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OpenStudy (anonymous):
prove sinh(-x)=-sinh(x)
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OpenStudy (zarkon):
\[\sinh(x)=\frac{e^x-e^{-x}}{2}\] \[\sinh(-x)=\frac{e^{-x}-e^{-(-x)}}{2}=\frac{e^{-x}-e^{x}}{2}=-\frac{e^{x}-e^{-x}}{2}=-\sinh(x)\]
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