# 15.2 Properties

## 15.2.1 Odd/even nature

Note that as with their analogous trigonometric functions 11 1 $\sin(x)$ is analogous to $\sinh(x)$ and $\cos(x)$ is analogous to $\cosh(x)$ , $\sinh(x)$ is an odd function, and $\cosh(x)$ is an even function.

[Proof that sinh(x) is an odd function]Proof that $\sinh(x)$ is an odd function

 $\displaystyle\sinh(-x)$ $\displaystyle=\frac{e^{(-x)}-e^{-(-x)}}{2}$ (15.5) $\displaystyle=\frac{e^{(-x)}-e^{-(-x)}}{2}$ (15.6) $\displaystyle=\frac{e^{-x}-e^{x}}{2}$ (15.7) $\displaystyle=-\sinh(x)$ (15.8)

Proof that $\cosh(x)$ is an odd function

 $\displaystyle\cosh(-x)$ $\displaystyle=\frac{e^{(-x)}+e^{-(-x)}}{2}$ (15.9) $\displaystyle=\frac{e^{-x}+e{x}}{2}$ (15.10) $\displaystyle=\cosh(x)$ (15.11)

Proof that $\tanh(x)$ is an odd function

22 2 This can be proved by rewriting $\tanh(x)$ in terms of $e^{x}$, but as we’ve already proved the even/odd nature of $\cos(x)$ and $\sin(x)$ it makes sense to use that!
 $\displaystyle\tanh(-x)$ $\displaystyle=\frac{\sinh(-x)}{\cosh(-x)}$ (15.12) $\displaystyle=\frac{-\sinh(x)}{\cosh(x)}$ (15.13) $\displaystyle=-\frac{\sinh(x)}{\cosh(x)}$ (15.14) $\displaystyle=-\tanh(x)$ (15.15)

## 15.2.2 Inverse functions

All the hyperbolic functions have inverses, although for some of them, it is necessary to restrict the domain.

As noted in the "functions" section (although I might not have uploaded that section yet?), functions can only have inverses if they are one-to-one (i.e. for every value the function outputs, there can only be one possible input which would have resulted in that output.) The graph of $y=\cosh(x)$ is not one-to-one for the whole domain (some outputs correspond to multiple inputs), so we have to restrict its value.

The inverse functions are in the formula booklet of most maths exam boards. Otherwise, they can be derived.

## 15.2.3 Inverse function of $\sinh(x)$

###### Example 15.2.1

Finding the inverse function of $\sinh(x)$.

From the definition of $\sinh(x)$ we know that

$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$ (15.16)

The right-hand side is actually a quadratic in $e^{x}$, which becomes more readily apparent33 3 In general, if we have an expression containing $x^{y}$, $x^{-y}$ and no other powers of $x$, it’s worth considering whether there is a hidden quadratic to be found. if we multiply the fraction through by $\frac{e^{x}}{e^{x}}$.

$\sinh(x)=\frac{e^{2x}-1}{2e^{x}}$ (15.17)

Here we can substitute $u=e^{x}$

$\sinh(x)=\frac{u^{2}-1}{2u}$ (15.18)

And then we can rearrange this to give a quadratic in terms of $u$.

$u^{2}-2u\sinh(x)-1=0$ (15.19)

By applying the quadratic formula, we can solve this for $u$

$u=\frac{-(-2\sinh(x))\pm\sqrt{(-2\sinh(x))^{2}-4(1)(-1)}}{2}$ (15.20)

We can then simplify this a bit

$u=\frac{2\sinh(x)\pm\sqrt{4}\sqrt{\sinh(x)^{2}+1}}{2}$ (15.21)

If we then reverse the substitution44 4 Arguably, the substitution didn’t help here, but substitutions do often make it easier to spot the structure of a problem by simplifying problems in a way which makes them look more like a previously seen problem., replacing $u$ with $e^{x}$.

$e^{x}=\sinh(x)\pm\sqrt{\sinh(x)^{2}+1}$ (15.22)

The next step is to take the natural logarithm of both sides

$x=\ln\left(\sinh(x)\pm\sqrt{\sinh(x)^{2}+1}\right)$ (15.23)

here, because $a<\sqrt{a^{2}+1}$ 55 5 This can be proved by considering the cases when $a>0$, $a=0$ and $a<0$ we cannot have the negative case (for the $\pm$), as the domain of the logarithm function requires that the input is greater than zero.

And we have found the inverse function!

$\operatorname{arsinh}(x)=\ln\left(x+\sqrt{x^{2}+1}\right)$ (15.24)