rational angles with sines expressible with radicals

We prove a standard constructibility result for angles in radians, then make the adaptation to degrees.

Theorem: Let $x=\frac{m}{n}$, where $m$ and $n$ are positive relatively prime integers. Then the $\frac{2\pi m}{n}$-radian angle is Euclidean constructible iff $n$ is a power of $2$ times a possibly empty product of distinct Fermat primes.

Note that the $\frac{2\pi m}{n}$-radian angle is constructible iff the $\frac{2\pi}{n}$-radian angle is constructible. One direction is obvious. For the other direction, since $m$ and $n$ are relatively prime, there exist integers $x$ and $y$ such that $xm+ny=1$. Multiply both sides by $\frac{2\pi}{n}$. We obtain $$x\frac{2\pi m}{n} +y(2\pi)=\frac{2\pi}{n}.$$ By assumption, the $\frac{2\pi m}{n}$-radian angle is constructible, and therefore so are $x$ copies of it. Clearly the $y(2\pi)$-radian angle is constructible, and therefore the $\frac{2\pi}{n}$-radian angle is constructible.

It is a standard fact about constructible regular polygons that the regular $n$-gon is constructible iff $n \ge 3$ is of the shape a power of $2$ times a product of distinct Fermat primes. This result takes care of everything but $n=1$ and $n=2$, which are obvious.

Adapting to Degrees: Let $a$ and $b$ be relatively prime positive integers. We ask for the possible values of $a$ and $b$ such that the $\frac{a}{b}$-degree angle is constructible. This is the case iff the $\frac{2\pi a}{360b}$-radian angle is constructible. Let $d=\gcd(a,360)$. So we are interested in the constructibility of the $\frac{2\pi m}{n}$-radian angle, where $$m=\frac{a}{d} \text{ and } n=\frac{360b}{d}.$$ By the result for radians, we have constructibility precisely if $\frac{360b}{d}$ is a power of $2$ times a product of distinct Fermat primes.

Suppose $d$ is not divisible by $3$. Then $\frac{360b}{d}$ is not of the right shape, since it is divisible by $3^2$. So for constructibility we need $3|a$. In addition, since we assumed that $a$ and $b$ are relatively prime, $b$ cannot be divisible by $3$.

The other problematic prime is $5$. If $a$ is not divisible by $5$, then $b$ cannot be divisible by $5$, else the Fermat prime $5$ would occur more than once in the factorization of $\frac{360b}{d}$. And if $a$ is divisible by $5$, again $b$ cannot be, since $a$ and $b$ are relatively prime. Thus in either case $5$ cannot divide $b$. We have proved:

Theorem: Let $x=\frac{a}{b}$, where $a$ and $b$ are positive relatively prime integers. Then the $\frac{a}{b}$-degree angle is Euclidean constructible iff (i) $a$ is a multiple of $3$ and (ii) $b$ is a power of $2$ times a possibly empty product of distinct Fermat primes greater than $5$.

Comment: Presumably the second result (about degrees) has been proved many times. The part about avoiding the primes $3$ and $5$ in the denominator is an "accident" caused by the choice of degree as the unit. If the Babylonians had decided to have a $340$-unit circle, $3$ would no longer be special, but $5$ and $17$ would be.