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If $e^{i t}=\cos t+i \sin t$ and $e^{-i t}=\cos t-i \sin t$ then $\cos \mathrm{h}$ $(x+i y)-\cosh (x-i y)=$
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$2 i \sin \mathrm{h} x \sin y$
$\sin \theta \cdot \cos \mathrm{h}(x+i y)=\cosh x \cos \mathrm{h}(i y)+\sinh \mathrm{x} \sin$ $\mathrm{h}(i y)$ and $\cos \mathrm{h}(x-i y)=\cosh x \cdot \cosh (i y)-\sinh x \cdot \sinh (i y)$ now $\cos \mathrm{h}(x+i y)-\cos \mathrm{h}(x-i y)=2 \sin \mathrm{h}(x) \sin \mathrm{h}(i y)$ $=2 i \sin \mathrm{h}(x) \sin y$
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