Sines, Cosines and Roots of Unity

December 19, 2022 Shreyas Comments 1 comment

1. Introduction

Today’s post will be more elementary in nature, as it is something that I first thought of many years ago but haven’t gotten around to sharing until now. Nevertheless, I think it is good to know and may come off as surprising to those that are not familiar with it. It has to do with the properties of a special set of complex numbers: the roots of unity. We can extract lots of information from these numbers; today’s post will focus on the values of trigonometric functions at certain special inputs.

2. On the Complex Plane

Take the equation $x=1$ . What are the solutions to this equation over the complex numbers?

Ok, that was easy. How about $x^2 = 1$ ?

Still easy? $1$ is an obvious solution, but so is $-1$ , and these are the only two numbers that work. Moving on.

How about $x^3 = 1$ ? $x = 1$ is still a solution, and you may be tempted to say it is the only one. But remember, we are working over the complex numbers, and there are two more hidden solution: $x = \frac{-1 + \sqrt{3}}{2}$ and $x = \frac{-1 - \sqrt{3}}{2}$ . Verify that their cubes are indeed $1$ . But where did these numbers come from? Let us plot these points on the complex plane:

They appear to form an equilateral triangle! Indeed, convince yourself as an exercise that the specified coordinates indeed do form an equilateral triangle. Is this just a coincidence? Lets try the next equation: $x^4 = 1$ . Which complex values of $x$ work? You may have been noticing a pattern: the first equation had one solution, the second had two, the third had three. Will this one have four?

Indeed, it does, and you may be able to find them more easily that the cube case by repeatedly taking square roots. Here they are on the complex plane:

It seems like the numbers $1, i, -1$ and $-i$ form another interesting shape: a square. Will this pattern continue? The answer is a resounding yes, and it has to do with an important property of complex numbers and their multiplication. Any complex number can be represented in polar form, with a magnitude (distance to the origin) and an argument (angle to the positive real axis). When two complex numbers are multiplied, the result has a magnitude equal to the product of the initial magnitudes and an argument equal to the sum of the initial arguments.

We apply this to the general case of $x^n = 1$ , where $n$ is a fixed natural number. Suppose $x$ has a magnitude $|x|$ and an argument $\theta$ . Then, $x^n$ will have a magnitude of $|x|$ multiplied by itself $n$ times, or $|x|^n$ . Since $x^n$ is also equal to $1$ , which has a magnitude of $1$ , we conclude that $|x|^n = 1 \rightarrow |x| = 1$ (note that there is only one solution for $|x|$ here since it is constrained to be a nonnegative real number). Similarly, the argument of $x^n$ will be $\theta$ added to itself $n$ times, or $n\theta$ . This is equal to the argument of $1$ , which is $0$ since it is aligned with the positive real axis. Hence, $n\theta = 0 \rightarrow \theta = 0$ . So is there only one solution?

We have forgotten something important: the argument of $1$ could be $0$ , but it could also be $2\pi,\ 4\pi,\ 6\pi…$ any multiple of $2\pi$ works! Hence, the possible arguments of $x^n$ are actually $\theta = \frac{2\pi*k}{n}$ for any integer $k$ . There seems to be infinitely many such solutions, but the arguments themselves cycle back: adding $n$ to any particular $k$ adds $2\pi$ to the total argument, effectively doing nothing. So there are indeed $n$ distinct roots; we have also shown that they are all a distance of $1$ from the origin and their angles to the $x$ axis increase by $\frac{2\pi}{n}$ for each new root. But this means they are equally spaced about the origin, which is exactly what we need to show that they form a regular $n$ -gon around the origin.

3. tan(36⁰)?

Now we turn to a convenient usage of our understanding of the complex roots of $x^n$ : if we can explicitly list these roots for some particular value of $n$ , then we can obtain lots of information about the values of trigonometric functions on the angle $\frac{2\pi}{n}$ . As a simple example, we take $n=3$ . We know $\frac{-1 + \sqrt{3}}{2}$ is a root, but we also know that it makes a particular angle with the positive x-axis: $\frac{2\pi}{3}$ . Hence, we know the tangent of $\frac{-1 + \sqrt{3}}{2}$ is the ratio of the vertical and horizontal coordinates of this point, specifically $-\sqrt{3}$ . Of course, this angle is just $120^\circ$ , and it probably isn’t that hard to find its tangent. But what about a harder angle?

