Difference between revisions of "Relationship between sinh and sin"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\sinh(z)=-i\sin(iz),$$ | $$\sinh(z)=-i\sin(iz),$$ | ||
where $\sinh$ is the [[sinh|hyperbolic sine]] and $\sin$ is the [[sine]]. | where $\sinh$ is the [[sinh|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== | ||
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Coth|next=Relationship between cosh and cos}}: $4.5.7$ | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
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): $4.5.7$
