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 $\cosh x = \frac{e^x + e^{-x}}{2}$?

Question

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?

Answer 1

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

Answer 2

Make it simple and start from the factorisation formula $$\cosh(u+v)+\cosh(u-v)=2\cosh u\cosh v,$$ readily deduced from the addition formulæ $$\cos(u\pm v)=\cosh u\cosh v\pm\sinh u\sinh, v$$ and solve the linear system \begin{cases} u+v=A, \\u-v=B. \end{cases}