Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations? First, let us recall some trigonometric values
$\sin(0)=0$
$\sin(\pi/6)=\frac{1}{2}$
$\sin(\pi/3)=\frac{\sqrt{3}}{2}$
$\sin(\pi/2)=1$
$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$
$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Here, we can observe that for some values of $\theta$, $\sin(\theta)$ can be expressed as a sum of finite radical terms. It is also easy to see that, there exists infinite such $\theta$'s. Say, I have given $\sin(\pi/14)$, can we determine whether it can also be expressed as sum of finite radical terms? What about the general case? Are there any criteria that $\theta$ must obey in order to do that?
 A: It is not so bad !
As @Doug M commented, it is one of the roots of
$$8x^3 - 4x^2 - 4x + 1 = 0$$
Using the trigonometric method for three real roots,we have
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6} \left(1+2 \sqrt{7} \sin \left(\frac{1}{3} \csc ^{-1}\left(2
   \sqrt{7}\right)\right)\right)$$
Otherwise, using radicals
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6}-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{6\ 2^{2/3} \sqrt[3]{-1+3 i
   \sqrt{3}}}-\frac{1}{12} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3
   i \sqrt{3}\right)}$$
The second and third terms are two of the roots of
$$46656 x^6-1512 x^3+343=0$$ which is a quadratic in $x^3$.
The solutions of
$$46656 X^2-1512 X+343=0$$ are
$$X_{1,2}=\frac{7}{432} \left(1\pm3 i \sqrt{3}\right)$$
A: $8x^3−4x^2−4x+1=0$
Where does this polynomial come from?
Suppose you take a regular polynomial and plot the coordinates.
Consider a n-gon, with one vertex on $(1,0)$ all the vertices one unit from the center, and the center at $(0,0).$
$(1,0)\\
(\cos \frac {2\pi}{n},\sin \frac {2\pi}{n})\\
(\cos \frac {4\pi}{n},\sin \frac {4\pi}{n})\\
\cdots\\
(\cos 2(n-1)\pi,\sin 2(n-1)\pi)$
If we average, or sum, all of these coordinates we get the center of the circle.
$1 + \cos\frac {2\pi}{n} + \cdots + \cos \frac {2(n-1)\pi}{n} = 0$
and similarly
$1 + \sin\frac {2\pi}{n} + \cdots + \sin {2(n-1)\pi}{n} = 0$
We could rotate the polygon to make one vertex associate with the angle $\frac {\pi}{14},$ but that isn't what I did.  Instead, I kept the same orientation and said:
$\sin \frac {\pi}{14} = \cos (\frac {\pi}{2}-\frac {\pi}{14}) = -\cos (\frac {\pi}{2}+\frac {\pi}{14}) = -\cos \frac {4\pi}{7}$
Working with:
$1 + \cos\frac {2\pi}{7} + \cos\frac {4\pi}{7} + \cdots + \cos \frac {12\pi}{7} = 0$
This polygon has symmetry about the x-axis.
$\cos \frac {2\pi}{7} = \cos \frac {12\pi}{7}\\
\cos \frac {4\pi}{7} = \cos \frac {10\pi}{7}\\
\cos \frac {6\pi}{7} = \cos \frac {8\pi}{7}$
$1 + 2\cos\frac {2\pi}{7} + 2\cos\frac {4\pi}{7} + 2\cos\frac {6\pi}{7}=0$
Now we use multiple-angle identities.
$1 + 2\cos\frac {2\pi}{7} + 2(2\cos^2\frac {2\pi}{7}-1) + 2(4\cos^3\frac {2\pi}{7} - 3\cos\frac {2\pi}{7})$
Say $\cos\frac {2\pi}{7} = x$
$8x^3 +4x^2 - 4x - 1 = 0$
This polynomial has 3 roots.  One equals $\cos\frac {2\pi}{7}$, another equals $\cos\frac {4\pi}{7}$ and the third equals $\cos\frac {6\pi}{7}$
But we wanted the negative of one of those roots.  Then replace $x$ with $-x$ in the equation above, remembering that $(-x)^n = x^n$ when $n$ is even and $-x^n$ when $n$ is odd.
$-8x^3 +4x^2 + 4x - 1 = 0\\
8x^3 -4x^2 - 4x + 1 = 0$
$\sin \frac {\pi}{14}$ is a root of $8x^3 -4x^2 - 4x + 1 = 0$