What is the tangent of $\frac{2\pi}{5}$ ?

To find the trigonometric functions of this angle, we first recognize that it is simply $\frac{2\pi}{5}$ in radians. Hence, we can draw a regular pentagon on the complex plane centered on the origin, as shown in the above diagram. From the previous section, we know that these points are also the complex solutions of $x^5=1$ . Our goal is to learn something about the real and imaginary parts of these points, so we plug $x=a+bi$ into the equation, expand with the binomial theorem and try to solve for $a$ and $b$ .

$\begin{equation*} (a+bi)^5=1 \rightarrow \end{equation*}$

$\begin{equation*} a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i=1 \rightarrow \end{equation*}$

$\begin{equation*} a^5-10a^3b^2+5ab^4+i(5a^4b-10a^2b^3+b^5)=1 \rightarrow \end{equation*}$

$\begin{equation*} a^5-10a^3b^2+5ab^4=1,\ 5a^4b-10a^2b^3+b^5=0 \end{equation*}$

Looking at the second equation, we see that it is actually homogeneous in $a,\ b$ : that is, all terms of the equation have the same combined degree of $5$ in $a$ and $b$ . Hence, if we divide both sides by $b^5$ , we will be able to obtain an equation in a single common variable, $r=\frac{a}{b}$ . However, before dividing, we must investigate the case of $b=0$ . This just corresponds to the case of real solutions, which we are not interested in as we know the only real solution is $x=1$ . Hence, we proceed with the division by $b^5$ :

$\begin{equation*} 5\left(\frac{a}{b}\right)^4-10\left(\frac{a}{b}\right)^2+1=0 \rightarrow \end{equation*}$

$\begin{equation*} 5r^4-10r^2+1=0 \end{equation*}$

Now, we realize we can do even better: substitute $u=r^2$ to make the equation a simple quadratic:

$\begin{equation*} 5u^2-10u+1=0 \rightarrow \end{equation*}$

$\begin{equation*} u=\frac{10 \pm \sqrt{10^2-4*5}}{10}=\frac{5 \pm 2\sqrt{5}}{5} \end{equation*}$

Which root do we choose? One root corresponds to the numbers $\cos(\frac{2\pi}{5}) \pm i\sin(\frac{2\pi}{5})$ , while the other corresponds to $\cos(\frac{4\pi}{5}) \pm i\sin(\frac{4\pi}{5})$ . It is clear from visual inspection that the point at $\frac{2\pi}{5}$ has the lower value of $\frac{a^2}{b^2}$ , so we will use the minus sign. Further, observe that the variable $u=\frac{a^2}{b^2}$ has a trigonometric meaning: it is $\cot^2\left(\frac{2\pi}{5}\right)$ . Therefore,

$\begin{equation*} \cot^2\left(\frac{2\pi}{5}\right)=\frac{5-2\sqrt{5}}{5} \rightarrow \tan^2\left(\frac{2\pi}{5}\right)=\frac{5}{5-2\sqrt{5}}=5+2\sqrt{5} \end{equation*}$

$\begin{equation*} \tan\left(\frac{2\pi}{5}\right)=\sqrt{5+2\sqrt{5}} \end{equation*}$

Thus, we have the answer. Although this is a valid expression, it may not seem as straightforward as quantities like $\tan(\frac{2\pi}{3})$ , since it contains a nested square root inside another square root. However, there is a related trigonometric quantity that happens to have a simpler expression:

$\begin{equation*} \csc^2\left(\frac{2\pi}{5}\right)=1+\cot^2\left(\frac{2\pi}{5}\right)=1+\frac{5-2\sqrt{5}}{5}=\frac{10-2\sqrt{5}}{5} \end{equation*}$

$\begin{equation*} \sin^2\left(\frac{2\pi}{5}\right)=\frac{5}{10-2\sqrt{5}}=\frac{5+\sqrt{5}}{8} \end{equation*}$

$\begin{equation*} \cos\left(\frac{\pi}{5}\right)=-\cos\left(\frac{4\pi}{5}\right)=-\left(1-2\sin^2\left(\frac{2\pi}{5}\right)\right)=2\sin^2\left(\frac{2\pi}{5}\right)-1 \rightarrow \end{equation*}$

$\begin{equation*} \cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{5}}{4} \end{equation*}$

Thus, we see that $\cos\left(\frac{\pi}{5}\right)$ also happens to be half of the golden ratio! This fact can also be demonstrated in simpler ways rooted in synthetic geometry, such as applying Ptolemy’s theorem to part of a regular pentagon.

