Question: In a hyperbolic right angled triangle, the two legs have hyperbolic lengths of $3$ and $4$. What is the hyperbolic length of the hypotenuse? Is this larger or smaller than $5$?
I'm having trouble with this problem mainly because I am not allowed to use a calculator. So I have to do everything manually.
I used the formula:
$$\cosh(c)=\cosh(a)\cosh(b) $$
where $a=3$ and $b=4$
I know $\cosh(x)=\frac{e^x+e^{-x}}{2}$
With that being said, by plugging the values of $a$ and $b$
$$\cosh(c)=\cosh(3)\cosh(4) $$
and substituting into $\cosh(x)$
$$\cosh(c)=\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}$$
I know:
$$\cosh^{-1}(x)=\ln(x+\sqrt{x^2-1})$$
In this case, my $x$ would be $\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}$
Plugging into the inverse, I would obtain:
$$c=\cosh^{-1}(x)=\ln\left(\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}+\sqrt{\left(\frac{e^3+e^{-3}}{2} \times \frac{e^4+e^{-4}}{2}\right)^2-1}\right)$$
Maybe my algebra is a bit rusty but how would I know if $c$ is less than or greater than $5$. The only way I could think is to simplify above but I'm not sure how to simplify above. It became way too messy. Any help?
If there is an alternate way, I'm open to hear that as well.
\begin{align*} \cosh\sqrt{a^2+b^2} &= \sum_{m=0}^\infty \frac{(a^2+b^2)^m}{(2m)!} \\ &= \sum_{m=0}^\infty \sum_{n=0}^m \binom mn \frac{a^{2(m-n)}b^{2n}}{(2m)!} \\ &= \sum_{n=0}^\infty \sum_{m=n}^\infty \binom mn \frac{a^{2(m-n)}b^{2n}}{(2m)!} \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{m+n}n \frac{a^{2m}b^{2n}}{(2(m+n))!} \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{\binom{m+n}n}{\binom{2(m+n)}{2n}} \cdot \frac{a^{2m}b^{2n}}{(2m)!\,(2n)!} \\ &\le \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{a^{2m}b^{2n}}{(2m)!\,(2n)!} \\ &= \sum_{m=0}^\infty \frac{a^{2m}}{(2m)!} \sum_{n=0}^\infty \frac{b^{2n}}{(2n)!} \\ &= \cosh a \cosh b \\ &= \cosh c \end{align*} which since $\cosh$ is increasing on $[0,\infty)$ yields $\sqrt{a^2+b^2}\le c$. We have equality iff $a=0$ or $b=0$.