Double and Half Angle Formulas for Hyperbolic Functions
Formulas
- Double Angle Formula:
$$ \begin{align} \sinh (2x) =&\ 2\sinh x \cosh x \label{1} \\ \cosh (2x) =&\ \cosh^{2} x + \sinh^{2} x = 2\cosh ^{2 } x -1 = 2\sinh ^{2} x +1 \\ \tanh (2x) =&\ \frac{2\tanh x}{1+\tanh^{2}x} \end{align} $$
- Half Angle Formula:
$$ \begin{align} \sinh^{2} \frac{x}{2} =&\ \frac{\cosh x -1 }{2} \\ \cosh^{2} \frac{x}{2} =&\ \frac{\cosh x +1 }{2} \\ \tanh ^{2} \frac{x}{2} =&\ \frac{\cosh x -1}{\cosh x +1} \end{align} $$
$$ \begin{align} \sinh \frac{x}{2}=&\ \frac{\sinh x}{\sqrt{2(\cosh x +1)}} \\ \cosh \frac{x}{2}=&\ \frac{\sinh x}{\mathrm{sgn}(x)\sqrt{2(\cosh x -1)}} \end{align} $$
$\mathrm{sgn}(x)$ is the sign function.
Explanation
As we proved the double angle and half angle formulas of trigonometric functions, we use the addition formula of hyperbolic functions for the proof. The process is not difficult. The proof of $(4)-(6)$ is immediately obtained from the double angle formula, hence we won’t prove it separately. Similar to the half angle formula of trigonometric functions, it is obtained directly by binomial expansion and rearrangement.
Proof
Proof of $(1)$
$$ \begin{align*} \sinh (2x)=&\ \sinh (x+x) \\ =&\ \sinh x \cosh x + \sinh x \cosh x \\ =&\ 2\sinh x \cosh x \end{align*} $$
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Proof of $(2)$
$$ \begin{align*} \cosh (2x)=&\ \cosh (x+x) \\ =&\ \cosh x \cosh x + \sinh x \sinh x \\ =&\ \cosh ^{2} x + \sinh ^{2} x \end{align*} $$
Also, since $\cosh^{2} x-\sinh^{2} x=1$,
$$ \cosh^{2} x +\sinh^{2} x = 2\cosh^{2}x -1 = 2\sinh x+1 $$
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Proof of $(3)$
$$ \begin{align*} \tanh (2x) =\tanh (x+x) =&\ \frac{\sinh(x+x)}{\cosh (x+x)} \\ =&\ \frac{2\sinh x \cosh x }{\cosh^{2} x + \sinh^{2} x} \\ =&\ \frac{2\frac{\sinh x}{\cosh x}}{1+\frac{\sinh^{2} x}{\cosh^{2} x}} \\ =&\ \frac{2\tanh x}{1+\tanh^{2}x} \end{align*} $$
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Proof of $(7)$
$$ \begin{align*} \sinh \frac{x}{2} =&\ \frac{e^{x}-e^{-x}}{2} \\ =&\ \frac{(e^{\frac{x}{2}}-e^{-\frac{x}{2}})(e^{\frac{x}{2}}+e^{-\frac{x}{2}} )}{2(e^{\frac{x}{2}}+e^{-\frac{x}{2}} )} \\ =&\ \frac{e^{x}-e^{-x}}{2}\frac{1}{e^{\frac{x}{2}}+e^{-\frac{x}{2}}} \\ =&\ \sinh x \frac{1}{\sqrt{(e^{\frac{x}{2}}+e^{-\frac{x}{2}})^{2}}} \\ =&\ \frac{\sinh x}{\sqrt{e^{x}+e^{-x}+2}} \\ =&\ \frac{ \sinh x}{\sqrt{2 \frac{e^{x}+e^{-x}}{2}+2}} \\ =&\ \frac{\sinh x}{\sqrt{2(\cosh x +1)}} \end{align*} $$
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Proof of $(8)$
$$ \begin{align*} \cosh \frac{x}{2} =&\ \frac{e^{x}+e^{-x}}{2} \\ =&\ \frac{(e^{\frac{x}{2}}+e^{-\frac{x}{2}})(e^{\frac{x}{2}}-e^{-\frac{x}{2}} )}{2(e^{\frac{x}{2}}-e^{-\frac{x}{2}} )} \\ =&\ \frac{e^{x}-e^{-x}}{2}\frac{1}{e^{\frac{x}{2}}-e^{-\frac{x}{2}}} \\ =&\ \sinh x \frac{1}{\mathrm{sgn}(x)\sqrt{(e^{\frac{x}{2}}-e^{-\frac{x}{2}})^{2}}} \\ =&\ \frac{\sinh x}{\mathrm{sgn}(x)\sqrt{e^{x}+e^{-x}-2}} \\ =&\ \frac{ \sinh x}{\mathrm{sgn}(x)\sqrt{2 \frac{e^{x}+e^{-x}}{2}-2}} \\ =&\ \frac{\sinh x}{\mathrm{sgn}(x)\sqrt{2(\cosh x -1)}} \end{align*} $$
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