Prerequisites

To understand this post, you’ll need to know $\sin \alpha+\beta = \sin\alpha\cos\beta+\cos\alpha\sin\beta$ and $\cos \alpha+\beta=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$ A good grasp on Algebra is also nice. This might be helpful, but knowing these are not necessary.

Preamble

One morning I was thinking of the following configuration:

Consider $\triangle OA_1A_2$ with $OA_1=3,$ $A_1A_2=4,$ and $OA_2=5.$ Then for all $n,$ construct $\triangle OA_nA_{n+1}$ such that $\triangle OA_nA_{n+1}\sim \triangle OA_1A_2.$

It doesn’t take long to figure out that this is very easy to express with polar coordinates (or complex, whatever flavor you desire). If $O=(0,0)$ and $A_0=(3,0),$ then in polar coordinates, $A_n=(3\cdot (\frac{5}{3})^n,n\theta)$ where $\theta=\arcsin\frac{3}{5}.$

Then the realization strikes; none of the triangles “overlap” each other! In other words, there is no pair of integers $p,q$ such that $O,A_p,A_q$ are collinear. What does this mean? Well, there are no integers $p,q$ such that $p\theta=q\theta+2\pi k$ for some integer $k.$ This basically means that there is no rational number $k$ such that $\theta = 2\pi k.$

The Meat

We claim that $\frac{\arcsin \frac{3}{5}}{\pi}$ is irrational. This is the same as claiming $\frac{\arccos \frac{4}{5}}{\pi}$ is irrational.

Suppose otherwise. Then there must be some positive integer $n$ such that $\cos(n\theta)=0.$ But notice that $\cos (n\theta)$ is a polynomial about $\cos \theta.$ Similarly, $\sin (n\theta)$ is a polynomial about $\sin \theta.$

We claim that $\cos (n+1)\theta = 2\cos \theta \cos n\theta-\cos(n-1)\theta$ for all $n\geq 2.$ Adding $\cos (n-1)\theta$ and expanding out using the sum-difference identities yields $\cos n\theta \cos \theta - \sin n\theta \sin \theta + \cos n\theta \cos \theta + \sin n\theta \sin \theta = 2\cos \theta,$ which is true.

Now we have a recurrence for $\cos n\theta!$ This general recurrence is called the Chebyshev Polynomial of the First Kind. In fact, generally, the Chebyshev Polynomial of the First Kind states the following:

1. $P_0(x)=1.$
2. $P_1(x)=x.$
3. $P_{n+1}(x)=2xP_{n}(x)-P_{n-1}(x).$

Now we have to prove that $\frac{4}{5}$ is not a root of any of these polynomials. What follows is probably the most ingenious yet simplest step of this problem (and the extension, Niven’s Theorem) – an application of the Rational Root Theorem. We claim that the coefficient of the leading term of $P_{n}(x)$ is $2^{n-1}.$ The proof is fortunately very obvious; we can discard the $P_{n-2}(x)$ and notice that the leading coefficient of $P_{n-1}(x)$ is multiplied by $2.$ By the Rational Root Theorem, $\frac{4}{5}$ CANNOT be a root of $\cos n\theta$ for ANY $n,$ because $5$ NEVER divides $2^{n-1}.$ Isn’t it beautiful?

Generalization

Now let’s talk about Niven’s Theorem. It states that the only non-negative values of $k$ such that $\frac{\arcsin k}{\pi}$ is rational are $0, \frac{1}{2}, 1.$ To do this, we need to expand the Chebyshev Polynomial of the First Kind a little bit more. The Rational Root Theorem will be more powerful if we have the leading coefficient and the constant value; let’s try to look for a pattern of the constant value.

$P_0(x)=1$

$P_1(x)=x$

$P_2(x)=2x^2-1$

$P_3(x)=4x^3-3x$

$P_4(x)=8x^4-8x^2+1$

$P_5(x)=16x^5-20x^3+5x$

$P_6(x)=32x^6-48x^4+18x^2-1$

Actually, let’s make it a bit more obvious.

$P_0(x)=\dots+1$

$P_1(x)=\dots+0$

$P_2(x)=\dots-1$

$P_3(x)=\dots+0$

$P_4(x)=\dots+1$

$P_5(x)=\dots-0$

$P_6(x)=\dots-1$

$P_7(x)=\dots+0$

$P_8(x)=\dots+1$

$P_9(x)=\dots+0$

$P_{10}(x)=\dots-1$

Again, this is very easy to prove using induction. So what does this mean? By the Rational Root Theorem, the only possible values of $k$ are either $0$ or of the form $\frac{1}{2^m}$ for even $n$ of $P_n(x),$ and for odd $n$ we can also have things of the form $\frac{n}{2^m}.$ (Remember $k$ is non-negative; negative values are symmetric anyways.) So now all that remains is to prove that these are not roots.

To utilize modular arithmetic, we’ll divide by $\frac{1}{x^n}$ for $P_n(x).$ If we let $y = \frac{1}{x},$ we are basically “reversing” the polynomial. Let’s rewrite the polynomials in hope we see a pattern:

$P_0(x)=x^0(1)$

$P_1(x)=x^1(1)$

$P_2(x)=x^2(2-y^2)$

$P_3(x)=x^3(4-3y^2)$

$P_4(x)=x^4(8-8y^2+y^4)$

$P_5(x)=x^5(16-20y^2+5y^4)$

$P_6(x)=x^6(32-48y^2+18y^4-y^6)$

Let $Q_{n}(x)=\frac{P_{n}(x)}{x^n}.$ Notice the recursion is $Q_{n+1}(x)=2Q_{n}(x)-y^2Q_{n-1}(x).$ We’ll write the first few $Q_{n}x.$

$Q_0(x)=1$

$Q_1(x)=1$

$Q_2(x)=-y^2+2$

$Q_3(x)=-3y^2+4$

$Q_4(x)=y^4-8y^2+8$

$Q_5(x)=5y^4-20y^2+16$

$Q_6(x)=-y^6+18y^4-48y^2+32$

The only possible roots are $2^0,2^1\dots 2^{n-1},2^n.$

Notice that for odd $n,$ the leading coefficient is $\pm n.$ This then causes an obvious problem $\pmod n,$ so it’s proven for odd $n.$ (It is obvious that this is impossible if $x=\frac{n}{2^m},$ or $y=\frac{2^m}{n},$ because we need everything to be an integer if we are to get rid of the constant.)

For even $n,$ notice that the constant term is of the form $2^{k+1}$ while the terms with a coefficient of $y$ are of the form $(2^{k})^{2j}),$ so it won’t work out either. Thus we have proven this is impossible for all rational numbers except for our three exceptions. (For $y=2,$ $Q_6(x)$ is an exception.)

And that’s Niven’s Theorem!

References

Stack Exchange: Is arccos 3/5 rational?

Stack Exchange: 1/pi arccos (1/3) is irrational.

Wikipedia: Niven’s Theorem

Wikipedia: Chebyshev Polynomials

Wolfram Mathworld: Chebyshev Polynomials of the First Kind

Brilliant: Chebyshev Polynomials