4. Constructibility

Let us now discuss a very interesting application regarding the exact values of sines/cosines/tangents of angles of the form $\frac{2\pi}{n}$ for integer $n$ : they directly tell us whether or not the regular $n$ -gon can be constructed with ruler and compass! As you may all remember from math class, shapes like the equilateral triangle, square and regular hexagon can be easily constructed with ruler and compass, but what about the regular pentagon? Given a prescribed length of one unit on a piece of paper, can we create a regular pentagon with a side length of one unit?

Observe that constructing a regular pentagon effectively boils down to constructing a length of $\cos\left(\frac{\pi}{5}\right)$ : once this has been accomplished, it is easy to construct an angle of $\frac{\pi}{5}$ using right triangles, and then one can easily generate the pentagon. Here is the important fact about ruler-and-compass constructions: given a length of one unit, it is possible to create lengths of any multiple $k$ of that unit, where $k$ is a real number that can be generated by a finite number of operations of addition/subtraction, multiplication/division and square roots of integers. Now, we see that since $\cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{5}}{4}$ , it fits the bill: it can be generated by applying a square root to the integer $5$ , then adding $1$ , and finally dividing by $4$ . Hence, the regular pentagon is indeed constructible! (Try it out yourself by first generating a length of $\cos\left(\frac{\pi}{5}\right)$ , then proceeding from there. There are more geometrically elegant methods, but this one seems to be the most failsafe method in the general case).

Unlike square roots, however, cube roots of integers don’t fare so well with ruler and compass constructions. Consider the case of a regular heptagon: can we construct a length relating to a trigonometric function of $\frac{2\pi}{7}$ ? It turns out that we can use a very similar method as in the pentagon case to find the following expression for $\cot^2\left(\frac{2\pi}{7}\right)$ (readers are strongly encouraged to try this out by expanding $(a+bi)^7=1$ on their own before looking ahead!)

$\begin{equation*} 7u^{3}-35u^{2}+21u-1=0,\ u=\cot^2\left(\frac{2\pi}{7}\right) \end{equation*}$

The fact that this is a cubic equation should already make us suspicious of the constructibility of the heptagon. A very important fact about this cubic is that it has integer coefficients, meaning that if there are any solutions involving square root radicals, then their radical conjugates must also be solutions. We can now make the following informal argument: it can be quickly checked with the Rational Root Theorem that there are no purely rational solutions, so if the solutions can be expressed using square root radicals, then they must pair up with their conjugates to form an even number of total solutions. But the given equation is a cubic which has three solutions, a contradiction. Hence, the solutions cannot be expressed with square roots, and the case for the heptagon is closed.

The regular heptagon cannot be constructed with ruler and compass 🙁

Due to the work of Gauss, we actually have a full characterization of all $n$ -gons that can be constructed with ruler and compass. On a final departing note, it may come as a great surprise to many that a number like $17$ , which seems at first glance to present nothing but odd deviations and inconveniences, would have the property of constructibility! Indeed, Gauss showed that this comes from the fact that $17=2^{2^2}+1$ is a Fermat prime.

Thanks for sticking through this long post, and see you next time!

Shreyas' blog – Ignited Minds

I love math and science. This is a blog about things I think of, or work on, in my spare time.

Sines, Cosines and Roots of Unity

December 19, 2022 Shreyas Comments 1 comment

1. Introduction

2. On the Complex Plane

3. tan(36⁰)?

4. Constructibility

One thought on “Sines, Cosines and Roots of Unity”

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1. Introduction

2. On the Complex Plane

3. tan(360)?

4. Constructibility

One thought on “Sines, Cosines and Roots of Unity”

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3. tan(36⁰)?