Please answer the following question?

cosB = cos(π-π/3-A) 
         = cos(2π/3)cosA + sin(2π/3)sinA
         = -(1/2)cosA +(√3/2)sinA
cos^2(A) + cos^2(B)
        = cos^2(A) + 
          (1/4)cos^2(A) + (3/4)sin^2(A) - (√3/2)sin(A)cos(A)
        = cos^2(A) + sin^2(A) +
          (1/4)cos^2(A) - (1/4)sin^2(A) -  (√3/2)sin(A)cos(A)
        = 1 + (1/4)cos(2A) - (√3/4)sin(2A)
        =  1 + (1/2)(cos(π/3)cos(2A) - sin(π/3)sin(2A))
        = 1 + (1/2)cos(2A+π/3)