TITLE: What is the most efficient way to calculate the sine of a rational number?
QUESTION [8 upvotes]: I'm happy  that we can use some trig identities like $$\sin\left(\frac{\theta}{2}\right) \equiv \pm \sqrt{\frac{1-\cos(\theta)}{2}}$$ and $$\sin(\alpha \pm\beta) \equiv \sin(\alpha) \cos(\beta)\pm \cos(\alpha)\sin(\beta)$$ (and many more) to calculate exactly, for example, $\sin(15^\circ)=\sin\left(\frac{30^\circ}{2}\right)$
and $\sin(75^\circ)=\sin(90^\circ-15^\circ)$, but it seems that the smaller the angle of which I want to find the sine, the more identities I have to use, so, to calculate, for example, $\sin(1^\circ)$, is very fiddly and tedious.
My question is:

What is the most efficient way to calculate the sine of any rational angle (in as few steps as possible)?
Is there some algorithm for doing this?


P.S. I'm aware of this, which gives $\sin(\theta)$ for $\theta \in[0^\circ,90^\circ] \cap \mathbb{Z}$, but I'd like to be able to calculate, for example, $\sin\left(\frac1{12}\right)$ (exactly).

REPLY [2 votes]: For positive integers $m$ and $n$, $(-1)^m = \cos(m \pi) = T_n(\cos(m \pi/n))$, where $T_n$ is the $n$'th Chebyshev polynomial of the first kind.  So $\cos(m \pi/n)$ is one of the roots of the polynomial $T_n(z) - (-1)^m$, and of course $\sin(m \pi/n) = \pm \sqrt{1 - \cos^2(m \pi/n)}$.