Question:
The equation $x^{2}-x+1=0$ has roots $\alpha$ and $\beta$. Show that $\alpha ^{n}+\beta ^{n}=2\cos\frac{n\pi }{3}$ for $n=1, 2, 3...$
Attempt:
$x^{2}=x-1 \Rightarrow x^{n}=x^{n-1}-x^{n-2}$ for $n=3, 4, 5...$
$\therefore \alpha^{n}=\alpha^{n-1}-\alpha^{n-2}$
$\therefore \alpha ^{n}+\beta ^{n}=\alpha ^{n-1}+\beta ^{n-1}-\alpha ^{n-2}-\beta ^{n-2}$
I don't see how I could link this with cosine.
Could you please go beyond answering the question and proving that $\alpha ^{n}+\beta ^{n}=2\cos\frac{n\pi }{3}$ and explain the question to me why this relation between the roots and trig happen?
The question can probably be done by induction but is there another way?
Thank you!
Hints:
$$x^2-x+1=0\implies x_{1,2}=\frac{1\pm\sqrt{-3}}2=\begin{cases}\frac{1-\sqrt3\,i}2=e^{-\frac{2\pi i}3}=\text{cis}\left(-\frac{2\pi}3\right)\\{}\\\frac{1+\sqrt3\,i}2=e^{\frac{2\pi i}3}=\text{cis}\left(\frac{2\pi}3\right)\end{cases}$$
Note thus that
$$x_1=\overline{x_2}=x_2^{-1}\implies x_1+x_2=2\text{Re}\,(x_1)=2\cos\frac{2\pi}3\implies x_1^n+x_2^n=\;\ldots\ldots$$