Is there a general formula for $\sin( {p \over q} \pi)$?

Question

Virtually everyone knows the basic values of the unit circle, $\sin(\pi) = 0; \ \ \sin({\pi \over 2}) = 1; \ \ \sin({\pi \over 3}) = {\sqrt{3} \over 2} \\$ And other values can be calculated through various identities, like $\sin({\pi \over 8}) =\frac{1}{2} \sqrt{2 - \sqrt{2}}\\$ Does there exist a general formula for $\sin({p\over q} \pi)$ for rational ${p \over q}$ as an algebraic number?

Answer 1

Too long for a comment ...

Perusing the "Exact Values [...] in Increments of 3 Degrees" list that @Claude referenced (the main idea is to use the known valus for $\sin\frac{\pi}{3}$, $\sin\frac{\pi}{4}$ and $\sin\frac{\pi}{5}$ and combining them using the sine addition formula), I found that the sines have this common (if not-necessarily-insightful) form: $$\sin \left(k\cdot 3^\circ\right) = \frac{\sqrt{2}}{4}\;\sqrt{\;4\;\pm_1\;\sqrt{\phi\,(a\phi+b\overline{\phi})}\;\pm_2\;\sqrt{\overline{\phi}\,(c\phi+d\overline{\phi})}\;}$$ where $\phi := \frac{1}{2}(\sqrt{5}+1)$ and $\overline{\phi} := \phi^{-1} = \frac{1}{2}(\sqrt{5}-1)$ are the golden ratio and its reciprocal.

$$\begin{array}{r|cccccc} \theta\; & \pm_1 & a & b & \pm_2 & c & d \\ \hline 0^\circ & - & 16 & 0 & + & 0 & 16 \\ 3^\circ & - & 3 & 0 & - & 1 & 1 \\ 6^\circ & - & 3 & 3 & - & 0 & 1 \\ 9^\circ & - & 4 & 4 & - & 0 & 0 \\ 12^\circ & - & 1 & 0 & - & 3 & 3 \\ 15^\circ & - & 12 & 0 & + & 0 & 12 \\ 18^\circ & - & 4 & 0 & + & 0 & 0 \\ 21^\circ & - & 1 & 1 & - & 0 & 3 \\ 24^\circ & - & 3 & 3 & + & 0 & 1 \\ 27^\circ & + & 0 & 0 & - & 4 & 4 \\ 30^\circ & - & 4 & 0 & + & 0 & 4 \\ 33^\circ & - & 3 & 0 & + & 1 & 1 \\ 36^\circ & + & 0 & 0 & - & 0 & 4 \\ 39^\circ & - & 1 & 1 & + & 0 & 3 \\ 42^\circ & + & 1 & 0 & - & 3 & 3 \\ 45^\circ & + & 0 & 0 & + & 0 & 0 \end{array}$$ $\sin(90^\circ-\theta)$ uses the same $a$, $b$, $c$, $d$ as $\sin \theta$, but each of $\pm_1$ and $\pm_2$ is inverted. For instance, $$\begin{align} \sin 24^\circ &= \frac{\sqrt{2}}{4}\;\sqrt{\;4\;-\;\sqrt{\phi\,(3\phi+3\overline{\phi})}\;+\;\sqrt{\overline{\phi}\,(0\phi+1\overline{\phi})}\;} \\[6pt] \sin 66^\circ &= \frac{\sqrt{2}}{4}\;\sqrt{\;4\;+\;\sqrt{\phi\,(3\phi+3\overline{\phi})}\;-\;\sqrt{\overline{\phi}\,(0\phi+1\overline{\phi})}\;} \end{align}$$

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For $\theta = 0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, these reduce to a commonly-known pattern:

$$\begin{align} \sin 0^\circ = \sqrt{\;\color{red}{0}\;}/2 \\ \sin 30^\circ = \sqrt{\;\color{red}{1}\;}/2 \\ \sin 45^\circ = \sqrt{\;\color{red}{2}\;}/2 \\ \sin 60^\circ = \sqrt{\;\color{red}{3}\;}/2 \\ \sin 90^\circ = \sqrt{\;\color{red}{4}\;}/2 \\ \end{align}$$

