Hyperbolic Cosine in terms of Cosine

Theorem

 * $\cos \left({ix}\right) = \cosh x $

where $\cos$ is the cosine, $\cosh$ is the hyperbolic cosine, and $i^2=-1$.

Proof
Recall the Cosine Exponential Formulation:


 * $ \displaystyle \cos x = \frac 1 2 \left({ e^{-ix} + e^{ix} }\right) $

Then: