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This is what I tried doing. Yes, $10$ as in $10$ degrees.

I used the formula $\sin{3x} = 3\sin{x} - 4\sin^3{x}$ to get the cubic equation $x^3-\frac{3}{4}x+\frac{1}{8}=0$, where $x$ is $\sin{10}$. When I tried using the cubic formula on it, it prompted the calculation of $\sqrt[3]{\frac{\sqrt{-3}-1}{2}}$. Putting that into WA tells me that this is equal to $e^{\frac{2}{9}i\pi}$. This leads to me having to calculate $(\sqrt[9]{i})^4$. To calculate the ninth root of $i$, I used the following technique.

$\sqrt[9]{i} = a+b i$
$0 + 1i = (a+b i)^9$
$0 + 1i = a^9-36a^7b^2+126a^5b^4-84a^3b^6+9ab^8+i(9a^8b-84a^6b^3+126a^4b^5-36a^2b^7+b^9)$

This gives me the simultaneous equations:
$a^9-36a^7b^2+126a^5b^4-84a^3b^6+9ab^8 = 0$ and $9a^8b-84a^6b^3+126a^4b^5-36a^2b^7+b^9=1$

Again, I tried using WA to solve this, but that didn't lead anywhere, since the correct answer had $i$ in another more complicated nested cube root.

How do I solve this problem?

avighnac
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  • Is $10$ meant as degrees ? 2. I think that an expression exactly calculating $\sin(10°)$ is ugly.
    • – Peter Dec 23 '22 at 11:09
  • Observe that $\sin 10^\circ = \text{Im}(e^{\frac{2\pi i}{36}})$. This suggests to me that your intermediate answer $e^{\frac{2}{9}i\pi}$ is somehow incorrect. – Charles Hudgins Dec 23 '22 at 11:11
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    Why do you want this? Closed forms for cubics are generally bad. WA can give you the radical form if you really want it. The one you want is the middle one. – lulu Dec 23 '22 at 11:11
  • @lulu Most questions in maths don't have an answer to this question except "just for fun". That's my answer too. – avighnac Dec 23 '22 at 11:12
  • For those who do not know what WA is : It stands for "Wolfram Alpha" and is a quite powerful mathematical online tool. – Peter Dec 23 '22 at 11:14
  • @CharlesHudgins the cube root is just one part of it, not the entire answer – avighnac Dec 23 '22 at 11:14
  • Maybe this satisfies you. The second expression is what you want. Click on "exact form" for the expression. – Peter Dec 23 '22 at 11:18
  • @Peter WA destroys quite a bit of beautiful symmetry in those exact forms when it insists for some inexplicable reason on having one of the radicals in a denominator. – Arthur Dec 23 '22 at 11:25
  • @avighnac I see. In any case you need to be careful asking WA to calculate the result. That expression has up to 6 values (2 for the square root, 3 for the cube root). Do you know which value you need? – Charles Hudgins Dec 23 '22 at 12:20
  • @Peter that expression has imaginary numbers even though the final answer has none; I have an expression with imaginary numbers simpler than that, so that doesn’t satisfy me. – avighnac Dec 23 '22 at 12:23
  • Then , google "casus irreducibilis" to see that this is impossible here , at least with algebraic methods. – Peter Dec 23 '22 at 12:29
  • Even if you use the cubic formula, you will cycle through cube roots of other complex numbers. – Тyma Gaidash Dec 23 '22 at 13:03
  • @Peter, sorry if this makes me sound a little foolish, but this just seems incredibly hard to comprehend, surely it can’t be impossible; if one can take the roots of unity then why can one not express a perfectly real number with no imaginary part without complex numbers?! – avighnac Dec 23 '22 at 13:16
  • It is a proven result that with this approach you cannot avoid non-real numbers here , there is also a geometrical approach, not sure whether this establishes what you want. – Peter Dec 23 '22 at 14:15
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    @svighnac Congratulations, you have discovered one of the biggest headaches of renaissance mathematicians. You need complex numbers. – Arthur Dec 23 '22 at 18:56
  • We can take $\sin$ of $10=15-15/2+15/4-15/8+15/16-...$ – Bob Dobbs Dec 23 '22 at 21:18