Q:

How do I prove that $\cosh A + \cosh B = 2\cosh \frac{A+B}{2} \cosh \frac{A-B}{2}$ using only the definition of $\cosh x = \frac{e^x + e^{-x}}{2}$?

How do I prove that $\cosh A + \cosh B = 2\cosh \frac{A+B}{2} \cosh \frac{A-B}{2}$ using only the definition of $\cosh x = \frac{e^x + e^{-x}}{2}$?
After starting from the RHS, I got down to:
$\cosh A + \cosh B = e^{\frac{A^2-B^2}{4}} + e^{\frac{-(A^2-B^2)}{4}}$
Can anyone give me a hint on how to continue?

A:

The R.H.S should instead be $$\frac{(e^{\frac{A+B}{2}} +e^{\frac{-A-B}{2}} )(e^{\frac{A-B}{2}}+e^{\frac{B-A}{2}})}{2} =\frac{e^A+e^B +e^{-A} +e^{-B}}{2} \\ = \frac{e^A+e^{-A}}{2} + \frac{e^B + e^{-B}}{2}\\ =\cosh A + \cosh B$$

