Degree of minimal polynomial for $\sin (\frac {2 \pi} 7)$ So I was playing around with trying to prove the regular 7-gon is not constructable under qualifier-exam conditions, so I didn't have a book open.  I got it down to having (If I didn't make any basic math errors, I was doing this on pen and paper while my class was taking a calculus test), that $\sin (\frac {2\pi} 7)$ would by a root of $f(x)=64x^6 -80x^4+24x^2 -1$  I then noticed that by setting $c=(2x)^2 $, this gave me a polynomial $g(c)=c^3 -5c^2 +6c-1$,  which is irreducable over the rationals by the rational root test and being a cubic.
So, the question is...does this make $f$ irreducible, so my algebraic degree is 6?  Or does since $g$ is irreducible and I'd have to extract a square root to get back to $x$,  does that then give me a field extension of degree 6?  In any case, obviously my goal is to prove it's algebraic of a degree that's not a power of 2. 
 A: As a general rule, $f(x)$ being irreducible does not mean $f(x^2)$ is irreducible.
For a simple example, $f(x)=x-1$. 
Another example: $f(x)=x^2+x+1$ then $f(x^2)=f(x)f(-x)$.
A: $\sin \left( \frac{2 \pi}{7}\right)$ is a zero of $f(x)=64x^6-112x^5+56x^3-7$. By Eisenstein's criterion with $p=7$, it is clear that this polynomial is irreducible.
A: Assuming everything you've done up to this point is correct, then based on your work, we know the roots of $f$ will be of the form $\displaystyle \frac{ \pm \sqrt{\alpha}}{2}$, where $\alpha$ is any of the three roots of that cubic.  
Now notice that $\displaystyle \frac{\pm \sqrt{\alpha}}{2} \in \mathbb{Q}[\sqrt{\alpha}]$, and we have a tower of fields $\mathbb{Q} \subset \mathbb{Q}[\alpha] \subset \mathbb{Q}[\sqrt{\alpha}]$.  Now consider the degrees of these extensions to finish up.

 Given the tower of fields, we know the degree of $\mathbb{Q}[\sqrt{\alpha}]$ over $\mathbb{Q}$ is equal to the product of $[\mathbb{Q}[\alpha]:\mathbb{Q}]=3$ and $[\mathbb{Q}[\sqrt{\alpha}]:\mathbb{Q}[\alpha]]$.  It follows that the degree of $\mathbb{Q}[\sqrt{\alpha}]$ over $\mathbb{Q}$ is divisible by $3$ and thus is not a power of $2$.  

