Show that $\mathbb{Q}(\sin\theta)$ is a field

Question

QUESTION:

Show that $F_{\theta}=\mathbb{Q}(\sin\theta); \theta\in\mathbb{R}$ is a field. Moreover, Show that $E_{\theta}=\mathbb{Q}(\sin\frac{\theta}{3}); \theta\in\mathbb{R}$ is a field extension of $F_{\theta}$ and find the maximum value of [$E_{\theta}:F_{\theta}$]

I am unable to show that for every $x\in F_{\theta}$, it has an inverse.
Also, I have shown that for some specific values of $\theta$, $E_{\theta}$ is a field extension of $F_{\theta}$.
But I can't prove for a general $\theta$.

Answer 1

Hint: $\forall x\in\mathbb R, \sin(3x)=3\sin(x)-4\sin^3(x)$.

Edit: Didn't see your first question concerning inverses in $F_\theta$, my bad. As Arthur said in comment, $\mathbb Q(\sin(\theta))$ should be by definition the smallest field containing $\mathbb Q$ and $\sin(\theta)$. Also for some values of $\theta$, $\sin(\theta)$ is not algebraic over $\mathbb Q$. Hence for such a value $\theta$, you can't express $\dfrac{1}{\sin(\theta)}$ as a $\mathbb Q$-linear combinaision of powers of $\sin(\theta)$.

Edit 2: (See Edit 3 for mistakes) The definition you gave in comments is the one for Q[sin(theta)]. It is only normal that you struggle to show that it is a field, because in fact it is a field if and only if $\sin(\theta)$ is algebraic over $\mathbb Q$, otherwise it's merely an integral domain. For example, if you take $\theta\in\mathbb R$ such that $\sin(\theta)=e^{-1}$, then knowing that $e^{-1}$ is not algebraic over $\mathbb Q$, then $\dfrac{1}{\sin(\theta)}=e\notin\mathbb Q[e^{-1}]$, because otherwise there exists a polynomial $P(X)$ with coefficients in $\mathbb Q$ such that $P(e^{-1})=e$, and by multiplying in both sides of the expression with a high enough power of $e$, you get $Q(e)=1$ for some polynomial $Q(X)\in\mathbb Q[X]$, which contradicts the transcendence of $e$ over $\mathbb Q$.

Edit 3: As Mark said in comment, the definition you gave doesn't even make the structure a ring, and the smallest ring containing it is $\mathbb Q[\sin(\theta)]:=\{P(\sin\theta)\mid P(X)\in\mathbb Q[X]\}$ where $\mathbb Q[X]$ is the ring of polynomials with coefficients in $\mathbb Q$. As for $\{a+b\sin\theta\mid a,b\in\mathbb Q\}$, all that can be said in a general matter about it ($\theta\in\mathbb R$ arbitrary) is that it is a $\mathbb Q$-vector space, of dimension $1$ if $\sin\theta\in\mathbb Q$, and $2$ otherwise.

Answer 2

$sin(\theta)=3sin(\theta/3)-4sin^3(\theta/3)$
so $[E_\theta:F_\theta]\leq 3$
to show that max = 3, let $\theta=30^\circ$
$1/2=sin(30^\circ)=3sin(10^\circ)-4sin^3(10^\circ)$
$8x^3-6x+1=0$ does not have rational roots by the rational root theorem, so it is irreducible.
$[E_{30^\circ}:F_{30^\circ}] = 3$