Hyperbolic Functions' Identities
Formulas
$$ \begin{align} \sinh(-x) =&\ -\sinh x \\ \cosh(-x) =&\ \cosh x \\ \tanh(-x) =&\ - \tanh x \\ \cosh x + \sinh x =&\ e^{x} \\ \cosh x - \sinh x =&\ e^{-x} \\ \cosh^{2}x -\sinh^{2}x =&\ 1 \end{align} $$
Explanation
There’s really no proof needed. This can be directly known from the definition.
Proof
Proof of $(1)$
$$ \begin{align*} \sinh(-x) =&\ \frac{e^{-x}-e^{x}}{2} \\ =&-\frac{e^{x}-e^{-x}}{2} \\ =&-\sinh x \end{align*} $$
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Proof of $(2)$
$$ \begin{align*} \cosh(-x) =&\ \frac{e^{-x}+e^{x}}{2} \\ =&\ \frac{e^{x}+e^{-x}}{2} \\ =&\ \cosh x \end{align*} $$
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Proof of $(3)$
$$ \tanh (-x)=\frac{\sinh (-x)}{\cosh (-x)}=\frac{-\sinh x }{\cosh x}=-\tanh x $$
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Proof of $(4)$
$$ \begin{align*} \cosh x + \sinh x =&\ \frac{e^{x}+e^{-x}}{2}+\frac{e^{x}-e^{-x}}{2} \\ =&\ e^{x} \end{align*} $$
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Proof of $(5)$
$$ \begin{align*} \cosh x - \sinh x =&\ \frac{e^{x}+e^{-x}}{2}-\frac{e^{x}-e^{-x}}{2} \\ =&\ e^{-x} \end{align*} $$
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Proof of $(6)$
$$ \begin{align*} \cosh^{2} x -\sinh^{2}x =&\ \frac{(e^{x}+e^{-x})^{2}}{4}-\frac{(e^{x}-e^{-x})^{2}}{4} \\ =&\ \frac{(e^{2x}+e^{-2x}+2)-(e^{2x}+e^{-x}-2)}{4} \\ =&\ \frac{4}{4} \\ =&\ 1 \end{align*} $$
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