This is so odd

With the identity $\cos(\theta)=-\cos(\pi-\theta)$ we can rewrite the expression:

$-\cos\left(\dfrac{10\pi}{11}\right)-\cos\left(\dfrac{2\pi}{11}\right)-\cos\left(\dfrac{8\pi}{11}\right)-\cos\left(\dfrac{4\pi}{11}\right)-\cos\left(\dfrac{6\pi}{11}\right)$ .

Now, let $w$ be a complex 11th root of unity. Then, $w^{11}=1$ and, factorizing: $w^{10}+w^9+\cdots+w+1=0$ and dividing by $w^5$ :

$w^5+w^{-5}+w^4+w^{-4}+w^3+w^{-3}+w^2+w^{-2}+w+w^{-1}=-1$ . But also. $w^k+w^{-k}=2\cos\left(\dfrac{2\pi k}{11}\right)$ , so we have:

$2\cos\left(\dfrac{10\pi}{11}\right)+2\cos\left(\dfrac{8\pi}{11}\right)+2\cos\left(\dfrac{6\pi}{11}\right)+2\cos\left(\dfrac{4\pi}{11}\right)+2\cos\left(\dfrac{2\pi}{11}\right)=-1$

And finally:

$-\cos\left(\dfrac{10\pi}{11}\right)-\cos\left(\dfrac{2\pi}{11}\right)-\cos\left(\dfrac{8\pi}{11}\right)-\cos\left(\dfrac{4\pi}{11}\right)-\cos\left(\dfrac{6\pi}{11}\right)=\boxed{\dfrac{1}{2}}$

$\begin{aligned} X & = \cos{\dfrac{\pi}{11}}\color{#3D99F6}{-\cos{\dfrac{2\pi}{11}}}+\cos{\dfrac{3\pi}{11}}\color{#3D99F6}{-\cos{\dfrac{4\pi}{11}}}+\cos{\dfrac{5\pi}{11}} \\ & = \cos{\dfrac{\pi}{11}}\color{#3D99F6}{+\cos{\dfrac{9\pi}{11}}}+\cos{\dfrac{3\pi}{11}}\color{#3D99F6}{+\cos{\dfrac{7\pi}{11}}}+\cos{\dfrac{5\pi}{11}} & \small {\color{#D61F06}\text{Note that }\sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi \right) = \frac 12 \text{ see Proof.}} \\ & = {\color{#D61F06}\frac 12} = \boxed{0.5} \end{aligned}$

Proof:

This is to prove that $\displaystyle \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi \right) = \frac 12$

$\begin{aligned} \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi \right) & = \Re \left \{ \sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k \pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \cdot \frac {1 - e^{\frac {2n \pi i}{2n+1}}}{1 - e^{\frac {2 \pi i}{2n+1}}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \cdot \frac {1 - e^{\frac {2n+1-1 \pi i}{2n+1}}}{1 - e^{\frac {2 \pi i}{2n+1}}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \cdot \frac {1 + e^{- \frac {\pi i}{2n+1}}}{1 - e^{\frac {2 \pi i}{2n+1}}} \right \} \\ & = \Re \left \{\frac {1 + e^{-\frac {\pi i}{2n+1}}}{e^{-\frac {\pi i}{2n+1}} - e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{\frac {1 + \cos \frac \pi{2n+1} - i\sin \frac \pi{2n+1}}{- 2i\sin \frac \pi{2n+1}} \right \} \\ &= \frac 12 \end{aligned}$