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, $$\cos(2^\circ)=2\cos^2(1^\circ)-1$$ we have that, $\cos(2^\circ)$ is also rational. Note that, we also have $$\cos(n^\circ +1 ^\circ)+\cos(n^\circ -1 ^\circ)=2\cos(1^\circ)\cdot 
\cos(n^\circ)$$ 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.

$\sin(1^\circ) = \cos(89^\circ)$, and since 89 is relatively prime to 360, the proof for $\cos 1^\circ$ works with almost no change.

More precisely: Assume that $\cos(89^\circ)$ is rational. Then, by the same induction as before with every $1^\circ$ replaced by $89^\circ$ we get that $\cos(89n^\circ)$ is rational for every $n\in\mathbb N$. In particular, since $150\times 89=37\times 360+30$, we get that $$\cos(150\times 89^\circ)=\cos(37\times 360^\circ+30^\circ)=\cos(30^\circ)$$ is rational, a contradiction.

For $\tan(1^\circ)$, a slight variant of the same proof works. Assume that $\alpha = \tan(1^\circ)$ is rational. Then $1+\alpha i$ is in $\mathbb Q[i]$, and then $\tan(n^\circ)$, being the ratio between the imaginary and real parts of $(1+\alpha i)^n$ is also rational. But $\tan(30^\circ)$ is not rational, so $\tan(1^\circ)$ cannot be either.
