Hyperbolic Cosine Function is Even/Proof 1

Theorem

$\map \cosh {-x} = \cosh x$


Proof

\(\ds \map \cosh {-x}\) \(=\) \(\ds \frac {e^{-x} + e^{-\paren {-x} } } 2\) Definition of Hyperbolic Cosine
\(\ds \) \(=\) \(\ds \frac {e^{-x} + e^x} 2\)
\(\ds \) \(=\) \(\ds \frac {e^x + e^{-x} } 2\)
\(\ds \) \(=\) \(\ds \cosh x\)

$\blacksquare$

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