About the definition of angle, from Geometric Algebra

In GA, bivectors represent  2-dimensional directions. Any simple bivector can be written as
\[
\theta e_1e_2
\]
with \(\theta \in \mathbb{R}\) and \(\{e_i\}\) unitary and orthogonal vectors. The unitary vectors specify the 2-direction itself, and \(\theta\) is the size (analogous to the length of a vector, that is a 1-dimensional direction).

Given two vectors \(a\) and \(b\) (that we take to be unitary, wlog) we say that the simple bivector \(\theta e_1e_2\) is the angle formed by them if
\[
ab=e^{\theta e_1 e_2}
\]
where we take as a definition
\[
e^A=1+A+A^2 /2+\cdots=\lim_N (1+\frac{A}{N})^N
\]



Real analysis let us assure that this expression is well defined when \(A\) is a 2-blade. In fact, since \(e_1e_2e_1e_2=-1\), we can write
\[
e^{\theta e_1 e_2}=cos(\theta)+e_1 e_2sin(\theta)
\]
where \(cos\) and \(sin\) are defined by their power series.


Oserve that \(b=a e^{\theta e_1 e_2}\) and then
\[
b=a(cos(\theta)+e_1 e_2sin(\theta))=a\cdot cos(\theta)+ ae_1e_2 \cdot sin(\theta)
\]
and from here we conclude that \(a\) is in the plane generated by \(e_1\) and \(e_2\), since otherwise we would have a trivector in the right hand side of the above equation. Of course, \(b\) is also inside the plane.


The fact that \(ab=e^{\theta e_1 e_2}\) can be interpreted as follows. We have that
\[
b=a e^{\theta e_1 e_2}\cong a(1+\frac{\theta e_1 e_2}{N})^N=a+\frac{\theta}{N}a_{\perp}+\cdots
\]
whose meaning is that we arrive to \(b\) from \(a\) by applying a special infinite process. We need such an infinite process because we are  producing a circle with straight lines operations, as you can observe at this gif:

From this starting point we can prove and define all the usual trigonometry. For example, observe that
\[
ab=a\cdot b+a\wedge b=
\]
\[
=\mu +\lambda e_1e_2
\]
and is satisfied that \(\mu=cos(\theta)\) and \(\lambda=sin(\theta)\). Since \(baab=1\) we can show that
\[
cos^2(\theta)+sin^2(\theta)=1
\]



In this context, we can define \(\pi\) as the lowest real positive number \(\theta \in \mathbb{R}\) such that
\[
e_1 e_2=e^{\frac{\theta}{2} e_1 e_2}
\]

Now, natural questions arise: could \(\pi\) be the same for every pair of unitary orthogonal vectors \(e'_1, e'_2\)? Moreover, is the only one?

We can deal with the unicity question by means of real analysis by studying the pair of equations
\[cos(\theta/2)=0\]
\[
sin(\theta/2)=1\]
and we will arrive to the usual conclusions.

On the other hand, suppose \(e'_1 e'_2\) are other unitary and orthogonal vectors. Then
\[
e^{\frac{\pi}{2}e'_1 e'_2}=cos(\pi/2)+e'_1 e'_2 sin(\pi/2)=e'_1 e'_2
\]

So \(\pi\) is universal for every 2-direction.

From here we can conclude usual facts like
\[
e^{\pi i}=-1
\]
where we have called \(i=e_1e_2\) to resemble the famous expression. To see it, take any unitary \(v\) spanned by \(e_1, e_2\). And then, since
\[
-v=ve_1e_2e_1e_2=ve^{\frac{\pi}{2} e_1 e_2}e^{\frac{\pi}{2} e_1 e_2}=ve^{\pi e_1e_2}
\]
and therefore
\[
e^{\pi i}=-1
\]

Observe that we have used that
\[
e^A e^B=e^{A+B}
\]
but this is true whenever \(AB=BA\).