Can we express $\sin 1^\circ$ in a real closed, not repetitive, radical forms? Can we express $\sin 1^\circ$ in a real closed, not repetitive radical forms? Any radical forms mean you can use any roots but without constants $\pi$, $e$ or other trigonometry functions.
 A: If we're allowed to use complex numbers, then sure. Note that $\sin(1 ^\circ)=\sin(\pi/180)=\frac{e^{i \pi / 180} - e^{-i \pi/180}}{2i}$. We can express $e^{i \pi/180} = (-1)^{1/180}$ and  $e^{-i \pi/180} = -(-1)^{179/180}$. So the result is $\sin(1^\circ) = -\frac{1}{2} (-1)^{1/2}((-1)^{1/180}+(-1)^{179/180})$
A: Since $(-1)^{89/180}=\cos(89^\circ)+i\sin(89^\circ)$, we get
$$
\sin(1^\circ)=\cos(89^\circ)=\frac12\left((-1)^{89/180}+(-1)^{-89/180}\right)\tag{1}
$$
However, I think more in keeping with the spirit of the question, if we start with $\sin(6^\circ)=\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}8$ and apply
$$
\sin\left(\frac x2\right)=\sqrt{\frac{1-\sqrt{1-\sin^2(x)}}2}\tag{2}
$$
and
$$
\sin\left(\frac x3\right)=\frac{\sqrt[\Large3]{-\sin(x)+\sqrt{\sin^2(x)-1}}+\sqrt[\Large3]{-\sin(x)-\sqrt{\sin^2(x)-1}}}2\tag{3}
$$
we get $\sin(1^\circ)$. Of course, we still get into the realm of complex numbers when applying $(3)$.
A: In principle, yes.
This paper gives a value for $\sin 3^{\circ}$:
$$\sin 3^{\circ} = \frac{1}{4} \sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}} \, .$$
Moreover, we have the triple-angle identity for $\sin$, which I will suggestively write as:
$$
4 \sin^3 \theta-3 \sin \theta + \sin 3\theta=0 \, .
$$
Combining these, you can see that $x=\sin 1^{\circ}$ is a root of the cubic polynomial
$$
4x^3-3x+ \frac{1}{4} \sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}} = 0\, .
$$
You could then use the cubic formula to find a closed-form expression of $\sin 1^\circ$ that uses only radicals.
Two caveats:


*

*When you use the cubic formula on a polynomial with three real roots, the radical expression you get must always involve complex numbers. This will be the case here, since $\sin 121^\circ$ and $\sin 241^\circ$ must also be roots of the same polynomial. So if you want to express $\sin 1^\circ$ in terms of radicals using only real numbers, you're out of luck.

*The expression you get will be so horrifically complicated as to be totally useless for any practical or computational purpose.

A: Well you can construct a regular pentagon and an equilateral triangle inscribed in a circle with ruler and compasses (equivalent to taking square roots) - that gets you angles of $72 ^\circ$ and $60^\circ$ - you can get sin and cos of both angles, so can get sin and cos of the difference $12^\circ$. You can halve that twice to get down to $3^\circ$, and then you need to solve a cubic, which is solvable by radicals.
A: Here's a table of exact values of the sine, cosine, tangent, and cotangent of integer multiples of $3^\circ$: http://en.wikipedia.org/wiki/Exact_trigonometric_constants
A: This page has closed formulas for $\sin(n^\circ)$ for all $n=1,\dots,90$.  Many of them involved $\sqrt{-1}$, not sure if that is avoidable but somebody in another answer on this page says it is not.
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
I would like to give credit to the person who made me aware of that list, but their post was deleted so I can't recover who it was.  But thank you, whoever you are.
A: Using the formula
$$
\sin(\pi/n) = \frac 12 i e^{-\pi i/n}-\frac 12 i e^{\pi i/n} = \frac 12 (-1)^{1/2} (-1)^{(n-1)/n}-\frac 12 (-1)^{1/2} (-1)^{1/n},
$$
one can write $\sin(\pi/n)$ using radicals for any positive integer $n$.
