monkeyman130yb

2022-03-30

$\alpha$ and $\beta$ are the roots of the quadratic equation.
$\omega \ne 1$ and ${\omega }^{13}=1$
If $\alpha =\omega +{\omega }^{3}+{\omega }^{4}+{\omega }^{-4}+{\omega }^{-3}+{\omega }^{-1}$
$\beta ={\omega }^{2}+{\omega }^{5}+{\omega }^{6}+{\omega }^{-6}+{\omega }^{-5}+{\omega }^{-2}$,
then quadratic equation, whose roots are $\alpha$ and $\beta$ is:
I tried finding the sum and product of roots and placing in the equation ${x}^{2}-\left(\alpha +\beta \right)x+\left(\alpha ×\beta \right)=0$ but I was not able to solve for it.

Lana Hamilton

Step 1
By the formula for a finite geometric series,
$\alpha +\beta =\sum _{i=-6}^{-1}{\omega }^{i}+\sum _{i=1}^{6}{\omega }^{i}$
$=\sum _{i=7}^{12}{\omega }^{i}+\sum _{i=1}^{6}{\omega }^{i}$
$=\frac{\left(1-{\omega }^{13}\right)}{1-\omega }-1$
$-1=-1$
For the product $\alpha \beta$, note that ${\omega }^{n}={\omega }^{-n}$
Taking $\omega$ as ${e}^{2i\frac{\pi }{13}}=\mathrm{cos}\left(2\frac{\pi }{13}\right)+i\mathrm{sin}\left(2\frac{\pi }{13}\right)$ for example, $\alpha =2\left(\mathrm{cos}\left(2\frac{\pi }{13}\right)+\mathrm{cos}\left(6\frac{\pi }{13}\right)+\mathrm{cos}\left(8\frac{\pi }{13}\right)\right)$ and you can do the same for $\beta$. Then you can use the product-to-sum formula
$\mathrm{cos}A\mathrm{cos}B=\frac{1}{2}\left(\mathrm{cos}\left(A-B\right)+\mathrm{cos}\left(A+B\right)\right)$
by expanding the brackets, there are nine terms in this form.

Tristatex9tw

Step 1
We have
$1+\alpha +\beta =1+\omega +{\omega }^{2}+\cdots +{\omega }^{12}=0$
and so $\alpha +\beta =-1$
Therefore, $\alpha \beta =-\left(\alpha +{\alpha }^{2}\right)=-3$
Unfortunately, I can't see shortcut to compute $\alpha +{\alpha }^{2}$ except by expanding it.

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