This is a subset of the cases in which $ab=cd=a-d=0$, which can be written in this form (noting that we happen to have $a=d=4n$): $$\frac{\sqrt{2}}{4} \sqrt{\;4\;\pm\;\sqrt{\phi\cdot 4n \phi}\;\mp\;\sqrt{\overline{\phi}\cdot 4n \overline{\phi}}\;} \;=\; \frac{1}{2}\sqrt{\;2\;\pm\;(\phi-\overline{\phi})\sqrt{n}\;} \;=\; \frac{1}{2}\sqrt{\;2\;\pm\;\sqrt{n}\;}$$ so that we have $$\begin{align} \sin 0^\circ &= \sqrt{\;2-\sqrt{\color{red}{4}}\;}\;/2 \\ \sin 15^\circ &= \sqrt{\;2-\sqrt{\color{red}{3}}\;}\;/2 \\ \color{blue}{\sin 22.5^\circ} &= \sqrt{\;2-\sqrt{\color{red}{2}}\;}\;/2 \\ \sin 30^\circ &= \sqrt{\;2-\sqrt{\color{red}{1}}\;}\;/2 \\ \sin 45^\circ &= \sqrt{\;2-\sqrt{\color{red}{0}}\;}\;/2 \\ \sin 60^\circ &= \sqrt{\;2+\sqrt{\color{red}{1}}\;}\;/2 \\ \color{blue}{\sin 67.5^\circ} &= \sqrt{\;2+\sqrt{\color{red}{2}}\;}\;/2 \\ \sin 75^\circ &= \sqrt{\;2+\sqrt{\color{red}{3}}\;}\;/2 \\ \sin 90^\circ &= \sqrt{\;2+\sqrt{\color{red}{4}}\;}\;/2 \\ \end{align}$$ with $22.5^\circ$ and $67.5^\circ$ thrown in to complete the pattern attributed to Ernesto La Orden on Ron Knott's page.

The multiples of $9^\circ$ (excluding $0^\circ$ and $90^\circ$) are characterized by having one or the other (or both) of the inner radicals vanish (that is, $(a+b)(c+d)=0$). These lead to more reductions from Knott's page, although the collection lacks the kind of "counting" pattern shown above.

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Update. Inspired by this answer, which lists cosines of multiples-of-$3^\circ$ without using an outer square root, I've devised this unified form:

$$\sin(k\cdot 3^\circ) = \frac{m \sqrt{n}}{4} \left(\; ( a\psi + b\overline{\psi} ) \sqrt{\phi\,( e\phi + f\overline{\phi} )} \;+\; ( c\psi + d\overline{\psi} ) \sqrt{\overline{\phi}\,( g\phi + h\overline{\phi} )} \;\right) $$ where $$ \{\phi,\overline{\phi}\} := \frac12\left(\sqrt{5}\pm 1\right) \quad \{\psi,\overline{\psi}\} := \frac12\left(\sqrt{3}\pm 1\right)$$ (with the overlined character using the "$-$").

