How to prove that $\cos\theta$ is even without using unit circle? The proofs I have come across on showing that $\cos \theta$ is even is something like this:

  
*
  
*In a unit circle, $\cos\theta$ gives you the $x$ coordinate after traveling $\theta$ radians counterclockwise.
  
*Since, moving $\theta$ radians counterclockwise and $\theta$ radians clockwise i.e $-\theta$ will give you the same x coordinate, we have:
$\cos(\theta)=\cos (-\theta)$

It is possible to prove this without relying on any diagrams/geometry? For example, to prove that $f(x)=x^2$ is even we do the following:

$f(-x)=(-x)^2=x^2=f(x)$

Can we do something similar for proving $\cos \theta$ too? If not, why?
 A: If you want to do without diagrams and geometry, incl. arc length, you have to use an analytic definition of cosine. The simplest way is via the function $t\mapsto e^{it}$ (see Robert Lewis' answer). Another way is studying the coupled ODE system
$$c'(t)=-s(t), \quad s'(t)=c(t); \quad c(0)=1,\ s(0)=0\ .$$
According to the existence and uniqueness theorem for such systems it has a unique solution valid in some neighborhood $|t|<h$ of the origin, and it is then easy to see that the function $c(\cdot)=:\cos$ is even. But  a lot of work is needed  to prove that this solution can be extended to all of ${\mathbb R}$ and that it is periodic with some positive fundamental period, denoted by $2\pi$. (The latter problem you also have with the function $t\mapsto e^{it}$.)
A: Well, if you allow
$e^{i \theta} = \cos \theta + i\sin \theta, \tag{1}$
then you can simply observe that
$e^{i(-\theta)} = \cos (- \theta) + i \sin(-\theta), \tag{2}$
and that
$e^{i(-\theta)} = (e^{i \theta})^{-1}, \tag{3}$
and recall that for unimodular complex numbers $z$ we have $z^{-1} = z^*$ to conclude that
$e^{i (-\theta)} = (e^{i \theta})^*, \tag{4}$
whence
$\cos (-\theta) + i \sin (-\theta) = \cos \theta - i\sin \theta, \tag{5}$
from which we infer that
$\cos (-\theta) = \cos \theta, \tag{6}$
I.e., $\cos$ is even.  You also get $\sin$ is odd thrown in for free.  
Hope this helps.  Cheers, 
and as always,
Fiat Lux!!!
