Trigonometric identity on $\cos \pi/7$ I found this in a book I used to train myself for the Math Olympics a bunch of years ago:


  Prove that $$\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac{1}{2} $$


I couldn't solve it then and I can't solve it now.
As far as I know, polynomials might be involved somehow since it was in the polynomials section of the book, and it's not suposed to use complex numbers since it's in the section previous to that.

Using the double and triple angle formulaes I got the LHS to:

$$4\cos^3\frac{\pi}{7} -2\cos^2\frac{\pi}{7} - 2\cos\frac{\pi}{7}+1 = \frac{1}{2}$$

Using the sum-to-product formulaes I got to the same result.

And I'm pretty much stuck at this point.
There is a related question on the site with the answer found through this link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&amp;sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908

Answer I found useful was post #7

Older, yet same question:
How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$
The trick here lies in the '7'.   These are chords of a heptagon, and so if you mix $\cos \frac{4\pi}7 $ into the mix, you can use that these are equal, and thus derive the desired result.