Prove (cosx+cosy) ^2 + (sinx-siny) ^2 = 2+2cos (x+y)

Here you go:

1) Applying (a + b) ² = a² + b² + 2ab,

(cosx + cosy) ² = cos²x + cos²y + 2cos (x) * cos (y) and

(sinx + siny) ² = sin²x + sin²y + 2sin (x) * sin (y)

2) = = > (cosx + cosy) ² + (sinx + siny) ² =

= (cos²x + sin²x) + (cos²y + sin²y) + 2{cos (x) cos (y) + sin (x) sin (y) }

= 2 + 2{cos (x - y) } = 2[1 + cos (x - y) ]

= 2*2cos²{ (x - y) / 2} [Multiple angle identity, 1 + cos (2A) = 2cos²A]

= 4*cos²{ (x - y) / 2} [Proved]