Comparing $\cos^2(t)+\sin^2(t)=1$ with $\cosh^2(t)-\sinh^2(t)=1$ Case 1:
I noticed that the parametrization of $x=\cosh(t)$ and $y=\sinh(t)$ only gives us the righthand side of the graph of $x^2-y^2=1$, not the lefthandside. 
I tried to de-parametrize (or whatever that is called) by squaring: $x^2=\cosh^2(t)$ and $y^2=\sinh^2(t)$. Therefore $\cosh^2(t)-\sinh^2(t)=x^2-y^2=1$. Of course, left-hand-side is introduced because of the squaring.
Case 2:
Now I wondered about the parametrization of $x=\cos(t)$ and $y=\sin(t)$. By squaring we get: $\cos^2(t)+\sin^2(t)=x^2+y^2=1$
Question: why doesn't squaring in the second case introduce new parts to the graph? In other words, why does $x^2-y^2=1$ give us more values than the underlying parametrization, while $x^2+y^2=1$ does not?
 A: With this change of parametrsation for every point $(x,y)$ fulfilling the first parametrsation the points $(-x,y)$, $(x,-y)$ and $(-y,-x)$ are added to the graph. This means that the graph from the first equation is mirroed first ad the $x$-axis and then the reault is mirrored ad the $y$-axis.
For the $\cosh / \sinh$ parametrisation this means that after the mirroring the right-hand-side two times the lefthand hand side appears.
But for the $\cos / \sin$ parametrisation the circle is mirrored into itself, so no new point appear.
A: I'd say that a parameterization is not "underlying" as you put it. Given a curve, there is no intrinsic way to paramaterize it. Here are three for the unit circle, that all have a "flaw":


*

*$(\cos(t),\sin(t))$ lets $\mathbb{R}$ cover it countably many times (which I think is a "flaw" akin to the hyperbola arms only being half-covered)

*$(\operatorname{sech}(t),\tanh(t))$ covers the right half of the unit circle

*$\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$ covers the whole unit circle except the point $(-1,0)$.


Meanwhile, for those hyperbolas, we have:


*

*$(\cosh(t),\sinh(t))$ overs the right arm

*$(\sec(t),\tan(t))$ covers both arms countably many times, if you skip over $t$ where this is undefined

*$\left(\frac{1+t^2}{1-t^2},\frac{2t}{1-t^2}\right)$ covers both arms except the point $(-1,0)$, if you skip over $t=\pm1$


Ultimately since these curves are not topologically equivalent to the real line, any parameterization will have some "flaw" like these. It's a matter of which "flaw"s you favor.
A: Because for a real parameter $t$ the functions $\cos t$ and $\sin t$ are periodic, but $\cosh t$ and $\sinh t$ are not periodic.
A: For small values of $t,$
$ x=\cosh t\approx 1 +t^2/2 ;\; y=\sinh t \approx t ;$
If you plot cosh on x -axis and sinh on y-axis you get not cosh function but its inverse $ \cosh^{-1}$function.
Eliminating $t, y= \pm \sqrt{x^2-1}$ which is not real for $x<1$ and  symmetric only w.r.t.x-axis
On the other hand $ x^2 +y^2 =1 $ has symmetry w.r.t. both x- and y- axes.
