cosh(x) + cosh(y) = 2cosh((x+y)/2)cosh((x-y)/2) Verify the identity.

cosh(x) + cosh(y) = 2cosh((x+y)/2)cosh((x-y)/2)
proof:
Taking RHS , let us solve the proof
 RHS=>2cosh((x+y)/2)cosh((x-y)/2)
=2(((e^((x+y)/2)+e^(-(x+y)/2))/2)* ((e^((x-y)/2)+e^(-(x-y)/2))/2))
its like 2((A+B)*(C+D))=2(AC+AD+BC+BD)
on multilication
=2[[(e^((x+y)/2)*(e^((x-y)/2)]+[(e^((x+y)/2)*(e^(-(x-y)/2)]+[(e^(-(x+y)/2)*(e^((x-y)/2)]+[(e^(-(x+y)/2)*(e^(-(x-y)/2)]]/4
=[[(e^((x+y)/2)*(e^((x-y)/2)))]+[(e^((x+y)/2)*(e^(-(x-y)/2))]+[(e^(-(x+y)/2)*(e^((x-y)/2))]+[(e^(-(x+y)/2)*(e^(-(x-y)/2))]]/2
As (e^((x+y)/2)*(e^((x-y)/2))) = e^((2x+y-y)/2)=e^x
similarly
(e^((x+y)/2)*(e^(-(x-y)/2)))=e^y
(e^(-(x+y)/2)*(e^((x-y)/2)))=e^-y
(e^(-(x+y)/2)*(e^(-(x-y)/2)))=e^-x
so,
[[(e^((x+y)/2)*(e^((x-y)/2))]+[(e^((x+y)/2)*(e^(-(x-y)/2)]+[(e^(-(x+y)/2)*(e^((x-y)/2)]+[(e^(-(x+y)/2)*(e^(-(x-y)/2)]]/2
=(e^x+e^y+e^-y+e^-x)/2
=(e^x+e^(-x)+e^y+e^(-y))/2
= (e^x+e^(-x))/2 +(e^y+e^(-y))/2
= cosh(x) + cosh(y)
 
And so , LHS=RHS
so ,
cosh(x) + cosh(y) = 2cosh((x+y)/2)cosh((x-y)/2)