Hyperbolic functions. Why are they named with trig functions? I don't really get why these hyperbolic functions are named after trig functions. Can someone enlighten me?

 A: An explanation can comes from the so-called Euler formula for $\cos$ and $\sin$, namely:
$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2},\sin(z)=\frac{e^{iz}-e^{-iz}}{2i},$$
you can probably see the similarities in the definitions of $\cosh$ and $\sinh$.
Another good explanation is that $\cosh$ and $\sinh$ play the roles of $\cos$ and $\sin$ in the so-called hyperbolic geometry.
A: Firstly, why are they called hyperbolic?
Let’s take an equilateral hyperbola $\;x^2-y^2=1\,,\;$ and so we have $\;y=\sqrt{x^2-1}\;$ for the upper parts of the hyperbola (above $x$-axis). 

From the figure above (where the hyperbola is at the right): 
$$the\; area\; AMB=\int^x_1{\sqrt{x^2-1}\;dx}$$
We should apply integration by parts here: $u=\sqrt{x^2-1}$, $\;dv=dx\,.\;$ So $\;du=\frac{xdx}{\sqrt{x^2-1}}\,$, $\;v=x\,.\;\;$ But I'm gonna skip integration steps here. So the end result 
$$\int^x_1{\sqrt{x^2-1}}\;dx=\bigg(\frac{x\sqrt{x^2-1}}{2}-\frac{1}{2}\ln\,(x+{\sqrt{x^2-1}})\bigg)\;\bigg\rvert^{x}_{1}=$$
$$=\frac{x\sqrt{x^2-1}}{2}-\frac{1}{2}\ln\,(x+{\sqrt{x^2-1}})=\frac{xy}{2}-\frac{1}{2}\ln\,(x+y)$$
because $\;y=\sqrt{x^2-1}\;.$
Now 
$$Area \; OMA=Area \;OMB - Area \;AMB=\frac{1}{2}xy-\int^x_1{\sqrt{x^2-1}\;dx}\;=$$
$$=\frac{1}{2}xy-\bigg(\frac{xy}{2}-\frac{1}{2}\ln\,(x+y)\bigg)=\frac{1}{2}\ln\,(x+y)$$
Now let's denote $\;2\cdot Area\;OMA=t\,.\;\,$($t\,$ is the hatched area in the figure above). Thus $\;t=\ln\,(x+y)\,,\;$ which means $\;e^t=x+y\,.$
But our equilateral hyperbola is $\;x^2-y^2=1\,.\;$ Now, dividing $\;e^t=x+y\;$ by $\;x^2-y^2=1\;\;$ 
we obtain:
$$\frac{e^t}{1}=\frac{x+y}{x^2-y^2}=\frac{1}{x-y}\;,\quad x-y=\frac{1}{e^t}$$
So we have $\;x-y=e^{-t}\;$ and we already obtained $\;e^t=x+y\,.\;\;$ By adding and subtracting:
$$x=\frac{e^t+e^{-t}}{2}\;,\quad y=\frac{e^t-e^{-t}}{2}\;$$
As you can see in the figure above $\;x=OB\;$ and $\;y=BM\,.\quad$Just like with the unit circle $\;x^2+y^2=1\;$ or with its upper part $\;y=\sqrt{1-x^2}\;$ where we use $\;x=\cos\,t\;$ and $\;y=\sin\,t\,,\;$ here with the hyperbola we also use $\;x=\cosh\,t\;$ and $\;y=\sinh\,t\,\;$in a very similar way.$\,$ In fact, we have an incredible similarity here.
The hyperbola $\;x^2-y^2=1\,,\;$ and the unit circle $\;x^2+y^2=1\,.\;$ 
Secondly, besides similarities above ($y=\sin{t}\;$ and $\;x=\cos{t}\;$ on the unit circle whereas on the hyperbola $\;x^2-y^2=1\,\;$ we $\,$have $\;x=\cosh{t}\;$ and $\;y=\sinh{t})\;$ hyperbolic functions are extremely similar in a multitude of other ways. Comparing formulas for $\;\sin\,(x+y)\;$ and $\;\cos\,(x+y)\;$ with those for $\,\sinh\,$ and $\,\cosh\,$ we can easily derive formulas with outrageous similarities! Just a few examples should suffice:
$$\sinh\,(x+y)=\frac{e^{x+y}-e^{-x-y}}{2}=\frac{e^{x}-e^{-x}}{2}\cdot\frac{e^{y}+e^{-y}}{2}+\frac{e^{x}+e^{-x}}{2}\cdot\frac{e^{y}-e^{-y}}{2}=$$
$$=\sinh{x}\,\cosh{y}+\cosh{x}\,\sinh{y}$$
You can check this formula by doing the multiplication. $\,$ It's also very easy to check that 
$$\cosh\,(x+y)=\cosh{x}\,\cosh{y}+\sinh{x}\,\sinh{y}$$
$$\sinh{2x}=2\sinh{x}\,\cosh{x}$$
$$\sinh{x}+\sinh{y}=2\sinh{\frac{x+y}{2}}\,\cosh{\frac{x-y}{2}}\,$$
Even $\;\sin^2{x}+\cos^2{x}=1\;$ has an equivalent: $\;\;\cosh^2{x}-\sinh^2{x}=1\,.\;$ The list of such formulas for trig and hyperbolic functions is huge and the resemblance there is staggering.
Now let's look at the Taylor series:
$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\quad\quad \sinh{x}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+...$$
$$\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\quad\quad \cosh{x}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...$$
The hyperbolic functions are also periodic in the field of complex numbers:
$$\sinh{(x+2k\pi\,i)}=\sinh{x}\,,\quad \cosh{(x+2k\pi\,i)}=\cosh{x}$$
And let's note that
$$\sin{ix}=i\sinh{x}\;,\quad \cos{ix}=\cosh{x}\;,\quad \tan{ix}=i\tanh{x}$$
$$\sinh{ix}=i\sin{x}\;,\quad \cosh{ix}=\cos{x}\;,\quad \tanh{ix}=i\tan{x}$$
$$\sinh^{-1}{(ix)}=i\sin^{-1}{x}\;,\quad \cosh^{-1}{(ix)}=\pm i\cos^{-1}{x}$$
Also, we should not forget about these two beautiful formulas:
$$e^x=\cosh{x}+\sinh{x}\;,\quad e^{ix}=\cos{x}+i\sin{x}$$
One more thing. The Gateway Bridge is basically the graph of cosh function. It's called a catenary. Power cables hanging between electric towers have the shape of catenaries, not parabolas or ellipses. 
