In many occasions, we may meet hyperbolic functions, as well as their combined ones. I want to simplify expressions like $$ \tanh\left( N\left(\textrm{acosh}~ a\right)\right) $$ and $$ \sinh\left( N\left(\textrm{acosh}~ a\right)\right) $$ for any positive integer $N$.


By the substitution $b=\textrm{acosh}~ a$, then $a=\cosh b$, and we have $$ \tanh\left( N\left(\textrm{acosh}~ a\right)\right) = \tanh\left( Nb\right) $$ If $N=1$, then $$ \tanh\left(\textrm{acosh}~ a\right) = \tanh\left(b\right)=\frac{\sqrt{a^2-1}}{a} $$ But how to do this for general $N$?


EDIT: please see my new question for inspiration.


We'll consider the case where the "outer" hyperbolic trigonometric function is $\sinh$, which contains the essential ideas of the general case. The key observation is that for any integer $N$ we can write $\sinh N x$ as some polynomial $P_N(\sinh x)$ in $\sinh x$ (of degree $N$), so that for any function $f(a)$ we have $$\sinh [N f(a)] = P_N(\sinh f(a)).$$

In the case that $f$ is an inverse trigonometric function, using either explicit computation or a hyperbolic reference triangle, we can write $\sinh f(a)$ as an algebraic expression in $a$, for example, $$\sinh \operatorname{arcosh} a = \sqrt{a^2 - 1} ,$$ and hence $$\sinh (N \operatorname{arcosh} a) = P_N(\sinh \operatorname{arcosh a}) = P_N\left(\sqrt{a^2 - 1}\right) .$$

As an example, we'll work out $P_3$ (again, this contains the essential ideas of the general case) and give an explicitly algebraic formula for $$\sinh (3 \operatorname{arcosh} a).$$

By definition, we have \begin{align} \sinh 3 x &= \frac{e^{3x} - e^{-3x}}{2}\\ &= \frac{1}{2}\left[(e^x - e^{-x})^3 + 3 (e^x - e^{-x})\right]\\ &= 4 \cdot \left(\frac{e^x - e^{-x}}{2}\right)^3 + 3 \left(\frac{e^x - e^{-x}}{2}\right)\\ &= 4 \sinh^3 x + 3 \sinh x . \end{align}

Substituting in the above formula we get $$\sinh (3 \operatorname{arcosh} a) = 4(a^2 - 1)^{3 / 2} + 3(a^2 - 1)^{1 / 2}.$$

One can of course work out general formulas for $P_N$. Likewise, for any integer $N$ there is some polynomial $Q_N$ of degree $N$ such that $\cosh N x = Q_N(\cosh x)$.

  • $\begingroup$ Hi Travis, thank you very much. If you wish, I suggest you to see my new question for fun. The link is presented in the body of the old question. $\endgroup$ – Roger209 Jan 23 '16 at 14:34
  • $\begingroup$ You're welcome, I hope you found my answer useful. I'll check out your new question soon. $\endgroup$ – Travis Willse Jan 23 '16 at 14:36
  • $\begingroup$ Yes, your answer is useful and quite general. BTW, I have solved my new question myself, see the link. Thank you for your attention. $\endgroup$ – Roger209 Jan 23 '16 at 14:54

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