By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) + \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.
I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.
Note that\begin{multline}\cos\left(\frac{2\pi}9\right)+\cos\left(\frac{4\pi}9\right)+\cos\left(\frac{6\pi}9\right)+\cos\left(\frac{8\pi}9\right)=\\=\frac12\left(e^{2\pi i/9}+e^{-2\pi i/9}+e^{4\pi i/9}+e^{-4\pi i/9}+e^{6\pi i/9}+e^{-6\pi i/9}+e^{8\pi i/9}+e^{-8\pi i/9}\right)\end{multline}But this is half the sum of all ninth roots of unity other than $1$. So, it's half the sum of the roots of$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x$$and that sum is $-1$.
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
The roots of $$\dfrac{z^9-1}{z-1}=0$$ are $e^{2\pi ir/9}; r=1,2,3,4,5,6,7,8$
Dividing both sides $z^4,$
$$z^4+\dfrac1{z^4}+z^3+\dfrac1{z^3}+z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$
$z+\dfrac1z=2\cos\dfrac{2\pi r}9=2y$(say)
Use $\left(z+\dfrac1z\right)^2=z^2+\dfrac1{z^2}+2$
$\left(z+\dfrac1z\right)^3=z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$
$z^4+\dfrac1{z^4}=\left(z^2+\dfrac1{z^2}\right)^2-2=\left(\left(z+\dfrac1z\right)^2-2\right)^2-2$
to form a bi-quadratic equation in $y$
Now use Vieta's formula
The roots of unity for $z^9=1$ are $e^{i \frac{2\pi k}9},\>k=0, \pm1,\pm2,\pm3,\pm4$ and their sum is equal to zero, i.e. $$0=1+ e^{i\frac{2\pi k}9} + e^{-i\frac{2\pi k}9} + e^{i\frac{4\pi k}9} + e^{-i\frac{4\pi k}9} + e^{i\frac{6\pi k}9} + e^{-i\frac{6\pi k}9} + e^{i\frac{8\pi k}9} + e^{-i\frac{8\pi k}9} $$ which leads to $$\cos\frac{2\pi}{9}+ \cos\frac{4\pi}{9}+ \cos\frac{6\pi}{9} + \cos\frac{8\pi}{9} =- \frac{1}{2}$$
