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 ℜ(u+u3+u5).
Then we have u7=−1 so u6−u5+u4−u3+u2−u+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(1−u)=1, or a=11−u.
So all we have to do is find ℜ(11−u).
11−u=11−cos(π/7)−isin(π/7)=1−cos(π/7)+isin(π/7)2−2cos(π/7)
so
ℜ(11−u)=1−cos(π/7)2−2cos(π/7)=12
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