Hyperbolic Cosine Function is Even

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Theorem

Let $\cosh: \C \to \C$ be the hyperbolic cosine function on the set of complex numbers.


Then $\cosh$ is even:

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


Proof 1

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

$\blacksquare$


Proof 2

\(\displaystyle \map \cosh {-x}\) \(=\) \(\displaystyle \map \cos {-i x}\) Hyperbolic Cosine in terms of Cosine
\(\displaystyle \) \(=\) \(\displaystyle \map \cos {i x}\) Cosine Function is Even
\(\displaystyle \) \(=\) \(\displaystyle \cosh x\) Hyperbolic Cosine in terms of Cosine

$\blacksquare$


Also see


Sources