$a=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right) \Rightarrow a=e^{\frac{i 8 \pi}{11}}$
$\Rightarrow a$ is 11 th root of unity and all roots are $1, a, a^2, \ldots, a^{10}$
Now, $\quad a^{10}=\frac{a^{10} \cdot a}{a}=\frac{a^{11}}{a}=\frac{1}{a}=\bar{a}$
Similarly, $a^9=\overline{a^2}, a^8=\overline{a^3}, a^7=\overline{a^4}, a^6=\overline{a^5}$,
We know that,
Sum of $n$ roots of unity $=0$
$$
\begin{gathered}
1+a^1+a^2+a^3+\ldots+a^{10}=0 \\
\Rightarrow a+a^2+a^3+a^4+a^5+a^6+a^7+a^8 \\
a^9+a^{10}=-1 \\
\Rightarrow(a+\bar{a})+\left(a^2+\overline{a^2}\right)+\left(a^3+\overline{a^3}\right) \\
+\left(a^4+\overline{a^4}\right)+\left(a^5+\overline{a^5}\right)=-1 \\
\Rightarrow 2 \operatorname{Re}(a)+2 \operatorname{Re}\left(a^2\right)+2 \operatorname{Re}\left(a^3\right)+2 \operatorname{Re}\left(a^4\right)+2 \operatorname{Re}\left(a^5\right) \\
=-1[z+\bar{z}=2 \operatorname{Re}(z)] \\
\Rightarrow 2 \operatorname{Re}\left(a+a^2+a^3+a^4+a^5\right)=-1 \\
\Rightarrow \operatorname{Re}\left(a+a^2+a^3+a^4+a^5\right)=-\frac{1}{2}
\end{gathered}
$$