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  "body": "\n## Proof\n\nLet $\\theta = \\pi/9$, let $c = \\cos\\theta$ and $s = \\sin\\theta$, and let $\\mathcal{C}$ be the set of the [[constructible numbers|0FMN2MN7TM]].\n\n1. By [[de Moivre's formula|03MTBVG3VE]], $\\cos 3\\theta + i \\sin 3\\theta = (c + i s)^3 = c^3 + 3 i c^2 s - 3 c s^2 - i s^3$.\n2. Comparing real parts, and using [[$s^2 = 1 - c^2$|03H7WZ4T3K]], we get $\\cos 3\\theta = c^3 - 3c(1 - c^2) = 4c^3 - 3c$.\n3. Since $\\cos 3\\theta = \\cos(\\pi/3) = 1/2$, we get $8c^3 - 6c - 1 = 0$. So $u = 2c$ is a root of $g = x^3 - 3x - 1$.\n4. By the [[rational root theorem|0819CQ1V8H]], a rational root of $g$ is $\\pm 1$. But $g(1) = -3$ and $g(-1) = 1$. So $g$ has no rational root.\n5. A cubic with no root in $\\Q$ is [[irreducible over $\\Q$|0VNE161QCK]]. So the [[minimal polynomial|0GK1WB9R98]] of $u$, which [[divides|06K7SARZKR]] the monic $g$, equals $g$.\n6. By the [[degree of a simple extension|0N5QW9K7MH]], $[\\Q(u) : \\Q] = 3$. Since $c = u/2$ and $u = 2c$, we have $\\Q(c) = \\Q(u)$, so $[\\Q(c) : \\Q] = 3$.\n7. The degree of a constructible number is a [[power of $2$|0SSSM1D475]], and $3$ is not. So $\\cos(\\pi/9) \\notin \\mathcal{C}$.\n\nNow we turn to the construction problem. The angle $\\pi/3$ has vertex $0$ and sides through $1$ and $\\omega = e^{i\\pi/3}$. A [[trisection|00J3DCCEK5]] would construct a point $w \\neq 0$ on the ray from $0$ at angle $\\pi/9$.\n\n8. By the [[field theorem|0PN9WMRWT7]], $\\mathcal{C}$ is a subfield of $\\C$ closed under square roots, and it is the smallest such subfield.\n9. By [[Euler's formula|04R3CY41VJ]], $\\omega = 1/2 + i\\sqrt{3}/2$. Here $i$ is a square root of $-1$ and $\\sqrt 3$ a square root of $3$, so $\\omega \\in \\mathcal{C}$. So the angle $\\pi/3$ can be drawn.\n10. Let $\\overline{\\mathcal{C}}$ be the set of the [[conjugates|0BJV24C4SY]] of the elements of $\\mathcal{C}$. Conjugation preserves the field operations, so $\\overline{\\mathcal{C}}$ is a subfield of $\\C$.\n11. If $w' = \\bar z$ with $z \\in \\mathcal{C}$ and $t^2 = w'$, then $\\bar t^2 = z$, so $\\bar t \\in \\mathcal{C}$ and $t \\in \\overline{\\mathcal{C}}$. So $\\overline{\\mathcal{C}}$ is closed under square roots.\n12. By minimality, $\\mathcal{C} \\subseteq \\overline{\\mathcal{C}}$. So each $z \\in \\mathcal{C}$ is $\\bar u$ for some $u \\in \\mathcal{C}$, and then $\\bar z = u \\in \\mathcal{C}$.\n13. Suppose that the point $w$ is constructed. The given points $0$, $1$ and $\\omega$ lie in $\\mathcal{C}$, so $w \\in \\mathcal{C}$. Then $\\bar w \\in \\mathcal{C}$, so [[$w \\bar w = \\abs{w}^2$|0ZB7J071WQ]] is in $\\mathcal{C}$. Its square root, the [[modulus|0KW2X0TFBX]] $\\abs{w}$, is in $\\mathcal{C}$.\n14. Since $w$ lies on the ray at angle $\\theta$, we have $w = \\abs{w} e^{i\\theta}$, so $e^{i\\theta} = w / \\abs{w} \\in \\mathcal{C}$.\n15. Then $\\cos\\theta = (e^{i\\theta} + \\overline{e^{i\\theta}})/2 \\in \\mathcal{C}$, by Euler's formula and step 12. This contradicts step 7.\n16. So the angle $\\pi/3$ cannot be trisected with straightedge and compass, and $\\cos(\\pi/9)$ is not constructible.\n\n## Idea\n\nThe triple angle formula makes $2\\cos(\\pi/9)$ a root of the irreducible cubic $x^3 - 3x - 1$. So its degree over $\\Q$ is $3$, which is not a power of $2$. A trisection would produce the point $e^{i\\pi/9}$, and then its real part would be constructible.\n",
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