Congratulations to the following members for their correct solutions:
1) MarkFL
2) anemone
Solution (from anemone): [sp]Let $\displaystyle P=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
Multiply the left and right side of the equation above by $\displaystyle 2\sin\frac{\pi}{7}$, we get:
$\displaystyle \left(2\sin\frac{\pi}{7}\right)P=2\sin\frac{\pi}{7}\cos\frac{ \pi }{7}-2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{ \pi}{7}\cos\frac{3\pi}{7}$
$\displaystyle \left(2\sin\frac{\pi}{7}\right)P=\sin\frac{2 \pi}{7}-\left(\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}\right)+\left(\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}\right)$
$\displaystyle \left(2\sin\frac{\pi}{7}\right)P=\sin\frac{4\pi}{7}-\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}$
$\displaystyle \left(2\sin\frac{\pi}{7}\right)P=\sin\left(\pi-\frac{3\pi}{7}\right)-\sin\frac{3\pi}{7}+\sin\frac{\pi}{7}$
$\displaystyle \left(2\sin\frac{\pi}{7}\right)P=\sin\frac{3\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{\pi}{7}$
$\displaystyle \left(2\sin\frac{\pi}{7}\right)P=\sin\frac{\pi}{7}$
$\displaystyle P=\frac{1}{2}$, i.e.
$\displaystyle \cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac{1}{2}$[/sp]
Bonus (from MarkFL): [sp]$\displaystyle \sum_{i=1}^{n}\cos \left( \frac{(2i-1)\pi}{2n+1} \right)=\frac{1}{2}$
Using the identity $\cos(\pi-\theta)=-\cos(\theta)$, the left side becomes:
$\displaystyle -\sum_{i=1}^{n}\cos \left( \frac{2(n+1-i)\pi}{2n+1} \right)$
Multiplying by $\displaystyle 1=\frac{2\sin\left(\frac{2\pi}{2n+1} \right)}{2\sin\left(\frac{2\pi}{2n+1} \right)}$ we obtain:
$\displaystyle -\frac{1}{2\sin\left(\frac{2\pi}{2n+1} \right)}\sum_{i=1}^{n}2\sin\left(\frac{2\pi}{2n+1} \right)\cos \left( \frac{2(n+1-i)\pi}{2n+1} \right)$
Using the identities $2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ and $sin(-\theta)=-\sin(\theta)$ the sum becomes:
$\displaystyle -\frac{1}{2\sin\left(\frac{2\pi}{2n+1} \right)}\sum_{i=1}^{n}\left[\sin\left(\frac{2\pi(n+2-i)}{2n+1} \right)-\sin\left(\frac{2\pi(n-i)}{2n+1} \right) \right]$
Discarding all the terms that add to zero in the telescoping series, we are left with:
$\displaystyle -\frac{\sin\left(\frac{2\pi(n+1)}{2n+1} \right)+\sin\left(\frac{2\pi n}{2n+1} \right)-\sin\left(\frac{2\pi}{2n+1} \right)}{2\sin\left(\frac{2\pi}{2n+1} \right)}$
Using the identity $\sin(\pi-\theta)=\sin(\theta)$ on the first two terms in the numerator, we have:
$\displaystyle -\frac{\sin\left(-\frac{\pi}{2n+1} \right)+\sin\left(\frac{\pi}{2n+1} \right)-\sin\left(\frac{2\pi}{2n+1} \right)}{2\sin\left(\frac{2\pi}{2n+1} \right)}$
Using the identity $\sin(-\theta)=-\sin(\theta)$ on the first term in the numerator this becomes:
$\displaystyle -\frac{-\sin\left(\frac{\pi}{2n+1} \right)+\sin\left(\frac{\pi}{2n+1} \right)-\sin\left(\frac{2\pi}{2n+1} \right)}{2\sin\left(\frac{2\pi}{2n+1} \right)}$
Collect like terms and distribute negative sign:
$\displaystyle \frac{\sin\left(\frac{2\pi}{2n+1} \right)}{2\sin\left(\frac{2\pi}{2n+1} \right)}$
Reduce:
$\displaystyle \frac{1}{2}$[/sp]