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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?

Sumanta
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Mathematical Curiosity
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    Have you ever heard of the notion of Galois theory? It answers precisely the question of 'when is a solution to a polynomial equation expressible using radicals?'. – Steven Stadnicki Dec 18 '20 at 06:35
  • @Steven Stadnicki. I am in high school. I didn't hear about Galois theory. Is it too much for a high schooler to understand? – Mathematical Curiosity Dec 18 '20 at 06:55
  • It would be one of the roots of the cubic equation. $8x^3 - 4x^2 - 4x + 1 = 0$ Which means that it can be represented in terms of radicals, but it is not nice. – Doug M Dec 18 '20 at 07:14
  • I suspect that if $x$ is a rational multiple of $\pi$ then $sinx$ can be expressed in terms of sum of finite radicals – Infinity_hunter Dec 18 '20 at 08:29
  • @Doug M. How did you construct that equation? – Mathematical Curiosity Dec 18 '20 at 14:16
  • Incidentally, you may be interested to know that the solvability in terms of roots of a cubic here means that the regular heptagon can be constructed via origami! See, for instance, http://origametry.net/papers/heptagon.pdf . Also, I admire your curiosity for a high schooler! Galois theory may be too much for you to understand right now, but keep an ear out. – Steven Stadnicki Dec 18 '20 at 22:16
  • @Steven Stadnicki, Thank you very much. – Mathematical Curiosity Dec 21 '20 at 17:43

2 Answers2

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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 $$X_{1,2}=\frac{7}{432} \left(1\pm3 i \sqrt{3}\right)$$

Claude Leibovici
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$8x^3−4x^2−4x+1=0$

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 $8x^3 -4x^2 - 4x + 1 = 0$

Doug M
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  • Once you write $\sin(\pi/14) = -\cos(4\pi/7)$, then you can also proceed by noting that $\theta = \frac{4\pi}{7}$ is a solution to $\cos(4\theta) = \cos(3\theta)$. But $\cos(4\theta) - \cos(3\theta)$ can be expressed as a quartic polynomial in $x := \cos \theta$ using Chebyshev polynomials; and this quartic polynomial is also divisible by $x-1$ since $\theta = 0$ is another solution. So, $\cos(4\pi/7)$ is a root of the polynomial $\frac{P_4(x) - P_3(x)}{x-1}$. – Daniel Schepler Dec 18 '20 at 21:07