# sin1∘\sin 1^\circ is irrational but how do I prove it in a slick way? And tan(1∘)\tan(1^\circ) is …..

In the book 101 problems in Trigonometry, Prof. Titu Andreescu and Prof. Feng asks for the proof the fact that $\cos 1^\circ$ is irrational and he proves it. The proof proceeds by contradiction and using the strong induction principle. (Problem on Pg:84, 70 in typeset; solution on Pg:126, 111 in Typeset). However, for Completeness, I’ll append it here:

## Proof of irrationality of $\cos (1^\circ)$

Assume for the sake of contradiction, that $\cos(1^\circ)$ is rational. Since, we have that, $\cos(2^\circ)$ is also rational. Note that, we also have By Strong induction principle, this shows that $\cos(n^\circ)$ is rational for all integers $n \geq 1$. But, this is clearly false, as for instance, $\cos(30^\circ)=\dfrac{\sqrt{3}}{2}$ is irrational, reaching a contradiction.

But, as my title suggests, $\sin(1^\circ)$ is irrational, (look at the following image for its value!)

Is there a proof as short as the above proof or can any of you help me with a proof that bypasses actual evaluation of the above value?

Image Courtesy: http://www.efnet-math.org/Meta/sine1.htm
This link explains how to evaluate this value.

My next question is

Is $\tan(1^\circ)$ rational and is there a short proof that asserts or refutes its rationality?

P.S.: This is not a homework question.

If it’s a slick proof you want, nothing beats this proof that the only cases where both $r$ and $\cos(r \pi)$ are rational are where $\cos(r \pi)$ is $-1$, $-1/2$, $0$, $1/2$ or $1$.
If $r=m/n$ is rational, $e^{i \pi r}$ and $e^{-i \pi r}$ are roots of $z^{2n} - 1$, so they are algebraic integers. Therefore $2 \cos(r \pi) = e^{i \pi r} + e^{-i \pi r}$ is an algebraic integer. But the only algebraic integers that are rational numbers are the ordinary integers. So $2 \cos(r \pi)$ must be an integer, and of course the only integers in the interval $[-2,2]$ are $-2,-1,0,1,2$.