30
$\begingroup$

For example, can $\sinh x$ be written as a function of $\sin x$?

Another question, are hyperbolic functions dependent of their trigonometric correspondence in any way?

$\endgroup$
68
$\begingroup$

Yes. For example \begin{align*} \sinh x &= -i \sin(ix) \\ \cosh x &= \cos(ix) \\ \tanh x &= -i \tan(ix) \\ \end{align*}

These identities come from the definitions,

$$ \sin x = \frac{e^{xi}-e^{-xi}}{2i} \text{ and } \sinh x = \frac{e^x - e^{-x}}{2} $$ and similar for cosine and tangent.

$\endgroup$
  • 1
    $\begingroup$ From this definition, can we say that a hyperbolic function is a trigonometric function with complex variable and with a complex coefficient for an odd function? $\endgroup$ – Leo Authersh Aug 9 '17 at 15:50
  • 3
    $\begingroup$ Might also be worth mentioning the link to Osborne's rule here. $\endgroup$ – Silverfish Aug 9 '17 at 21:35
  • 1
    $\begingroup$ @Typhon not sure what you mean, $$ -i\sin(ix) = \frac 1 i \frac{e^{xi^2}-e^{-xi^2}}{2i} = \frac{e^x-e^{-x}}{2} = \sinh x $$ $\endgroup$ – Dando18 Aug 10 '17 at 1:37
  • 1
    $\begingroup$ @LeoAuthersh I get what you're saying, but I would not word it like that. Trig functions are already defined for complex variables, i.e. $\cosh z = \cos iz$ for $z\in\mathbb C$. $\endgroup$ – Dando18 Aug 10 '17 at 3:42
  • 1
    $\begingroup$ @LeoAuthersh no. Simply think of their definitions $\cosh z \equiv 1/2(e^z + e^{-z}),\ \ z\in\mathbb C$ and $\cos z \equiv 1/2(e^{zi}+e^{-zi}),\ z\in\mathbb C$. Both are defined for complex input, but we just have this nice identity that $\cosh z = \cos(iz)$. $\endgroup$ – Dando18 Aug 11 '17 at 14:00
46
$\begingroup$

Besides the connections between hyperbolic and circular functions which arise from substitutions involving imaginary arguments the functions can also be related using only real arguments via the Gudermannian function defined as $$\text{gd}(x)=\int_0^x\text{sech}\,t\,dt$$ This leads to identities such as $\sinh x = \tan (\text{gd}\,x)$ and $\sin x = \tanh( \text{gd}^{-1}\, x)$.

$\endgroup$
  • $\begingroup$ Now this is something quite unusual. $\endgroup$ – Simply Beautiful Art Aug 10 '17 at 23:08
12
$\begingroup$

This may not be what you mean, but the phrase "as a function of" sometimes has a narrow and precise meaning. Specifically, "$x$ can be written as a function of $y$" means that there is a function $f$ such that $x = f(y)$.

To answer that narrow question, $\sinh x$ cannot be written as a function of $\sin x$ because $\sin$ is periodic along the real axis but $\sinh$ is not.

More specifically:

  1. Assume $\forall{x \in R},~ \sinh(x) = f(\sin(x))$
  2. Therefore, $\sinh(0) = f(\sin(0))$ and $\sinh(2\pi) = f(\sin(2\pi))$
  3. Therefore, $\sinh(0) = f(0) = \sinh(2\pi)$

The last equation is not true.

$\endgroup$
  • 3
    $\begingroup$ In the field of complex numbers, both $\sin$ and $\sinh$ are periodic. $\endgroup$ – Pedro A Aug 9 '17 at 22:55
  • 1
    $\begingroup$ But I see your point. Writing $\sinh x$ strictly as a function of $\sin x$ is indeed not possible, as you showed. But I think OP wouldn't mind something like $\sin kx$. I will upvote your answer if you clarify this fact so that your answer doesn't appear to conflict with the other answers. $\endgroup$ – Pedro A Aug 9 '17 at 22:59
  • $\begingroup$ @Hamsteriffic this doesn't conflict as the other answer do not satisfy the conditions put forth in this answer. $\endgroup$ – The Great Duck Aug 10 '17 at 1:27
  • 2
    $\begingroup$ @Typhon I do know that. What I'm saying is that this answer could be more explicit on why it says "no" while the others say "yes". $\endgroup$ – Pedro A Aug 10 '17 at 1:30
  • $\begingroup$ @Hamsteriffic Reading it does make it explicit. It is explicit in what it considers to be criteria. The others also do. Reading both it is obvious why they differ. One allows for arbitrary equations, whereas this one requires literal dependency. $\endgroup$ – The Great Duck Aug 10 '17 at 2:46

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.