I am struggling to understand a step in the accepted solution of How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$.
I have been new to this site, I can not figure out where to ask this question.
Can someone please enlighten me why $u^5+u^3+u=a$ is equivalent to $ua+1=a$?
With algebra, I just get stuck. I also can't find any property that establishes the equivalence.
Thanks in advance.
If you plot the points $(\cos \frac {(2n+1)\pi}7,\sin\frac {(2n+1)\pi}7)$ with $n\in\{0,\cdots, 6\}$ they will form a regular $7-gon.$
If you average these $7$ points together, they will average to the center of the unit circle i.e. $(0,0)$
$\cos \frac {\pi}{7} + \cos \frac {3\pi}{7}+\cos \frac {5\pi}{7}+\cos \frac {7\pi}{7}+\cos \frac {9\pi}{7}+\cos \frac {11\pi}{7}+\cos \frac {13\pi}{7} = 0$
$\cos \frac{\pi}{7} = \cos \frac{13\pi}{7}$ and pair-off the rest.
$2\cos \frac {\pi}{7} + 2\cos \frac {3\pi}{7}+2\cos \frac {5\pi}{7}+\cos \frac {7\pi}{7}= 0\\ \cos \frac {7\pi}{7} = -1\\ \cos \frac {\pi}{7} + \cos \frac {3\pi}{7}+\cos \frac {5\pi}{7}= \frac 12$
$u^5 + u^3 + u = a$ alone does not imply $ua + 1 = a$.
But when we are given that $u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$ we can conclude that:
$(u^5 + u^3 + u)*u + 1 = (u^5 + u^3 + u)$.
For short hand we are defining $a$ to be $a:= u^5 + u^3 + u$
and as a result we have $[a]u + 1 = [a]$ where $a$ is shorthand for $u^5 + u^3 + u$.
So $a = u^5 + u^3 + u$ !!!AND!!!! $au + 1 = a$ $\iff u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$
