Question

Prove the following identities:
$\cosh A-\cosh B=2\sinh \left(\frac{A+B}{2}\right)\sinh \left(\frac{A-B}{2}\right)$

Answer 1

$\cosh A-\cosh B=2\sinh \left(\frac{A+B}{2}\right)\sinh \left(\frac{A-B}{2}\right)$

Let's solve above problem by proving LHS=RHS

RHS $=2\sinh \left(\frac{A+B}{2}\right)\sinh \left(\frac{A-B}{2}\right)$

$=2\left[\frac{e^{\left(\frac{A+B}{2}\right)}-e^{-\left(\frac{A+B}{2}\right)}}{2}\right]\left[\frac{e^{\left(\frac{A-B}{2}\right)}-e^{-\left(\frac{A-B}{2}\right)}}{2}\right]$

$=\frac{e^A-e^B-e^{-B}+e^{-A}}{2}$

$=\left(\frac{e^A+e^{-A}}{2}\right)-\left(\frac{e^B+e^{-B}}{2}\right)$

$=\cosh A+\cosh B=$LHS

$\therefore$ LHS =RHS

Hence, $\cosh A-\cosh B=2\sinh \left(\frac{A+B}{2}\right)\sinh \left(\frac{A-B}{2}\right)$