If $\displaystyle \alpha = e^{i\frac{8\pi}{11}}$ . Then find $\displaystyle Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)=$
Where $\displaystyle Re = $ Real part of $\displaystyle \alpha$
Let us denote $\displaystyle \beta=8\pi/11$ , we can use the following method:
$\displaystyle \alpha+\alpha^2+\ldots+\alpha^5=\dfrac{\alpha(\alp ha^5-1)}{\alpha -1}=\dfrac{e^{i\beta}(e^{5\beta i}-1)}{e^{\beta i} -1}=$
$\displaystyle e^{i\beta}\cdot \dfrac{e^{-i\beta/2}}{e^{i\beta/2}-e^{-i\beta/2}}\cdot \dfrac{1}{e^{-i5\beta/2}}\cdot (e^{i5\beta/2}-e^{-i5\beta/2})=$
$\displaystyle e^{3i\beta}\cdot \dfrac{1}{2i\sin (\beta/2)}\cdot (2i\sin (5\beta/2))=(\cos 3\beta+i\sin 3\beta)\cdot \dfrac{\sin (5\beta/2)}{\sin (\beta/2)}=\ldots$
The rest is easy.
Fernando Revilla
You can still do quite a bit of simplifying.
Use some identities: $\displaystyle \cos(-\theta)=\cos(\theta)$ $\displaystyle ,\ \cos(\pi\pm\theta)=-\cos(\theta)$ $\displaystyle ,\ \cos(2{\pi}n\pm\theta)=\cos(\theta)$
$\displaystyle \cos({{8\pi}\over{11}})=\cos(\pi-{{3\pi}\over{11}})=-\cos({{3\pi}\over{11}})$
$\displaystyle \cos({{16\pi}\over{11}})=\cos(\pi+{{5\pi}\over{11} })=-\cos({{5\pi}\over{11}})$
$\displaystyle \cos({{24\pi}\over{11}})=\cos(2\pi+{{2\pi}\over{11 }})=\cos({{2\pi}\over{11}})$
$\displaystyle \cos({{32\pi}\over{11}})=\cos({{10\pi}\over{11}})= \cos(\pi-{{\pi}\over{11}})=-\cos({{\pi}\over{11}})$
$\displaystyle \cos({{40\pi}\over{11}})=\cos(4\pi-{{4\pi}\over{11}})=\cos({{4\pi}\over{11}})$
This result is quite interesting.
Now look at the sum to product identity: $\displaystyle \displaystyle \cos(\varphi)+\cos(\theta)=2\cos\!\left({{\varphi+ \theta}\over{2}}\right)\cos\!\left({{\varphi-\theta}\over{2}}\right)\,.$ This also can give an interesting result.