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?

  • $\begingroup$ Can you use $i=e^{i\frac{\pi}{2}} = (e^{i\frac{\pi}{6}})^3= (\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})^3$? $\endgroup$
    – Clement C.
    Jan 25, 2017 at 14:51
  • $\begingroup$ @ClementC. No, this is strictly real-variable analysis. $\endgroup$ Jan 25, 2017 at 14:53
  • 2
    $\begingroup$ can you use trigonometric relations such as $\sin(a+b)=\sin a \cos b + \sin b \cos a$? $\endgroup$
    – Arnaldo
    Jan 25, 2017 at 14:55
  • $\begingroup$ @Arnaldo Yes, absolutely $\endgroup$ Jan 25, 2017 at 14:56
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 25, 2017 at 19:21

1 Answer 1



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

  • $\begingroup$ This is really neat and actually makes a lot of the results about algebraic trigonometric values make sense. $\endgroup$ Jan 25, 2017 at 15:31

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