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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.

Darth Geek
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    http://math.stackexchange.com/questions/347112/how-to-prove-cos-left-pi-over7-right-cos-left2-pi-over7-right-cos-left – lab bhattacharjee Aug 01 '14 at 08:40
  • Should I delete the question then? – Darth Geek Aug 01 '14 at 08:54
  • @Darth: I think if you are just trying to solve the quoted question, you should let it be closed as a duplicate. But if you want to ask if one can make your particular approach work, you should ask that specifically in your question, and it can be left open. –  Aug 01 '14 at 08:55

2 Answers2

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There is a related question on the site with the answer found through this link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908

Answer I found useful was post #7

Older, yet same question: How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$

Community
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Researcher_Witty
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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.

wendy.krieger
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