I'm a high school calculus/precalculus teacher, so forgive me if the question is a little basic. One of my (very gifted) students recently came up with a construction yielding a quartic, one of whose roots was sin(80º) — which led me to the startling discovery that this (and, indeed, all rational values of sine/cosine (in degrees; that is, rational multiples of π)) are algebraic.
I've come across a number of proofs that the numbers are algebraic since, which, as I understand it, goes back to complex roots of unity. What I -haven't- seen, and would very much like to see/understand, is some general method for generating/constructing polynomials (w/ integer coefficients) whose roots are sine/cosine of rational values (in degrees). (My student's method only works for 80º/10º, 70º/20º, and 75º/15º, unfortunately). Would much appreciate…
Best Answer
If you like composing polynomials and Fermat's little theorem (and don't care about getting huge polynomials), you can use the facts that $\cos(2x) = 2\cos(x)^2 - 1$ and that if $x$ is a rational multiple of $\pi$, there will always be exponents $n$ and $m$ such that $\cos(2^n x) = \cos(2^m x)$.
Let $P(X) = 2X^2-1$.
If $x = \frac {2a \pi}b$ with odd $b$, then $2^{\phi(b)} = 1 \pmod {b}$, so that $2^{\phi(b)}x = x \pmod {2\pi}$, which means that $P^{(\phi(b))}(X) - X = P \circ P \circ \ldots \circ P (X) - X$ has $\cos(x)$ as a root.
And if $x = \frac {2a \pi}{b2^c}$ with odd $b$, you only have to pre-compose the above polynomial by $P$ $c$ times, so you get that $P^{(\phi(b)+c)}(X) - P^{(c)}(X)$ has $\cos(x)$ as a root.
And all the roots of those polynomials are some cosines of rational multiples of $\pi$ too.