# Niven's Theorem

## Theorem

Consider the angles $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.

The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are rational are:

$\theta = 0: \sin \theta = 0$
$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
$\theta = \dfrac \pi 2: \sin \theta = 1$

## Proof

We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then:

$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$

### Lemma

For any integer $n \ge 1$, there exists a polynomial $F_n \left({x}\right)$ such that:

$F_n \left({2 \cos t}\right) = 2 \cos n t$

$\deg F_n = n$

and $F_n$ is a monic polynomial with integer coefficients.

### Proof of Lemma

The proof proceeds by induction.

For $n = 1$, it is seen that:

$F_1 \left({x}\right) = x$

fulfils the propositions.

For $n = 2$:

$F_2 \left({x}\right) = x^2 - 2$

For $n > 2$:

 $\displaystyle 2 \cos \left({n - 1}\right) t \cos t$ $=$ $\displaystyle \cos n t + \cos \left({n - 2}\right) t$ $\displaystyle \implies \ \$ $\displaystyle 2 \cos n t$ $=$ $\displaystyle \left({2 \cos \left({n - 1}\right) t}\right) \left({2 \cos t}\right) - 2 \cos \left({n - 2}\right) t$ $\displaystyle$ $=$ $\displaystyle 2 \cos t F_{n - 1} \left({2 \cos t}\right) - F_{n - 2} \left({2 \cos t}\right)$

so:

$F_n \left({x}\right) = x F_{n - 1} \left({x}\right) - F_{n - 2} \left({x}\right) \in \Z \left[{x}\right]$

will fulfil:

$F_n \left({2 \cos t}\right) = 2 \cos n t$

Because $\deg F_{n - 1} = n - 1$ and $\deg F_{n - 2} = n - 2$, we can conclude that:

$\deg F_n = \deg \left({x F_{n - 1} \left({x}\right) - F_{n - 2} \left({x}\right)}\right) = n$

In addition, the leading coefficient of $F_n$ is equal to the leading coefficient of $F_{n - 1}$, which is $1$.

Hence the lemma.

$\Box$

Suppose that $\dfrac \theta \pi$ is rational, meaning:

$\theta = \dfrac {2 \pi k} n$

where $k, n \in \Z$ and $n \ge 1$.

Suppose also that $\cos \theta \in \Q$.

Denoting $c = 2 \cos \theta \in \Q$, we get:

$F_n \left({c}\right) = F_n \left({2 \cos \dfrac {2 \pi k} n}\right) = 2 \cos \left({2 \pi k}\right) = 2$

So $c$ is a rational root of $F_n \left({x}\right) - 2$, which is a monic polynomial with integer coefficients.

By Rational Root Theorem, $c$ must be an integer.

But:

$\left\vert{c}\right\vert = \left\vert{2 \cos \theta}\right\vert \le 2$

so:

$c \in \left\{ {-2, -1, 0, 1, 2}\right\}$

Assuming that $0 \le \theta \le \dfrac \pi 2$, we get that:

$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$

Thus for any $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$ such that both $\dfrac \theta \pi$ and $\cos \theta$ are rational, then:

$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$

Instead of the above, suppose that:

$0 \le \alpha \le \dfrac \pi 2$

and both of $\dfrac \alpha \pi$ and $\sin \alpha$ are rational.

Then we can denote $\theta = \dfrac \pi 2 - \alpha$ and get that:

$0 \le \theta \le \dfrac \pi 2$
$\dfrac \theta \pi \in Q$
$\cos \theta \in \Q$

So:

$\dfrac \pi 2 - \alpha = \theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$

therefore:

$\alpha \in \left\{ {0, \dfrac \pi 6, \dfrac \pi 2}\right\}$

$\blacksquare$

## Source of Name

This entry was named for Ivan Morton Niven.

It is suspected that this result is considerably older, and may date back as far as Charles Hermite.