Problem: Prove that $\mathbb Q(\sin(θ/3))$ is a finite field extension of $\mathbb Q(\sin(θ))$ and compute its degree $[\mathbb Q(\sin(θ/3)):\mathbb Q(\sin(θ))]$ in terms of $θ ∈ \mathbb R$.
I can prove that it is a field extension, because $\sin(θ) = 3\sin(θ/3)-4\sin^3(θ/3)$. I can't really prove that it is finite, or what the basis should be for $\mathbb Q(\sin(θ/3))$ with coefficients in $\mathbb Q(\sin(θ))$. We define $F(S)$ as the smallest field containing $F$ and $S$, or as $\operatorname{Frac}(F[S])$.
You found a polynomial equation of degree $3$ for $\sin(\theta/3)$ over the field $\mathbf Q(\sin\theta)$, so why don't you see that $\sin(\theta/3)$ is algebraic over $\mathbf Q(\sin\theta)$?
Note that the field degree will depend on $\theta$.
If $\theta = 0$, then $\mathbf Q(\sin\theta)$ and $\mathbf Q(\sin(\theta/3))$ are both just $\mathbf Q$.
If $\theta = \pi$ then $\mathbf Q(\sin\theta) = \mathbf Q$ and $\mathbf Q(\sin(\pi/3)) = \mathbf Q(\sqrt{3}/2) = \mathbf Q(\sqrt{3})$, which has degree $2$ over $\mathbf Q(\sin\theta)$.
If $\theta = \pi/6$, then $\mathbf Q(\sin\theta) = \mathbf Q$ and $1/2 = \sin(\theta) = 3\sin(\theta/3) - 4\sin^3(\theta/3)$, so $\sin(\theta/3)$ is a root of $4x^3 - 3x + 1/2$, which is irreducible over $\mathbf Q$ (it's irreducible mod $5$), so $\mathbf Q(\sin(\theta/3))$ has degree $3$ over $\mathbf Q(\sin\theta)$.