Alternative definition of hyperbolic cosine without relying on exponential function Ordinary trigonometric functions are defined independently of exponential function, and then shown to be related to it by Euler's formula. 
Can one define hyperbolic cosine so that the formula
$$\cosh{x}=\dfrac{e^x+e^{-x}}{2}$$
becomes something to be proven?
 A: Well, that is usually simply taken to be the definition, but given that
$$\cos x=\cosh ix$$
you may be asking for a proof that 
$$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$
From Taylor's theorem, we know that 
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
So
$$e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\cos x+i\sin x$$
Using $e^{ix}=\cos x+i\sin x$, express $e^{ix}+e^{-ix}$ in terms of $\cos x$, noting that the cosine function is even and the sine function is odd.
A: The more-geometrically-minded of us take $\cosh u$ and $\sinh u$ to be defined via the "unit hyperbola", $x^2 - y^2 = 1$, in a manner directly analogous to $\cos\theta$ and $\sin\theta$. Specifically, given $P$ a point on the hyperbola with vertex $V$, and defining $u$ as twice(?!) the area of the hyperbolic sector $OVP$, then $\cosh u$ and $\sinh u$ are, respectively the $x$- and $y$-coordinates of $P$.

Just as in circular trig, we can assign measures $u$ (in "hyperbolic radians") to angles ---from the flat angle (when $u=0$) to half a right angle (when $u=\infty$)--- and associate those measures with the lengths of the corresponding $\cosh$ and $\sinh$ segments. And, just as in circular trig (prior to the advent of imaginary numbers), we might be forgiven for suspecting that the correspondences $u \leftrightarrow \cosh u$ and $u \leftrightarrow \sinh u$ are "non-arithmetical", which is to say: that no arithmetical formula converts angle measures to their associated trig values.
However, it turns out that the correspondences are not non-arithmetical; to find the appropriate arithmetical conversion formula, all we need is a bit of calculus ...

Edit. (Two years later!) Check the edit history for an inelegant argument that I now streamline with the help of this trigonograph, in which lengths from the unit hyperbola have been scaled by $\sqrt{2}$ (and, thus, areas by $2$):

Because the hyperbola is rectangular, we have that $|OX|\cdot|XY|$ is a constant (here, $1$), which guarantees that the regions labeled $v$ have the same area (namely, $1/2$), and therefore that the regions labeled $u$ have the same area (namely, $u$). Now, the bit of calculus I promised, to evaluate $u$ as the area under the reciprocal curve:
$$u = \int_1^{|OX|}\frac{1}{t}dt = \ln|OX| \quad\to\quad |OX| = e^{u} \quad\to\quad |XY| = \frac{1}{e^u}$$
With that, we clearly have
$$2\,\sinh u \;=\; e^{u}- e^{-u} \qquad\qquad 2\,\cosh u \;=\; e^{u} + e^{-u}$$
as desired. Easy-peasy!
End of edit.

That hyperbolic radians are defined via doubling the area of a hyperbolic sector may seem at odds with the common definition of circular radians in terms of arc-length, but it's hard to argue with success, given the elegance of the formulas above. Even so, the hyperbolic twice-the-sector-area definition can be seen as directly analogous to the circular case, since circular radians are also definable as "twice-the-sector-area": In the unit circle, the sector with angle measure $\pi/2$ radians has area $\pi/4$ (it's a quarter-circle), the sector with angle measure $\pi$ radians has area $\pi/2$ (it's a half-circle), and the "sector" with angle measure $2\pi$ radians has area $\pi$ (it's the full circle); in these, and all other, cases, the angle measure is twice the sector area.
A: This is often taken as the definition of $\cosh$ so it can't really be proved.
