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.

Solution 1:

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.

Solution 2:

FWIW, the minimal polynomial of $\sin(\pi/180)$ over the rationals is $$ 281474976710656\,{z}^{48}-3377699720527872\,{z}^{46}+18999560927969280 \,{z}^{44}-66568831992070144\,{z}^{42}+162828875980603392\,{z}^{40}- 295364007592722432\,{z}^{38}+411985976135516160\,{z}^{36}- 452180272956309504\,{z}^{34}+396366279591591936\,{z}^{32}- 280058255978266624\,{z}^{30}+160303703377575936\,{z}^{28}- 74448984852135936\,{z}^{26}+28011510450094080\,{z}^{24}- 8500299631165440\,{z}^{22}+2064791072931840\,{z}^{20}-397107008634880 \,{z}^{18}+59570604933120\,{z}^{16}-6832518856704\,{z}^{14}+ 583456329728\,{z}^{12}-35782471680\,{z}^{10}+1497954816\,{z}^{8}- 39625728\,{z}^{6}+579456\,{z}^{4}-3456\,{z}^{2}+1 $$

EDIT: the explicit expression obtained from Micah's cubic is not all that bad: $$ \frac{-1-\sqrt{3}i}{4} v^{1/3} + \frac{-1+\sqrt{3}i}{4} v^{-1/3} $$ where $$v=-\frac{1}{4}\sqrt {8-\sqrt {3}-\sqrt {15}-\sqrt {10-2\,\sqrt {5}}}+\frac{i}{4} \sqrt {8+\sqrt {3}+\sqrt {15}+\sqrt {10-2\,\sqrt {5}}} $$