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Proof of the Addition Formulas for Hyperbolic Functions 📂Functions

Proof of the Addition Formulas for Hyperbolic Functions

Formula

sinh⁡(x±y)= sinh⁡xcosh⁡y±sinh⁡ycosh⁡xcosh⁡(x±y)= cosh⁡xcosh⁡y±sinh⁡xsinh⁡ytanh⁡x±y=tanh⁡x±tanh⁡y1±tanh⁡xtanh⁡y \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)(1)

sinh⁡(x+y)= ex+y−e−x−y2= 2ex+y−2e−x−y4= ex+y−e−x+y+ex−y−e−x−y4+ex+y+e−x+y−ex−y−e−x−y4= (ex−e−x)ey+(ex−e−x)e−y4+ey(ex+e−x)−e−y(ex+e−x)4= (ex−e−x)(ey+e−y)4+(ey−e−y)(ex+e−x)4= (ex−e−x2)(ey+e−y2)+(ey−e−y2)(ex+e−x2)= sinh⁡xcosh⁡y+sinh⁡ycosh⁡x \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\sinh (-x)=-\sinh x and cosh⁡(−y)=cosh⁡y\cosh (-y)=\cosh y,

sinh⁡(x−y)= sinh⁡xcosh⁡(−y)+sinh⁡(−y)cosh⁡x= sinh⁡xcosh⁡y−sinh⁡ycosh⁡x \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)(2)

cosh⁡(x+y)= ex+y+e−x−y2= 2ex+y+2e−x−y4= ex+y+e−x+y+ex−y+e−x−y4+ex+y−e−x+y−ex−y+e−x−y4= (ex+e−x)ey+(ex+e−x)e−y4+(ex−e−x)ey−(ex−e−x)e−y4= (ex+e−x)(ey+e−y)4+(ex−e−x)(ey−e−y)4= (ex+e−x2)(ey+e−y2)+(ex−e−x2)(ey−e−y2)= cosh⁡xcosh⁡y+sinh⁡xsinh⁡y \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,

cosh⁡(x−y)= cosh⁡xcosh⁡(−y)+sinh⁡xsinh⁡(−y)= cosh⁡xcosh⁡y−sinh⁡xsinh⁡y \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)(3)

tanh⁡(x+y)= sinh⁡(x+y)cosh⁡(x+y)= sinh⁡xcosh⁡y+sinh⁡ycosh⁡xcosh⁡xcosh⁡y+sinh⁡xsinh⁡y= sinh⁡xcosh⁡ycosh⁡xcosh⁡y+sinh⁡ycosh⁡xcosh⁡xcosh⁡y1+sinh⁡xsinh⁡ycosh⁡xcosh⁡y= tanh⁡x+tanh⁡y1+tanh⁡xtanh⁡y \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\tanh (-x)=-\tanh x ,

tanh⁡(x−y)= tanh⁡x+tanh⁡(−y)1+tanh⁡xtanh⁡(−y)= tanh⁡x−tanh⁡y1−tanh⁡xtanh⁡y \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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