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?

• Can you use $i=e^{i\frac{\pi}{2}} = (e^{i\frac{\pi}{6}})^3= (\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})^3$? Jan 25, 2017 at 14:51
• @ClementC. No, this is strictly real-variable analysis. Jan 25, 2017 at 14:53
• can you use trigonometric relations such as $\sin(a+b)=\sin a \cos b + \sin b \cos a$? Jan 25, 2017 at 14:55
• @Arnaldo Yes, absolutely Jan 25, 2017 at 14:56
• @CuriousKid7: all right, then to actually use Novati's solution you just need to prove the addition and duplication formulas through the series definition, that is pretty standard. Jan 25, 2017 at 19:21

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: $$4y^3-3y+1=(y+1)(2y-1)^2=0$$