`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)`