Exercise 1.2

Question

Prove that, for $n 
eq 0$, $\cos(2\pi / n)$ is an algebraic number.

Abdullah Zubair · Accepted Answer

Note that
\begin{align*}
	\cos\left( n\cdot 2\pi / n ight) &= \operatorname{Re}\left( \left( \cos(2\pi / n + \sin(2\pi / n)i ight)^n  ight)\
		1 &= \operatorname{Re}\left( \sum_{k = 0}^{n}        {n \choose k}\cos^{n - k}(2 \pi / n)(\sin (2 \pi / n)i)^{k}  ight)\
	1 &= \sum_{l = 0}^{\left\lfloor n / 2 ightfloor} {n \choose k}\cos^{n - k}(2 \pi / n)(\sin (2 \pi / n)i)^{2k}\
	0 &= \sum_{l = 0}^{\left\lfloor n / 2 ightfloor} {n \choose k}\cos^{n - k}(2 \pi / n)(\sin (2 \pi / n))^{2k} - 1\
0 &= \sum_{l = 0}^{\left\lfloor n / 2 ightfloor} {n \choose k}\cos^{n - k}(2 \pi / n)\left( 1 - \cos^{2}(2\pi / n) ight)^{k}\left( -1 ight)^{k} - 1
.\end{align*}
If we let $\alpha = \cos(2 \pi / n)$, then the expression above explicitly defines a polynomial over $\mathbb Z$ which evaluated at $\alpha$ is $0$. This shows that $\cos(2 \pi / n)$ is algebraic, as desired.

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

\begin{proof} 
If $\alpha = \cos(2\pi/n)$, then
\begin{align*}
\cos(n \cdot 2\pi/n) &= \mathrm{Re}((\cos(2\pi/n) + \sin(2\pi/n)i)^n)\
1&=\mathrm{Re}\left(\sum_{k=0}^n \binom{n}{k} \left(\cos\left(\frac{2\pi}{n}ight)ight)^{n-k}\left(\sin \left(\frac{2\pi}{n}ight) i ight)^k ight)\
0 &= -1 +\sum_{l=0}^{\lfloor n/2 floor} (-1)^l \binom{n}{2l} \cos^{n-2l}\left(\frac{2\pi}{n}ight) \sin^{2l}\left(\frac{2\pi}{n}ight) \
 &=-1+ \sum_{l=0}^{\lfloor n/2 floor} (-1)^l \binom{n}{2l} \cos^{n-2l}\left(\frac{2\pi}{n}ight) \left(1 - \cos^2\left(\frac{2\pi}{n}ight)ight)^{l} \
&= -1+\sum_{l=0}^{\lfloor n/2 floor} (-1)^l \binom{n}{2l}\alpha^{n-2l} (1- \alpha^2)^{l}
\end{align*}
Then $P(x) =  -1+\sum\limits_{l=0}^{\lfloor n/2 floor} (-1)^l \binom{n}{2l}x^{n-2l} (1- x^2)^{l} \in \Z[x]$ is such that $P(\alpha) = 0$.
Moreover $P(x) 
e 0$, since the coefficient of $x^n$ is $\sum\limits_{l=0}^{\lfloor n/2 floor} \binom{n}{2l} 
e 0$.

Therefore $\alpha = \cos(2\pi/n)$ is an algebraic number.
\end{proof}

Exercise 1.2

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Exercise 1.2

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