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. $$ \blacksquare $$