# Relationship between sinh and sin

The following formula holds: $$\sinh(z)=-i\sin(iz),$$ where $\sinh$ is the hyperbolic sine and $\sin$ is the sine.
By definition, $$\sinh(z) = \dfrac{e^{z}-e^{-z}}{2},$$ and so by the definition of $\sin$ and the reciprocal of i, we see $$-i\sinh(iz)=\dfrac{e^{iz}-e^{-iz}}{2i},$$ as was to be shown. █