Hyperbolic Sine in terms of Sine

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Theorem

Let $z \in \C$ be a complex number.

Then:

$i \sinh z = \map \sin {i z}$

where:

$\sin$ denotes the complex sine
$\sinh$ denotes the hyperbolic sine
$i$ is the imaginary unit: $i^2 = -1$.


Proof

\(\ds \map \sin {i z}\) \(=\) \(\ds \frac {e^{i \paren {i z} } - e^{i \paren {-i z} } } {2 i}\) Euler's Sine Identity
\(\ds \) \(=\) \(\ds \paren {-i} \frac {e^{-z} - e^z} 2\) $i^2 = -1$
\(\ds \) \(=\) \(\ds i \frac {e^z - e^{-z} } 2\) $i^2 = -1$
\(\ds \) \(=\) \(\ds i \sinh z\) Definition of Hyperbolic Sine

$\blacksquare$

Also presented as

This identity is also seen in the form:

$\sinh z = -i \map \sin {i z}$

which can be seen to follow from the other form by multiplication by $-i$.


Also see


Sources

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