Friday, October 28, 2016

trigonometry - How Can One Prove cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2




Reference: http://xkcd.com/1047/



We tried various different trigonometric identities. Still no luck.



Geometric interpretation would be also welcome.



EDIT: Very good answers, I'm clearly impressed. I followed all the answers and they work! I can only accept one answer, the others got my upvote.


Answer




Hint: start with eiπ7=cos(π/7)+isin(π/7) and the fact that the lhs is a 7th root of -1.



Let u=eiπ7, then we want to find (u+u3+u5).



Then we have u7=1 so u6u5+u4u3+u2u+1=0.



Re-arranging this we get: u6+u4+u2+1=u5+u3+u.



If a=u+u3+u5 then this becomes ua+1=a, and rearranging this gives a(1u)=1, or a=11u.




So all we have to do is find (11u).



11u=11cos(π/7)isin(π/7)=1cos(π/7)+isin(π/7)22cos(π/7)



so



(11u)=1cos(π/7)22cos(π/7)=12


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