Trigonometric functions and rational multiples of pi

Recall that a real number is algebraic if it is the root of a polynomial with integer coefficients and that it is transcendental otherwise. For example ${\sqrt{3}/3}$ is algebraic because it is a root of the polynomial ${f(x)=9x^{2}-3}$ , but ${\pi}$ is transcendental because it is not the root of any such equation. (On a recent blog post I proved that ${e}$ is a transcendental number.)

Today I would like to prove that a certain large class of numbers is algebraic.

We know that ${\cos(\pi/3)=1/2}$ , ${\sin(\pi/4)=\sqrt{2}/2}$ , and ${\tan(\pi/6)=\sqrt{3}/3}$ are algebraic numbers. It may not be surprising, then, that when a rational multiple of ${\pi}$ is the argument of a trigonometric function we obtain an algebraic number.

Theorem. If ${\theta}$ is a rational multiple of ${\pi}$ , then ${\cos\theta}$ , ${\sin\theta}$ , ${\tan\theta}$ , ${\sec\theta}$ , ${\csc\theta}$ , and ${\cot\theta}$ are algebraic numbers (if they are defined).

It turns out that there is a nice proof of this fact that uses two of the most celebrated results in complex analysis: Euler’s identity

${e^{\pi i}=-1}$

and DeMoivre’s formula

${(\cos\theta+i\sin\theta)^{n}=\cos(n\theta)+i\sin(n\theta).}$

Let ${\theta}$ be a rational multiple of ${\pi}$ . For simplicity of calculations we will write ${\theta}$ as a rational multiple of ${2\pi}$ :

${\displaystyle\theta=\frac{2\pi m}{n}.}$

We use Euler’s identity and DeMoivre’s formula to obtain this string of equalities.

${(\cos\theta+i\sin\theta)^{n}=\cos(n\theta)+i\sin(n\theta)}$

${=e^{in\theta}}$

${=e^{in\frac{2m\pi}{n}}}$

${=e^{2m\pi i}}$

${=(e^{\pi i})^{2m}}$

${=(-1)^{2m}}$

${=1}$

(Note: this argument shows that ${\cos\theta+i\sin\theta}$ is an ${n}$ th root of unity in ${\mathbb{C}}$ ; that is, ${\cos\theta+i\sin\theta}$ is a root of the polynomial ${f(z)=z^{n}-1}$ . This shows that ${\cos\theta+i\sin\theta}$ is an algebraic complex number. We could simply use the theorem that if ${a+bi}$ is an algebraic complex number, then ${a}$ and ${b}$ are algebraic real numbers—a result that is not difficult to prove—to obtain a quick proof that ${\cos\theta}$ and ${\sin\theta}$ are algebraic. But the following trigonometric argument is too nice to skip. Plus the proof constructs the polynomials which have our numbers as roots.)

The idea of the proof is to multiply out ${(\cos\theta+i\sin\theta)^{n}}$ , set the real part equal to 1 and the imaginary part equal to 0. Then apply some trigonometric identities to obtain the polynomial relations.

We will illustrate with an example, but the proof of the general case is identical. Consider ${\theta=2\pi /5}$ . Using the relationship from above we have

${1=(\cos\theta+i\sin\theta)^{5}}$

${=(\cos^{5} \theta -10 \cos^{3} \theta \sin^{2} \theta+5\cos \theta \sin^{4} \theta)}$

${+i(5\cos^{4} \theta\sin \theta -10\cos^{2} \theta \sin^{ 3}\theta+\sin^{5} \theta)}$

Setting the real parts equal we obtain

${\cos^{5} \theta -10 \cos^{3} \theta \sin^{2} \theta+5\cos \theta \sin^{4} \theta=1.}$

Notice that all of the exponents of ${\sin\theta}$ are even (this will always happen because ${i^{k}\sin^{k}\theta}$ is real if and only if ${k}$ is even). We know that ${\sin^{2}\theta+\cos^{2}\theta=1}$ , so we may replace all instances of ${\sin^{2}\theta}$ with ${1-\cos^{2}\theta}$ to obtain

${16\cos^{5}\theta-20\cos^{3}\theta+5\cos\theta=1.}$

In other words, ${\cos(2\pi /5)}$ is a root of the polynomial

${f(x)=16x^{5}-20x^{3}+5x-1.}$

Thus ${\cos(2\pi/5)}$ is algebraic. This identical argument works for the cosine of any rational multiple of ${\pi}$ .

Now consider the imaginary part of the equation. Setting both sides equal we obtain

${5\cos^{4} \theta\sin \theta -10\cos^{2} \theta \sin^{ 3}\theta+\sin^{5} \theta=0.}$

Notice that every term is the product of five trigonometric functions (that is, the sum of the exponents of sines and cosines is 5). If we divide through by ${\cos^{5}\theta}$ we obtain the following expression with tangents

${5\tan\theta-10\tan^{3}\theta+\tan^{5}\theta=0.}$

Thus ${\tan(2\pi /5)}$ is a root of the polynomial

${g(x)=x^{5}-10x^{3}+5x,}$

and we conclude that ${\tan(2\pi/5)}$ is algebraic. Again, this same trick (dividing the imaginary part by ${\cos^{n}\theta}$ ) works for the tangent of any rational multiple of ${\pi}$ .

What about ${\sin(2\pi/5)}$ ? Here we use the identity ${\sin\theta=\cos(\theta-\pi/2)}$ . So

${\displaystyle\sin(\frac{2\pi}{5})=\cos(\frac{2\pi}{5}-\frac{\pi}{2})=\cos(-\frac{\pi}{10}),}$

which we have shown is algebraic.

Finally, since the set of algebraic numbers is a field, we know that ${\sec\theta=1/\cos\theta}$ , ${\csc\theta=1/\sin\theta}$ , and ${\cot\theta=1/\tan\theta}$ are algebraic. (We could also have used this field property to show that ${\sin\theta}$ is algebraic, since ${\sin\theta=\tan\theta\cos\theta}$ .)

5 Comments

Rick Meese says:

October 30, 2010 at 9:59 am

Just a basic question. Why do you subtract the 10x part of the expansion rather than add?
1. Dave Richeson says:
  
  November 1, 2010 at 9:45 am
  
  Are you referring to 10’s in the expansion of $(\cos x+i\sin x)^5$ ? It is because one of them will have an $i^2$ in front, which is -1 and the other has $i^3$ which is $-i$ .
  1. Rick Meese says:
    
    November 5, 2010 at 8:53 pm
    
    Thanks, I get it.
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