# Proving that $\sin1$(radian) is irrational without using Taylor Series Expansion.

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. $$\blacksquare$$

• I upvoted this post for this great method, which is also readily generalisable to $\sin (n)$ and the like. – Vim Aug 11 '17 at 17:38

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.

• And so I figured, that is much more direct indeed. Thanks! – HavelTheGreat Jan 29 '15 at 3:15
• Of course, to check if this is really "without using Taylor Series", someone would need to check to see if Taylor series are used anywhere in all the preliminaries leading up to the Lindemann--Weierstrass theorem. – GEdgar Jan 29 '15 at 3:21
• Naturally, really I just wanted to be sure there was a less 'brute force' driven proof than my own I suppose. – HavelTheGreat Jan 29 '15 at 3:31
• @Elizion: Your argument is elementary, so in that sense better than using the big hammer. – André Nicolas Jan 29 '15 at 3:36

I believe you can use Generalized Continued Fractions and a (sufficient) irrationality criterion for those.

The generalized CF for $\sin 1$, can by found by putting $x=1$ in the CF for $\sin x$ in robjohn's answer: https://math.stackexchange.com/a/298666 giving us

$$\cfrac{1}{1 + \cfrac{1}{5 + \cfrac{6}{19 + \cfrac{20}{\dots}}}}$$

which satisfies the irrationality criterion given in the wiki page.

It is probably just hiding the Taylor series though...

• btw, if it was $\sin 1^{\circ}$, the proof would be easy: $\sin 45^{\circ}$ is a polynomial in $\sin 1^{\circ}$ with rational (in fact, integer) coefficients! Since $\sin 45^{\circ}$ is irrational... – Aryabhata Jan 29 '15 at 4:04