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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?

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    Can you use $i=e^{i\frac{\pi}{2}} = (e^{i\frac{\pi}{6}})^3= (\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})^3$? – 2017-01-25
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    @ClementC. No, this is strictly real-variable analysis. – 2017-01-25
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    can you use trigonometric relations such as $\sin(a+b)=\sin a \cos b + \sin b \cos a$? – 2017-01-25
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    @Arnaldo Yes, absolutely – 2017-01-25
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    @Rohan I'm looking for a purely analytic proof – 2017-01-25
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    The answer depends on your definition of $\pi$, too. According to your definition, have you already proved that $\sin(\pi)=0$? – 2017-01-25
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    @JackD'Aurizio Yes, I have. Definition is $2\inf \{x>0 : \cos(x)=0\}$. – 2017-01-25
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    @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. – 2017-01-25

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

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    This is really neat and actually makes a lot of the results about algebraic trigonometric values make sense. – 2017-01-25
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In $\triangle ABC$, $AB=BD=DA$, $AC=CD$ enter image description here

enter image description here $$\sin\frac{\pi}6=\frac{BC}{AB}=\frac{a}{2a}=\frac12$$

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    The OP is "working with the series definitions for sine and cosine." It's not obvious *a priori* how to link that to the geometric interpretation you are using. – 2017-01-25
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    Right, this is not at all the kind of proof i'm looking for. I can't refer directly to trigonometry or geometry. – 2017-01-25