Cosx+cosy=2cos(x+y)/2 cos(x-y)/2 proof

i) By trigonometric sum identity,

cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) and
cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)

ii) Adding both above, cos(A+B) + cos(A-B) = 2cos(A)*cos(B)

iii) Now, let A = (x + y)/2 and B = (x - y)/2

==> A + B = (x + y + x - y)/2 = 2x/2 = x

and A - B = (x + y - x + y)/2 = 2y/2 = y

iv) Thus replacing these in (ii) above,

cos(x) + cos(y) = 2cos[(x + y)/2]*cos[(x - y)/2] [Proved]

Cosx-cosy=-2sin(x+y)/2 sin(x-y)/2 Is the same way?

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cos(a+b) = cosa cosb -sinasinb
cos(a-b) = cosacosb+sinasinb

now consider the rhs

2{ cos(x+y)/2cos(x-y)/2}

2{ (cosxcosy/2 - sinxsiny/2 )* (cosxcosy/2+sinxsiny/2)}

this is like (a+b)(a-b)= a^2-b^2 where a =cosxcosy/2 and b = sinxsiny/2

2{ (cosxcosy/2}^2 - (sinxsiny/2)^2}