Is this a typo, or am I missing something?

Question

I have a handout for my precalc II class. It says $\sinh(-x) = -\sin(x)$ It should be $\sinh(-x) = -\sinh(x)$ right? I don't see how a negative input could make a hyperbolic function circular.

Answer 1

The handout is in error. It should say $$\sinh(-x)=-\sinh x$$ as you suggest.

You may be interested to know that there is a relationship between circular and hyperbolic functions. It is true that $$\sin( \pm ix )= \pm i\sinh x.$$ You may find it fun to determine all of the similar relationship using the other functions ($\cos$ with $\cosh$, etc.).

Moreover, this formula (and friends) can be used with the familiar sum formula to immediately give cool formulas like $$\sin(x\pm iy)=\sin x \cosh y \pm i\cos x \sinh y.$$

Answer 2

Now $\sinh x:=\frac{e^x-e^{-x}}{2}$ by definition.

Hence $\sinh(-x)=\frac{e^{-x}-e^{x}}{2}=-\frac{e^x-e^{-x}}{2}=-\sinh x$.

So you're right: a negative input doesn't change the hyperbolic nature of the function. But if you work with complex numbers things begin to be more intriguing and such a case could happen, for example if $z\in\mathbb C$ you have $\sinh z=-i\sin(iz)$. See the related Wikipedia page.