A geometry problem by Ankit Kumar Jain

Let $z = cos\left(\dfrac{\pi}{7}\right) + isin\left(\dfrac{\pi}{7}\right)$ .

By de moivre's theorem , $z^7 = cos\left(\pi\right) + isin\left(\pi\right) = -1$

$\Rightarrow z^7 + 1 = 0$

$\Rightarrow \left(z + 1\right)\left(z^6 - z^5 + z^4 - z^3 + z^2 - z +1\right) = 0$

But since the first factor $\neq 0$ , therefore the second factor $= 0$ .

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By de movier's theorem,

$cos\left(\dfrac{\pi}{7}\right) + cos\left(\dfrac{3\pi}{7}\right) + cos\left(\dfrac{5\pi}{7}\right)$

$= \dfrac{1}{2} \left( \left( z + \dfrac{1}{z}\right) + \left(z^3 + \dfrac{1}{z^3}\right) + \left(z^5 + \dfrac{1}{z^5} \right) \right)$

$= \dfrac{1}{2} \left( \dfrac{z^6 + z^4 + z^8 + z^2 + z^{10} + 1}{z^5} \right)$ .

$= \dfrac{1}{2} \left( \dfrac{z^6 + z^4 + z^8 + z^2 + z^{10} + 1}{z^5} \right) + \dfrac{1}{2} - \dfrac{1}{2}$ .

Using $z^7 = -1$ to simplify the expression gives ,

$\dfrac{1}{2}\left(\dfrac{z^6 - z^5 + z^4 - z^3 + z^2 - z + 1}{z^5}\right) + \dfrac{1}{2}$

Using , $\left(z^6 - z^5 + z^4 - z^3 + z^2 - z +1\right) = 0$ .

We get the answer as $\boxed{\dfrac{1}{2}}$ .

In general, $\displaystyle \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi \right) = \frac 12$ , consider the following.

$\begin{aligned} S & = \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi \right) \\ & = \sum_{k=0}^{n-1} \Re \left \{ e^{\frac {2k+1}{2n+ 1} \pi i} \right \} & \small \color{#3D99F6} \text{where } \Re \{ z \} \text{ is the real part of complex number }z \\ & = \Re \left \{\sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k \pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \left( \frac {1 - e^{\frac {2n \pi i}{2n+1}}}{1 - e^{\frac {2\pi i}{2n+1}}} \right) \right \} \\ & = \Re \left \{\frac {e^{\frac {\pi i}{2n+1}} + 1}{1 - e^{\frac {2\pi i}{2n+1}}} \right \} \\ & = \Re \left \{\frac 1{1 - e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{\frac 1{1 - \cos {\frac \pi{2n+1}} - i \sin {\frac \pi{2n+1}}} \right \} \\ & = \Re \left \{\frac {1 - \cos {\frac \pi{2n+1}} + i \sin {\frac \pi{2n+1}}}{\left(1 - \cos {\frac \pi{2n+1}}\right)^2 + \sin^2 {\frac \pi{2n+1}}} \right \} \\ & = \Re \left \{\frac {1 - \cos {\frac \pi{2n+1}} + i \sin {\frac \pi{2n+1}}}{2 - 2\cos {\frac \pi{2n+1}}} \right \} \\ & = \boxed{\dfrac 12} \end{aligned}$

Directly use the formula of summation of cosines when the angles are in AP .

To get 1/2