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An angle x is rational when measured in degrees. sin(x) is can be written using radicals. What are the conditions on x? If nested square roots are allowed?

What I know so far: If sin(x) can be expressed in radicals then so can $\cos{x}=\sqrt{1-\sin^{2}{x}}$, and $\tan{x}=\sin{x}/\cos{x}$. As can sin{x/2} and sin{x/3}, because I can use trig identities to get them in terms of sin{x} and the resultant quadratics and cubics can be solved.

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You seem to be allowing any kind of radical, not only (possibly iterated) square roots. If that is so, then for all "rational" angles $x$, $\sin x$ is expressible by radicals. If we restrict to real radicals (so we can take any kind of root, but only of a real number) then we can do no more with arbitrary roots than we can with possibly iterated square roots. And the answer for square roots is standard: it involves the Fermat primes. – André Nicolas Sep 20 '11 at 15:57
@André, can you refer to a proof that taking real odd radicals doesn't help? It certainly gets us more numbers, such as $\sqrt[3]{2}$ which cannot be made with square roots, but of course it is possible that those new numbers don't include any new roots of unity. (Experimentally it does seem that you're right. For example when I try to compute $\cos(\pi/18)$ this way, I end up with $x^3-\frac{3}{4}x-\frac{\sqrt3}{8}=0$, and plugging that into Cardano's formula leads to needing $\sqrt[3]{\frac{\sqrt{3}+i}{16}}$, which is exactly the trisection of $\pi/6$ that I wanted in the first place). – Henning Makholm Sep 20 '11 at 16:37
Sorry, have to go out, quick reference. I have seen better! – André Nicolas Sep 20 '11 at 17:24
But if you only have square roots available, you can't expect to solve cubics. – Henning Makholm Sep 20 '11 at 17:42
@Angela Richardson: Here measuring in degrees is good, or alternately in radians, but asking for *rational multiples of $2\pi$. Then the problem is essentially the same as the classical problem of constructing regular polygons by straightedge and compass. Don't really want to write out an answer until it is clear what the problem is. I assume you allow things like $\sqrt{\sqrt{5}-1}$. – André Nicolas Sep 20 '11 at 18:56
up vote 5 down vote accepted

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.

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@ André Nicolas: For the degree theorem: Keith Robin McLean, Constructing rational angles--(1), Math. Gazette 67 #440 (June 1983), 127-128; David H. Armitage, Constructing rational angles--(2), Math. Gazette 67 #440 (June 1983), 128-129. Also, every trig value for a rational degree angle is an algebraic number that can be expressed using radicals (easy) and a trig value for a rational degree angle is constructible if and only if the value can be expressed using real radicals (several references are at the following URL). – Dave L. Renfro Sep 21 '11 at 19:29
@Dave L. Renfro: Thank you! I searched briefly, kept getting usual trisection stuff, gave up too soon. Will check to see how much cleaner the Math Gazette solutions are. – André Nicolas Sep 21 '11 at 22:18
@Dave L. Renfro: Planning not to be at library until Friday. Do you have a web source for Armitage and McLean papers? – André Nicolas Sep 21 '11 at 22:28
@ André Nicolas: Sorry, but I don't have access to JSTOR (other than getting a temporary account set up at the local university library, which local residents can do here) and I do not know if either paper is on the internet anywhere other than by googling (which I haven't tried, since I assume you have). A somewhat related item you might be interested in is the following manuscript I wrote last April. You'll note that I tried to mostly pick references that are freely available on the internet (several of which are largely unknown, BTW). – Dave L. Renfro Sep 22 '11 at 15:56

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