Defining the Cosine Function from First Principles, intuitively Throughout most of my mathematics education, the cosine function has been defined formally either using its series expansion, as $\mathfrak{Re}(\exp i\theta)$, or as the unique solution to $y+y''=0$ with $y(0)=1$ and $y'(0) = 0$. 
Although these definitions make the mathematics more convenient, I've always been interested to see what it would be like to try to formalise it directly with the geometric intuition from the unit circle, assuming only the notion of distance (i.e. that we have a complete metric space with metric $d\colon\mathbb R^2 \times \mathbb R^2 \to \mathbb R$).
Since, by geometric intuition, we have that the cosine function is the $x$-coordinate obtained by walking along the unit circle, my idea was to take  a line segment from the point $(1,0)$ to another point on the circle such that the distance is $\theta$, and split it into two line segments whose sum of lengths is also $\theta$. We then split into 3 line segments, and so on, in a limiting manner, 
in such a way that the sum of lengths is always $\theta$, as illustrated below with $\theta = 2\pi/5$. 

[Desmos link]
When this procedure converges, the $x$- and $y$-coordinates of the limiting point should be the cosine and sine of $\theta$ respectively. The problem is I can't find an easy way to express the $n$th coordinate of this procedure (assuming we are splitting the line into $n-1$ segments) so that I can take the limit.
I was going to try and do things the other way round, that is, define $\cos^{-1}\colon [-1,1] \to [0,\pi]$ by approaching the circle from below and finding the length. I did manage to do this, and got that 
\begin{align*}
    \cos^{-1}(x)  &= \sum_{k=1}^\infty d\left[\left(\frac{(k-1)x}{n}, \sqrt{1-\left(\frac{(k-1)x}{n}\right)^2}\right), \left(\frac{kx}{n}, \sqrt{1-\left(\frac{kx}{n}\right)^2}\right)\right]\\[4pt]
                  &= \sum_{k=1}^\infty \sqrt{2
   \left(1-\sqrt{1-\frac{(k-1)^2
   x^2}{n^2}}
   \sqrt{1-\frac{k^2
   x^2}{n^2}}\right)-\frac{2
   k(k-1)  x^2}{n^2}},
\end{align*}
and I suppose one can extrapolate from here to obtain the $\cos$ and $\sin$ functions and extend them appropriately. 
But it feels like the other way, that is, finding that limiting point, should be possible. I appreciate any assistance with this.

Plot the unit circle on a Cartesian plane. For purposes of exposition, let’s use the usual $x,y$ coordinates and measure angles by walking the circle counterclockwise from $(x,y)=(1,0)$ as in your figures. 
Then the point at arc distance $t$ from $(1,0)$ has coordinates 
$(\cos t, \sin t)$ by the unit-circle definition of the sine and cosine. 
We can also view $\mathbf x(t) = (\cos t, \sin t)$ as a parameterized position vector of the point at arc distance $t$ around the unit circle.
Since making a rigorous measurement of the length of the arc seems inevitably to involve some kind of limiting process, let's use the kind of limiting process that allows us to compute $\mathbf x'(t),$ 
the derivative of $\mathbf x(t)$ with respect to $t.$
Now, $\mathbf x'(t)$ will be in the direction of the tangent to the circle at
$\mathbf x(t),$ and $\lvert\mathbf x'(t)\rvert = 1,$ so it's a unit vector at a right angle counterclockwise from $\mathbf x(t).$
In particular, since $\mathbf x(0) = (1,0),$ it follows that
$\mathbf x'(0) = (0, 1).$
Differentiating a second time, $\mathbf x''(t)$ is a unit vector at a right angle counterclockwise from $\mathbf x'(t),$
exactly opposite to $\mathbf x(t).$
Therefore $\mathbf x(t) + \mathbf x''(t) = 0.$
Breaking this out into components, letting $\mathbf x(t) = (f_1(t), f_2(t)),$
we have $\cos t = f_1(t)$ where $f_1(t) + f_1''(t) = 0.$
We also have $f_1(0) = 1$ and $f_1'(0) = 0.$
Show that the solution to the differential equation with those boundary conditions is unique,
then rename $f_1$ to $y,$
and you have shown that the cosine function is the unique solution to $y+y''=0$ with $y(0)=1$ and $y'(0)=0,$
using the geometric intuition of the unit circle.
