## Calculus 8th Edition

$\cosh(-x)= \cosh x$
Since, we know $\cosh(x) = \dfrac{e^{x} + e^{-x}}{2}$ Thus, $\cosh(-x) =\dfrac{e^{-x} + e^{-(-x)}}{2}$ or, $=\dfrac{e^{-x} + e^{x}}{2}$ or, $= \cosh x$ Hence, it has been proved that $\cosh(-x)= \cosh x$