A geometry problem by Jim chale

Let $S=cos{\dfrac π 7}-cos{\dfrac {2π} 7} + cos{\dfrac {3π} 7} = cos{\dfrac π 7} + cos{\dfrac {3π} 7} + cos{\dfrac {5π} 7}$

The key is using:

$2sin{a}cos{b}=sin{(a+b)}-sin{(b-a)}$

Then we evaluate $2Ssin{\dfrac π 7}$

(SPOILER ALERT. It telescopes...)

We have : $\displaystyle 2Ssin{\dfrac π 7}=sin{\dfrac {2π} 7}+sin{\dfrac {4π} 7} - sin{\dfrac {2π} 7} + sin{\dfrac {6π} 7} - sin{\dfrac {4π} 7}= sin{\dfrac {6π} 7}=sin{\dfrac {π} 7} \Rightarrow S= \dfrac 1 2$

$\begin{aligned} x & = \cos \frac \pi 7 - {\color{#3D99F6}\cos \frac {2\pi}7} + \cos \frac {3\pi} 7 & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = \cos \frac \pi 7 + {\color{#3D99F6}\cos \frac {5\pi}7} + \cos \frac {3\pi} 7 \\ & = \frac 12 = \boxed{0.5} & \small \color{#3D99F6} \text{See proof below} \end{aligned}$

Click to see the proof .