Hyperbolic Cosine Function is Even/Proof 1
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Theorem
- $\map \cosh {-x} = \cosh x$
Proof
| \(\ds \map \cosh {-x}\) | \(=\) | \(\ds \frac {e^{-x} + e^{-\paren {-x} } } 2\) | Definition of Hyperbolic Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {e^{-x} + e^x} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {e^x + e^{-x} } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cosh x\) |
$\blacksquare$