In university last semester I was asked to prove that $\sin1$ (1 radian that is) is irrational, and ended up simply using the Taylor Series Expansion. This method provides a very quick solution, but I am curious as to whether anyone has a method for proving this without making use of the Taylor Series Expansion. I feel as though doing so must be possible using some number theory, but am low on ideas as to an alternative approach to the question.
Note:
If anyone is interested in my solution using Taylor Series Expansion (although it is not the focus of my question), here it is :
From
$$ \sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \dots$$
We see that
$$ \alpha = \sin 1 = 1 – \frac{1^3}{3!} + \frac{1^5}{5!} – \dots
$$
Given integers $a$ and $b$, if $\alpha = \frac{a}{b}$ then it follows that $b!\alpha \in \mathbb{Z}$, and $b!\alpha = C + D$ where $C \in \mathbb{Z}$ and we have :
$$
D = \begin{cases}
\pm(\frac{1}{b+1} – \frac{1}{(b+1)(b+2)(b+3)} + \dots) \text{ if $b$ is even}\\
\pm(\frac{1}{(b+1)(b+2)} – \frac{1}{(b+1)\dots(b+4)} + \dots ) \text{ if $b$ is odd}\\
\end{cases} $$
In each case we can see that $0 < D < 1$, giving us a contradiction.
Thus, we have that $\sin 1 $ is irrational.
$$ \blacksquare $$
Best Answer
An overkill proof.
By the Lindemann--Weierstrass Theorem, $e^i$ is transcendental. I can get $e^i$ from $\sin(1) = (e^i-e^{-i})/(2i)$ by solving a quadratic equation. Thus, if $\sin(1)$ were algebraic, then also $e^i$ would be algebraic.