91Ó°ÊÓ

To verify the identity \(\cosh(t+s) = \cosh t \cosh s + \sinh t \sinh s\), we followed these steps: 1. Expressed \(\cosh(t+s)\) using the definition of \(\cosh\): \[\cosh (t+s) = \frac{e^t e^s + e^{-t} e^{-s}}{2}\] 2. Expressed \(\cosh t \cosh s\) and \(\sinh t \sinh s\) using the definitions of \(\cosh\) and \(\sinh\): \[ \begin{aligned} \cosh t \cosh s &= \left(\frac{e^{t} + e^{-t}}{2}\right)\left(\frac{e^{s} + e^{-s}}{2}\right),\\ \sinh t \sinh s &= \left(\frac{e^{t} - e^{-t}}{2}\right)\left(\frac{e^{s} - e^{-s}}{2}\right) \end{aligned} \] 3. Showed that the sum of these expressions is equivalent to the expression obtained in step 1: \[ \cosh (t+s) = \cosh t \cosh s + \sinh t \sinh s \] Thus, the given identity is verified.