cos^2(A/2)+cos^2(B/2)+cos^2(C/2)=2+2sin(A/2)*sin(B/2)*sin(C/2)

There must be more to the question than this. Note that if A=B=C=0, the left hand side is equal to 3 while the right hand side is equal to 2. This makes me think there must be an aditional condition, such as A+B+C = 180 degrees. Please confirm.

Upate: if A+B+C =180, then this relationship can be shown to be true. Angle C = 180-(A+B), so on the left hand side of the equation you can replace cos(C/2) with cos(90-A/2-B/2) which equals sin(A/2+B/2) = sin(A/2)cos(B/2)+cos(A/2)sin(B/2). Also on the left side replace cos^2(A/2) with 1- sin^2(A/2), and replace cos^2(B/2) with 1 - sin^2(B/2). On the right hand side proceed in a similar way: replace sin(C/2) with cos(A/2+B/2) = cos(A/2)cos(B/2)-sin(A/2)sin(B/2). Now some simple manipulations will bring it home.