Difference between revisions of "Relationship between sinh and sin"

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==Proof==
 
==Proof==
 +
By definition,
 +
$$\sinh(z) = \dfrac{e^{z}-e^{-z}}{2},$$
 +
and so by the definition of $\sin$ and the fact that $-i=\dfrac{1}{i}$, we see
 +
$$-i\sinh(iz)=\dfrac{e^{iz}-e^{-iz}}{2i},$$
 +
as was to be shown. █
  
 
==References==
 
==References==
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[[Category:Theorem]]
 
[[Category:Theorem]]
[[Category:Unproven]]
+
[[Category:Proven]]

Revision as of 01:04, 25 June 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 fact that $-i=\dfrac{1}{i}$, we see $$-i\sinh(iz)=\dfrac{e^{iz}-e^{-iz}}{2i},$$ as was to be shown. █

References