## Irrationality of basic trigonometric and hyperbolic functions at rational values

A few days ago, while rummaging through my shelves, I stumbled upon a book by the number theorist Ivan Niven entitled “Irrational Numbers”-a relatively unknown book in my opinion. Obscured by stationery and other paraphernalia, the edge of the book peeked out of my drawer-its cover shrouded in dust and its title almost indiscernible if not for the striking glow of morning sunlight from my large window revealing its dark red tint. I wiped off the dust and peeked into its contents.

This was one of the first ‘real’ math textbooks that I ever read in high school along with Apostol’s books on Calculus and Lang’s Linear Algebra. I distinctly remember reading it for the first time when I was 13. It was undoubtedly quite difficult at the time, considering that I had little exposure to proof-based mathematics. Though the book is rather drab in its exposition, it undoubtedly had some interesting and simple mathematical facts about irrationality which you wouldn’t naturally find anywhere else except for some old number theory papers.

One of those little mathematical gems is the irrationality of trigonometric and hyperbolic functions at rational arguments. I shall soon prove that $cos(x)$ is irrational if $x \in \mathbb{Q} \backslash \{ 0 \}$.

### Lemma 1

If $h \in \mathbb{Z}[x]$ and $f(x)=\frac{x^{n}h(x)}{n!}$ is a polynomial in $\mathbb{Q}[x]$, then $f^{(k)}(0) \in \mathbb{Z}$ for$k \in \mathbb{Z^{+}}$ and $f^{(k)}(0)$ is divisible by $n+1$ if $k \neq n$. However, it is divisible by $n+1$ at $k=n$ if $h(0)=0$.

I’ll leave the proof to the reader since its quite simple and involves nothing more than playing around with the polynomials.

### Lemma 2

Let $f \in \mathbb{R}[x]$ and define $F(x)$ as such:
$F(x)=f((r-x)^{2})$ where $r \in \mathbb{R}$
Then, $f^{(k)}(r)=0$ if $k$ is an odd integer.

Again, I’ll leave the trivialities of calculations and polynomial manipulations to the reader.

Now, to the main theorem.

### Theorem

$sin(x),cos(x),tan(x)$ are all irrational for non-trivial rational values of $x$.

#### Proof:

Let $q=\frac{a}{b}$ where $a,b \in \mathbb{Q}$. Define a function $f(x)$ as follows:

$f(x)=\frac{x^{p-1}(a-bx)^{2p}(2a-bx)^{p-1}}{(p-1)!}$

where $p$ is an odd prime.
Substituting the expression for $q$,

$f(x)=\frac{(r-x)^{2p}(r^{2}-(r-x)^{2})^{p-1}b^{3p-1}}{(p-1)!}$                              (1)

Notice how we obtained a polynomial in $(r-x)^{2}$ and of the form presented in Lemma 1. Also, observe that $f(x)>0$ for all values of $x \in \mathbb{R}$($p-1$ is even) .

The next is to obtain an upper bound for $f(x)$ when $x \in (0,r)$.

## Using Legendre Polynomials in irrationality proofs and establishing irrationality measures

The Legendre polynomials are a set of polynomial solutions to the Legendre’s Differential equation:

$(1-x^{2})y^{''}-2xy^{'}+n(n+1)y=0$

The edifice of most irrationality arguments use the fact the repeated integration by parts can be used on generating functions involving the polynomial.Another useful property is that these polynomials have integer coefficients.

Now,I’ll present a few important examples which display this property.

## Refreshing problem

There isn’t too much going in this question.Nonetheless,I just liked it for some bloody reason.

• Let $a$ and $n$ be positive integers where $a>1$.If $a^{n}+1$ is prime,prove that $a$ is even and n is of the form $2^{m}$ where $m \in \mathbb{N}$

## The irrationality of e

This is the post excerpt.

The irrationality of e can be proved using the infinite series expansion.

$e=\sum_{k=0}^{\infty} \frac{1}{k!}$

Assume that e is rational.i.e $e=\frac{p}{q}$ where $p \in \mathbb{Z}$ and $q \in \mathbb{Z}-{0}$.