Easier than it looks

Use $\sum_{k=1}^{11}\cos\left(\frac{2k\pi}{11}\right) = 0$ and that $cos (\theta) = \cos(2\pi - \theta)$ thereby value at k = 1 equals that at k = 10, k= 9 equals that at k = 2, k = 8 same as k = 3, k = 6 same as k = 5, and k = 4 same as k = 7. The given sum S is then $2S = \sum_{k=1}^{11}\cos\left(\frac{2k\pi}{11}\right) + cos \left(\frac {2 . 11 . \pi}{11}\right) = 0 + 1 = 1\\ \Rightarrow S = 0.5$

Let $\zeta=e^{2\pi.i/11}$ , so it is a primitive 11th root of unity.Thus we are looking for the real part of the 1st, 2nd,3rd,5th,7th and elventh power of $\zeta$ .But $\zeta^{11}=1$ , and the conjugate of $\zeta^i$ is $\zeta^{11-i}$ , i.e. they have the same real part and cancelling imaginary parts.So ifwe take the conjugates of our previous sum and add the together, we would get 1, but that is twice the real part of the sum in question, so the answer is $0.5$ .

Note:We used De Moivre and the identity

$1+\zeta+\zeta^2+...+\zeta^{10}=0$

$\sum _{ k=1 }^{ 11 }{ cos(\frac { 2\pi k }{ 11 } ) } =0\quad \quad \quad \quad \quad \quad ....(1)\\ Consider\quad the\quad following\quad identities:\\ cos(\frac { 8\pi }{ 11 } )=-cos(\frac { 3\pi }{ 11 } )=cos(\frac { 14\pi }{ 11 } )\\ cos(\frac { 12\pi }{ 11 } )=-cos(\frac { \pi }{ 11 } )=cos(\frac { 10\pi }{ 11 } )\\ cos(\frac { 16\pi }{ 11 } )=-cos(\frac { 5\pi }{ 11 } )=cos(\frac { 6\pi }{ 11 } )\\ cos(\frac { 18\pi }{ 11 } )=-cos(\frac { 7\pi }{ 11 } )=cos(\frac { 4\pi }{ 11 } )\\ cos(2\pi )=1.\\ \\ Apply\quad the\quad identities\quad to\quad simplify\quad (1)\quad and\quad \\ get\quad the\quad answer.\\$

Identity 1 comes from the fact that the equation $x^n+1=0$ has all complex no. solutions of the form $e^{ik\theta}$ , where k goes from 1 to n; and using Vietta's, we can immediately see that the sum of roots equals 0. The real part of $e^{i\theta}=\cos(\theta)$ and its sum over all k must be equal to zero. Hence identity (1).