Question

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

Answer 1

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

Let we solve above problem by proving LHS=RHS

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

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

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

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

$=\sinh A+\sinh B=$LHS

$\therefore$ LHS $=$ RHS

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