Proof of the Addition Formulas for Hyperbolic Functions
Formula
$$ \begin{align} \sinh (x\pm y) =&\ \sinh x \cosh y \pm \sinh y \cosh x \\ \cosh (x \pm y) =&\ \cosh x \cosh y \pm \sinh x \sinh y \\ \tanh{x \pm y}&=\frac{\tanh x \pm \tanh y}{1 \pm \tanh x \tanh y} \end{align} $$
Description
Thinking about the relationship between hyperbolic and trigonometric functions makes it natural that their forms are similar to the addition theorem of trigonometric functions.
Proof
Proof of $(1)$
$$ \begin{align*} \sinh (x+y) =&\ \frac{e^{x+y}-e^{-x-y}}{2} \\ =&\ \frac{2e^{x+y}-2e^{-x-y}}{4} \\ =&\ \frac{e^{x+y} \color{red}{-e^{-x+y}} \color{blue}{+e^{x-y}}-e^{-x-y}}{4} +\frac{e^{x+y} \color{red}{+e^{-x+y}} \color{blue}{-e^{x-y}}-e^{-x-y}}{4} \\ =&\ \frac{(e^{x} -e^{-x})e^{y}+(e^{x}-e^{-x})e^{-y}}{4} +\frac{e^{y}(e^{x}+e^{-x})-e^{-y}(e^{x}+e^{-x})}{4} \\ =&\ \frac{(e^{x} -e^{-x})(e^{y}+e^{-y})}{4} +\frac{(e^{y}-e^{-y})(e^{x}+e^{-x})}{4} \\ =&\ \left( \frac{e^{x} -e^{-x}}{2} \right)\left( \frac{e^{y} +e^{-y}}{2} \right) + \left( \frac{e^{y} -e^{-y}}{2} \right)\left( \frac{e^{x} +e^{-x}}{2} \right) \\[1em] =&\ \sinh x \cosh y +\sinh y \cosh x \end{align*} $$
Since $\sinh (-x)=-\sinh x$ and $\cosh (-y)=\cosh y$,
$$ \begin{align*} \sinh (x-y) =&\ \sinh x \cosh (-y)+\sinh (-y) \cosh x \\ =&\ \sinh{x} \cosh y -\sinh y \cosh x \end{align*} $$
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Proof of $(2)$
$$ \begin{align*} \cosh (x+y) =&\ \frac{e^{x+y}+e^{-x-y}}{2} \\ =&\ \frac{2e^{x+y}+2e^{-x-y}}{4} \\ =&\ \frac{e^{x+y} \color{red}{+e^{-x+y}} \color{blue}{+e^{x-y}}+e^{-x-y}}{4} +\frac{e^{x+y} \color{red}{-e^{-x+y}} \color{blue}{-e^{x-y}}+e^{-x-y}}{4} \\ =&\ \frac{(e^{x} +e^{-x})e^{y}+(e^{x}+e^{-x})e^{-y}}{4} +\frac{(e^{x}-e^{-x})e^{y}-(e^{x}-e^{-x})e^{-y}}{4} \\ =&\ \frac{(e^{x} +e^{-x})(e^{y}+e^{-y})}{4} +\frac{(e^{x}-e^{-x})(e^{y}-e^{-y})}{4} \\ =&\ \left( \frac{e^{x} +e^{-x}}{2} \right)\left( \frac{e^{y} +e^{-y}}{2} \right) +\left( \frac{e^{x} -e^{-x}}{2} \right) \left( \frac{e^{y} -e^{-y}}{2} \right) \\[1em] =&\ \cosh x \cosh y +\sinh x \sinh y \end{align*} $$
Furthermore,
$$ \begin{align*} \cosh (x-y) =&\ \cosh x \cosh (-y) + \sinh x \sinh (-y) \\ =&\ \cosh x \cosh y -\sinh x \sinh y \end{align*} $$
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Proof of $(3)$
$$ \begin{align*} \tanh (x+y) =&\ \frac{\sinh (x+y)}{\cosh (x+y)} \\[1em] =&\ \frac{\sinh x \cosh y + \sinh y \cosh x}{\cosh x \cosh y +\sinh x \sinh y} \\[1em] =&\ \frac{\frac{\sinh x \cosh y }{\cosh x \cosh y}+ \frac{\sinh y \cosh x}{\cosh x \cosh y}}{1 +\frac{\sinh x \sinh y}{\cosh x \cosh y}} \\[1em] =&\ \frac{\tanh x +\tanh y}{1+\tanh x \tanh y} \end{align*} $$
Furthermore, since $\tanh (-x)=-\tanh x $,
$$ \begin{align*} \tanh (x-y) =&\ \frac{\tanh x +\tanh (-y)}{1+\tanh x \tanh (-y)} \\[1em] =&\ \frac{\tanh x -\tanh y}{1-\tanh x \tanh y} \end{align*} $$
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