# Problem of the week #48 - February 25, 2013

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#### Jameson

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Show that $$\displaystyle \cos \left( \frac{\pi}{7} \right)-\cos \left( \frac{2\pi}{7} \right)+\cos \left( \frac{3\pi}{7} \right)=\frac{1}{2}$$

Bonus:
Show that the general form of $$\displaystyle \sum_{i=1}^{n}\cos \left( \frac{(2i-1)\pi}{2n+1} \right)$$ is always one-half.

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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone

Solution (from anemone):
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.

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

Bonus (from MarkFL):
$\displaystyle \sum_{i=1}^{n}\cos \left( \frac{(2i-1)\pi}{2n+1} \right)=\frac{1}{2}$

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:

$\displaystyle \frac{1}{2}$

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