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]