# 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.

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Related question. I gave a radical expression for $\sin 1^\circ$ in my answer there. – J. M. Apr 9 '13 at 3:10

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.
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In my understanding only two are complex conjugates. – kiss my armpit Nov 2 '12 at 15:34

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 $$v=-\frac{1}{4}\sqrt {8-\sqrt {3}-\sqrt {15}-\sqrt {10-2\,\sqrt {5}}}+\frac{i}{4} \sqrt {8+\sqrt {3}+\sqrt {15}+\sqrt {10-2\,\sqrt {5}}}$$

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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 $\sin(1^\circ) = -\frac{1}{2} (-1)^{1/2}((-1)^{1/180}+(-1)^{179/180})$

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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 $(3)$.

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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.

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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

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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.

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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 $n$.

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How do you evaluate $(-1)^{1/n}$ as $a+bi$, where $a$ and $b$ are real and are written using radicals? – Michael Hardy Nov 4 '12 at 2:01