2
$\begingroup$

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 $N$.


My WAY:

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 $N$?

Tks.

EDIT: please see my new question for inspiration.

$\endgroup$
2
$\begingroup$

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 $\cosh N x = Q_N(\cosh x)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.