2

=> cos(A+B+C) + cos(B) + cos(A) + cos(-C)

use the relation cos x = cos(-x)

=> cos A + cos B + cos C + cos (A + B + C)

This proves that cos A + cos B + cos C + cos(A+B+C) = 4*cos(A+B)/2*cos(B+C)/2*cos(C+A)/2