Cosine Deviation

Let $\theta = 2020$ . Then we have:

$\begin{aligned} X & = \cos^2\left(\frac \pi 9 + \theta \right) + \cos^2 \left(\frac {7\pi}9 + \theta\right) + \cos^2 \left(\frac {13\pi}9 + \theta\right) \\ & = \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) + \cos \left(\frac {14\pi}9 + 2\theta\right) + \cos \left(\frac {26\pi}9 + 2\theta\right)\right) \\ & = \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) - \cos \left(\frac {5\pi}9 + 2\theta\right) + \cos \left(\frac {8\pi}9 + 2\theta\right)\right) \\ & = \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) - \cos \left(\frac {5\pi}9 + 2\theta\right) + \cos \left(\frac \pi 3 + \frac {5\pi}9 + 2\theta\right)\right) \\ & = \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) - \cos \left(\frac {5\pi}9 + 2\theta\right) + \frac 12 \cos \left(\frac {5\pi}9 + 2\theta\right) - \frac {\sqrt 3}2 \sin \left(\frac {5\pi}9 + 2\theta\right)\right) \\ & = \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) - \frac 12 \cos \left(\frac {5\pi}9 + 2\theta\right) - \frac {\sqrt 3}2 \sin \left(\frac {5\pi}9 + 2\theta\right)\right) \\ & = \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) - \frac 12 \cos \left(\frac \pi 3 + \frac {2\pi}9 + 2\theta\right) - \frac {\sqrt 3}2 \sin \left(\frac \pi3 + \frac {2\pi}9 + 2\theta\right)\right) \\ & = \small \frac 12 \left(3+\cos \left(\frac {2\pi}9 + 2\theta\right) - \frac 14 \cos \left(\frac {2\pi}9 + 2\theta\right) + \frac {\sqrt 3}4 \sin \left(\frac {2\pi}9 + 2\theta\right) - \frac {\sqrt 3}4 \sin \left(\frac {2\pi}9 + 2\theta\right) - \frac 34 \cos \left(\frac {2\pi}9 + 2\theta\right) \right) \\ & = \frac 32 = \boxed{1.5} \end{aligned}$

The given sum is $1.5+\dfrac{1}{2}\left (\cos \left (\dfrac{2π}{9}+4040\right )+\cos \left (\dfrac{14π}{9}+4040\right )+\cos \left (\dfrac{26π}{9}+4040\right )\right ) =1.5+\dfrac{1}{2}\left (2\cos \left (\dfrac{14π}{9}+4040\right )\cos \left (\dfrac{12π}{9}\right) +\cos \left (\dfrac{14π}{9}+4040\right )\right) =1.5+0=\boxed {1.5}$ .