Difference between revisions of "Relationship between sinh and sin"
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| Line 7: | Line 7: | ||
By definition, | By definition, | ||
$$\sinh(z) = \dfrac{e^{z}-e^{-z}}{2},$$ | $$\sinh(z) = \dfrac{e^{z}-e^{-z}}{2},$$ | ||
| − | and so by the definition of $\sin$ and the | + | and so by the definition of $\sin$ and the [[reciprocal of i]], we see |
$$-i\sinh(iz)=\dfrac{e^{iz}-e^{-iz}}{2i},$$ | $$-i\sinh(iz)=\dfrac{e^{iz}-e^{-iz}}{2i},$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
Latest revision as of 03:49, 8 December 2016
Theorem
The following formula holds: $$\sinh(z)=-i\sin(iz),$$ where $\sinh$ is the hyperbolic sine and $\sin$ is the sine.
Proof
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. █
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.5.7$
