# An angle $\theta$ can be trisected if and only if $4x^3-3x+\cos\theta$ is reducible over $\mathbb{Q}(\cos\theta)$

I was given the above problem for homework. There is (what seems to be) a relevant proof in my textbook regarding the impossibility of trisecting $\pi/3$. In this proof, the identity

$$\cos 3\theta = 4\cos^3 \theta -3\cos\theta$$

is used. Rearranging, we get $0 = 4\cos^3 \theta-3\cos\theta-\cos 3\theta$. I know if the given equation were $4x^3-3x-\cos\theta$, my homework problem would be relatively easy. At this point, however, I'm not sure where to go. A push in the right direction would be very appreciated.

Edit: The question isn't actually out of the textbook (Galois Theory by Stewart). It's on a worksheet my teacher typed up, which makes me think it might be a typo as well. In fact, the textbook asks the analogous question for $4x^3-3x-\cos\theta$.

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Perhaps you could tell us the name of the textbook? (I guess it might be a typo.) –  Martin Sleziak Oct 10 '12 at 15:40
Did you look: en.wikipedia.org/wiki/Angle_trisection ? –  Giovanni De Gaetano Oct 10 '12 at 15:43
If $\eta$ trisects $\theta$, i.e. $3\eta=\theta$, and $x=\cos\eta$, then $4x^2-3x-\cos\theta$ $=4\cos^3\eta-3\cos\eta-\cos\theta$ $=\cos(3\eta)-\cos\theta=0$. Is it possible that "$-$" should be where "$+$" appears in your question? –  Michael Hardy Oct 10 '12 at 15:55

An angle $\theta$ can be trisected if and only if $4x^3-3x-\cos \theta$ is reducible over $\mathbb Q(\cos \theta)$. It can be proved easily in following manner.
Assume $\phi=\frac{\theta}3$. $\phi$ can be constructed from $\theta$ iff $\cos \phi$ can be constructed from $\cos \theta$. Now use $$\cos \theta = \cos 3\phi=4 \cos^3 \phi-3\cos \phi$$ Hence $\cos \phi$ is a root of $f(x)=4x^3-3x-\cos \theta$.Now if $f(x)$ is reducible over $\mathbb Q(\cos \theta)$,the $\cos \phi$ is a root of polynomial of degree $1$ or $2$ over $\mathbb Q(\cos \theta)$, thus $[\mathbb Q(\cos \theta,\cos \phi):\mathbb Q(\cos \theta)]=1 \ or\ 2$. Hence $\cos \phi$ is constructible from $\mathbb Q(\cos \theta)$.
If $f(x)$ is irreducible over $\mathbb Q(\cos \theta)$, then $[\mathbb Q(\cos \theta,\cos \phi):\mathbb Q(\cos \theta)]=3$ and thus $\cos \phi$ cannot be contructed from $\mathbb Q(\cos \theta)$.
EDIT: As pointed out by Andre Nicholas, $\phi$ is constructible iff $-\phi$ is. Also $4x^3-3x-a$ is irreducible over a field $F$ iff $4x^3-3x+a$ as can be seen by putting change of variable $t=-x$.
Sign makes no difference. Angle $\phi$ is constructible iff $-\phi$ is. And $4x^3-3x+a$ is irreducible over field $F$ iff $4x^3-3x-a$ is, just make the change of variable $t=-x$. –  André Nicolas Oct 10 '12 at 16:35