Wikipedia makes the claim:
"Though a complex task, the analytical expression of $\sin 1°$ can be obtained by analytically solving the cubic equation $\sin 3° =3\sin 1° -4 \sin^3 1°$ from whose solution one can analytically derive trigonometric functions of all angles of integer degrees."
Where did this equation come from? I did a quick google search and I didn't find much.
P.S. If possible do not answer this using series.
On
See the trigonomtric triple angle formula for $\sin 3\theta$:
$$\begin{align}\sin 3\theta & = 3 \cos^2 \theta\sin\theta-\sin^3 \theta \\ \\ & = 3(1 - \sin^2\theta)\sin\theta - \sin^3 \theta \\ \\ & = 3\sin \theta - 4 \sin^3 \theta \\ \\ \end{align}$$ Now replace all occurrences of $\theta$ with $1^\circ$.
On
You can derive the identity $$\sin(3x) = 3\sin(x)\cos^2(x) - \sin^3(x)$$ by applying the angle sum rules $$\begin{align*} \cos(a + b) &= \cos(a)\cos(b) - \sin(a)\sin(b) \\ \sin(a+b) &= \sin(a)\cos(b) + \sin(b)\cos(a)\end{align*}$$
as follows:
$$\begin{align*} \sin(3x) &= \sin(2x)\cos(x) + \cos(2x)\sin(x) \\ &= \sin(x)\cos^2(x) + \sin(x)\cos^2(x) + \cos^2(x)\sin(x) - \sin^3(x) \\ &= 3\sin(x)\cos^2(x) - \sin^3(x)\end{align*}$$
The angle sum formulas are also how one can obtain $\sin(n), \cos(n)$ for all other integers once $\sin(1)$ is known, as $$\begin{align*} \cos(n+ 1) &= \cos(n)\cos(1) - \sin(n)\sin(1) \\ \sin(n+1) &= \sin(n)\cos(1) + \sin(1)\cos(n)\end{align*}.$$
(If you have some experience with complex numbers, you should figure out why the angle sum formulas are just a consequence of $$e^{i\theta} = \cos(\theta) + i\sin(\theta),$$ so all this comes from just one identity!)
It comes from the identity $\sin 3x=3\sin x \cos^2 x-\sin^3 x$ by applying $\cos^2 x=1-\sin^2 x$. Nothing special about $1^\circ$ or $3^\circ$