$$\begin{array}{r|cc|cc:cc|cc:cc} \theta\; & m & n & a & b & c & d & e & f & g & h \\ \hline 0^\circ & 2 & 1 & + & - & - & + & 0 & + & + & \color{red}{0} \\ \hline 3^\circ & 1 & 2 & 0 & - & + & 0 & + & + & \color{red}{0} & + \\ 6^\circ & 1 & 1 & - & + & + & + & + & \color{red}{0} & + & + \\ 9^\circ & 1 & 2 & + & - & - & + & + & \color{red}{0} & + & + \\ 12^\circ & 1 & 1 & + & - & - & - & + & + & \color{red}{0} & + \\ \hdashline \star\;15^\circ & 1 & 2 & 0 & + & 0 & + & 0 & + & + & \color{red}{0} \\ \hline 18^\circ & 2 & 1 & 0 & 0 & + & - & + & + & \color{red}{0} & + \\ \hline 21^\circ & 1 & 2 & 0 & - & + & 0 & + & \color{red}{0} & + & + \\ 24^\circ & 1 & 1 & + & + & - & + & + & \color{red}{0} & + & + \\ 27^\circ & 1 & 2 & + & - & - & + & + & + & \color{red}{0} & + \\ \hdashline \star\; 30^\circ & 1 & 1 & + & - & + & - & 0 & + & + & \color{red}{0} \\ \hdashline 33^\circ & 1 & 2 & 0 & + & + & 0 & + & + & \color{red}{0} & + \\ \hline 36^\circ & 2 & 1 & 0 & 0 & + & - & + & \color{red}{0} & + & + \\ \hline 39^\circ & 1 & 2 & + & 0 & 0 & - & + & \color{red}{0} & + & + \\ 42^\circ & 1 & 1 & + & + & - & + & + & + & \color{red}{0} & + \\ \hdashline \star\;45^\circ & 1 & 2 & + & - & + & - & 0 & + & + & \color{red}{0} \\ \hdashline 48^\circ & 1 & 1 & + & - & + & + & + & + & \color{red}{0} & + \\ 51^\circ & 1 & 2 & 0 & + & + & 0 & + & \color{red}{0} & + & + \\ \hline 54^\circ & 2 & 1 & + & - & 0 & 0 & + & \color{red}{0} & + & + \\ \hline 57^\circ & 1 & 2 & + & 0 & 0 & - & + & + & \color{red}{0} & + \\ \hdashline \star\; 60^\circ & 1 & 1 & + & + & + & + & 0 & + & + & \color{red}{0} \\ \hdashline 63^\circ & 1 & 2 & + & - & + & - & + & + & \color{red}{0} & + \\ 66^\circ & 1 & 1 & + & - & + & + & + & \color{red}{0} & + & + \\ 69^\circ & 1 & 2 & + & 0 & 0 & + & + & \color{red}{0} & + & + \\ \hline 72^\circ & 2 & 1 & + & - & 0 & 0 & + & + & \color{red}{0} & + \\ \hline \star\; 75^\circ & 1 & 2 & + & 0 & + & 0 & 0 & + & + & \color{red}{0} \\ \hdashline 78^\circ & 1 & 1 & + & + & + & - & + & + & \color{red}{0} & + \\ 81^\circ & 1 & 2 & + & - & + & - & + & \color{red}{0} & + & + \\ 84^\circ & 1 & 1 & + & + & + & - & + & \color{red}{0} & + & + \\ 87^\circ & 1 & 2 & + & 0 & 0 & + & + & + & \color{red}{0} & + \\ \hline 90^\circ & 2 & 1 & + & - & + & - & 0 & + & + & \color{red}{0} \end{array}$$

Some observations:

• $n$ alternates $1$ and $2$.
• $m=2$ at, and only at, multiples of $18^\circ$. Everywhere else, $m=1$.
• There are only three $(e,f,g,h)$ patterns: $\alpha := (0++0)$, $\beta:=(++0+)$, $\gamma := (+0++)$. They cycle $\alpha\beta\gamma\gamma\beta\alpha\beta\gamma\gamma\beta\ldots$. (Observe the zig-zagging red "$0$" in the $(e,f,g,h)$ columns.)
• In particular, the $(e,f,g,h)$ patterns of complementary angles match.
• Except as marked with $\star$, $(a,b)$ and $(c,d)$ for $\theta$ are $\pm(c,d)$ and $\pm(a, b)$ for $90^\circ-\theta$, with one "$+$" and one "$-$". For those marked $\star$ (aka, the non-extreme multiples of $15^\circ$), there's no clear pattern. Is there an unambiguous, unified relation between expressions for complementary angles? (It was so easy in the earlier table!)
• Beware: I already caught (and fixed) one typo in the table.