Non-trigonometric Proof for values of $\sin(\frac{\pi}{6})$ and $\cos(\frac{\pi}{6})$ I'm looking for non-trigometric (also, purely real analysis) proofs for the following facts. (For reference, I'm working with the series definitions for sine and cosine.)

$\sin(\frac{\pi}{6})= \cos(\frac{\pi}{3})=\frac{1}{2}$.

$\cos(\frac{\pi}{6})= \sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$.

I've proven the values of $\sin$ and $\cos$ at $0, \pi, \frac{\pi}{2}, 2\pi $and $
\frac{\pi}{4}$ as well as the the standard summation / double-angle formulas. But I'm still having trouble. My only strategy so far has been to write something like the following and perhaps expand out with the summation formulas.

$\cos(2(\frac{\pi}{3})+\frac{\pi}{3})= \cos(\pi)=-1$ and $\sin(2(\frac{\pi}{3})+\frac{\pi}{3})=\sin(\pi)=0$. 

Since I obviously don't know the value of these functions at $\frac{\pi}{3}$, I'm not sure if this will get me anywhere.
Could someone explain to me how to solve this problem?
Hint:

from $\cos(2(\frac{\pi}{3})+\frac{\pi}{3})= \cos(\pi)=-1$, using summation and double-angle formulas we have:
$$
\left(2\cos^2(\pi/3)-1 \right)\cos(\pi/3)-2\left(1-\cos^2(\pi/3)\right)\cos(\pi/3)+1=0
$$
that for $\cos(\pi/3)=y$ becomes:
$$
4y^3-3y+1=(y+1)(2y-1)^2=0
$$
In $\triangle ABC$, $AB=BD=DA$, $AC=CD$



$$\sin\frac{\pi}6=\frac{BC}{AB}=\frac{a}{2a}=\frac12$$