What is the algebraic degree of $\tan(\pi/180)$?
Every now and again, I skim through Conjecture and Proof by Miklos Laczkovich and attempt some of the exercises. In chapter 5, on algebraic and transcendental numbers, the bulk of the chapter is spent showing the transcendence of $e$, but the beginning page or so (the chapters are quite short) asserts that if $r$ is rational, $\cos(r \pi)$ is algebraic. (This is somewhat clear a priori by the angle-addition formulas.) This makes me buy the claim that if $r=k/n$, the degree of $\cos(r \pi)$ is $\phi(n)/2$, $n$ odd and $\phi(n)$, $n$ even (here $\phi$ is Euler's totient function).
Now $180=2^2\cdot 3^2\cdot 5$. By the introduction, we should expect its minimal polynomial to have degree $2\cdot 6\cdot 4=48$; a gruesome calculation bears this out. (Aside: the exercise after this one has you write $\cos(\pi/15)$ by radicals, which I did, and got its minimal poly degree $4$; then $180/15=12$ and $4\cdot 12=48$, so I buy it.) So at this point, I could just continue on and try to get the minimal polynomial of $\cos(\pi/180)$, although I'm not sure how that would help. The tangent is throwing me for a loop: even if I knew the minimal polynomials of $\alpha=\cos(\pi/180)$ and $\beta=\sin(\pi/180)$, I'm not sure how I would compute the minimal polynomial of $\alpha^{-1}\beta$.
Another related question: when are the minimal polynomials of $\cos(r\pi)$ and $\sin(r\pi)$ the same, or when do they have the same degree? For instance, when $r=1/30$, they have degrees $8,4$ respectively; when $r=1/15$, they have degrees $4,8$; when $r=1/12$ they are identical of degree $4$.
Best Answer
Let $t_n=\tan(2\pi/n)$. Then $$t_n=\frac{\zeta-\zeta^{-1}}{i(\zeta+\zeta^{-1})}\in\Bbb Q(i,\zeta)\cap\Bbb R$$ where $\zeta=\exp(2\pi i/n)$.
We are in the case $n=360$ where $4\mid n$. Let's suppose then that $4\mid n$. For $\gcd(a,n)=1$, let $\sigma_a$ be the automorphism of $K=\Bbb Q(\zeta)$ with $\sigma(\zeta)=\zeta^a$. Then $\sigma_{-1}(t_n)=t_n$ (as $\sigma^{-1}$ is complex conjugation). For $\gcd(a,n)=1$, $a$ must be odd, and then $$\sigma_a(t_n)=\sigma_{-a}(t_n)=(-1)^{(a-1)/2} \frac{\zeta^a-\zeta^{-a}}{i(\zeta^a+\zeta^{-a})}=(-1)^{(a-1)/2} \tan\frac{2\pi a}{n}.$$ This equals $t_n$ iff either $a\equiv\pm1\pmod n$ or if $a\equiv(1+n/2)\pmod n$ and $n/4$ is even. If $8\mid n$ then $t_n$ has degree $\phi(n)/4$ over $\Bbb Q$ while if $4\mid n$ but $4\nmid n$ then $t_n$ has degree $\phi(n)/2$ over $\Bbb Q$.
In the case $n=360$, $8\mid n$ and $\phi(n)=96$ so that $t_{360}$ has degree $24$ over $\Bbb Q$.