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Could I just do this?

Proof: if we want to show $\arccos\left(\frac{-12}{15}\right)$ is constructible, can't I just say, take $x_0=\cos(\theta)=-\frac{12}{17}$ implies $17x_0+12=0$ which says that $f(x_0)=17x+0+12$ is a polynomial with root $x_0$. But $f(x)$ is irreducible by Eisenstein's criterion with $p=3$. So $f(x)$ is the minimal polynomial with $x_0$ as a root, thus $\left[\mathbb{Q}[x]:\mathbb{Q}\right]=1$ which is equal to $2^m$ for $m=0$.

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  • $\begingroup$ but couldn't we refine the argument to say that if, theta is constructible, then cos(theta) is constructible? $\endgroup$ – Mr.Fry Jul 21 '13 at 20:25
  • $\begingroup$ A brain fart, sorry! I was "certain" that the question was about constructibility of a regular 17-gon :_) $\endgroup$ – Jyrki Lahtonen Jul 21 '13 at 20:25
  • $\begingroup$ haha, it's cool. $\endgroup$ – Mr.Fry Jul 21 '13 at 20:26
  • $\begingroup$ Take $\theta=1$ for a contradiction, since $\cos 1$ is transcendental. $\endgroup$ – pre-kidney Jul 21 '13 at 20:31
  • $\begingroup$ You have to be very careful what you mean by 'constructible' here, because it means subtly different things in different contexts. If you mean the angle which is one angle of a right triangle with side 12 and hypotenuse 17 (which the statement using $\theta$ certainly suggests), then as Andre's answer gives this is quite straightforward - but the linear quantity that corresponds to arccos(-12/17) is certainly not constructible; it's not even algebraic. $\endgroup$ – Steven Stadnicki Jul 21 '13 at 20:51
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There is a more concrete way: using Eisenstein to prove the irreducibility of a polynomial of degree $1$ seems like overkill.

We can explicitly construct a right triangle with one leg equal to $12$ and hypotenuse $17$. So we have constructed an angle whose cosine is $12/17$. Now it is easy to produce $\arccos(-12/17)$.

Remark: For any rational $r$ with $-1\le r\le 1$, the angle $\arccos(r)$ is compass and straightedge constructible.

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  • $\begingroup$ 5,12,17 is not a triple. $\endgroup$ – Mr.Fry Jul 21 '13 at 20:33
  • $\begingroup$ Oops! Indeed it isn't! But the triangle is still explicitly constructible. $\endgroup$ – André Nicolas Jul 21 '13 at 20:39
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    $\begingroup$ @Crypto Let $AB=12$, you can construct a perpendicular $(d)$ to $(AB)$ at $B$ and a circle $(\Gamma)$ of radius $17$ centered at $A$. Then $C\in\Gamma\cap(d)$ so that $ABC$ is a right triangle $\endgroup$ – Paracosmiste Jul 21 '13 at 20:44
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    $\begingroup$ Regarding the remark at the bottom of your "answer": take $r=1$ when $\cos^{-1}1=\frac{\pi}{4}$ is transcendental, hence not algebraic, hence not constructible. Furthermore, there is a dense set of counterexamples, so the statement is rarely true. $\endgroup$ – pre-kidney Jul 21 '13 at 21:05
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    $\begingroup$ @pre-kidney There's a difference between a constructible number (i.e. the length of a straight line segment) and a constructible angle. $\endgroup$ – Paracosmiste Jul 21 '13 at 21:15
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The way I would do it is with a triangle whose three sides are $51$, $92$, and $133$. Then Law of Cosines.

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  • $\begingroup$ thanks for your comment. The above proof is correct though right? $\endgroup$ – Mr.Fry Jul 21 '13 at 22:24

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