Difference between revisions of "Relationship between sin and sinh"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\sin(x)=-i \sinh(ix),$$ whe...") |
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
| − | $$\sin( | + | $$\sin(z)=-i \sinh(iz),$$ |
| − | where $\sin$ denotes the [[ | + | where $\sin$ denotes the [[sine]] and $\sinh$ denotes the [[sinh|hyperbolic sine]]. |
| − | + | ||
| − | + | ==Proof== | |
| − | + | From the definition of $\sin$ and $\sinh$ and the [[reciprocal of i]], | |
| − | + | $$-i\sinh(iz) = \dfrac{e^{iz}-e^{-iz}}{2i} =\sin(z),$$ | |
| + | as was to be shown. | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 05:17, 8 December 2016
Theorem
The following formula holds: $$\sin(z)=-i \sinh(iz),$$ where $\sin$ denotes the sine and $\sinh$ denotes the hyperbolic sine.
Proof
From the definition of $\sin$ and $\sinh$ and the reciprocal of i, $$-i\sinh(iz) = \dfrac{e^{iz}-e^{-iz}}{2i} =\sin(z),$$ as was to be shown.
