Show that if $[\mathbb Q(\cos\theta,\cos\frac{\theta}{3}):\mathbb Q(\cos \theta)]=1$, then $\theta$ can be trisected using straightedge and compass.
Proof: Clearly $\mathbb Q(\cos\theta,\cos\frac{\theta}{3})=\mathbb{Q}(\cos\theta)$, hence $\cos\frac{\theta}{3}\in\mathbb{Q}(\cos\theta)$, meaning $\cos\frac{\theta}{3}$ can be obtained from the elements of $\mathbb{Q}\cup \{\cos\theta\}$ by a finite sequence of additions, subtractions, multiplications and taking inverses. Screeching halt.
My question: Clearly I want to show that $\cos\frac{\theta}{3}$ is an element of $\mathbb{Q}(P)^{py}$, where $P:=\{0 ,1, \cos\theta\}$ and $\mathbb{Q}(P)^{py}$ is the Pythagorean closure of $\mathbb{Q}(P)$, because then it follows immediately that $\cos\frac{\theta}{3}$ is constructible from $P$; but the following theorem
"A complex number $\alpha$ is an element of $\mathbb{Q}(P)^{py}$ IFF there is a tower of field extensions $\mathbb{Q}(P)=K_0\subset K_1 \subset \dots \subset K_n=\mathbb{Q}(\alpha)$ such that $[K_{j+1}: K_j]=2$ for $0\leq j\leq n-1$"
clearly contradicts the result I wish to prove. What did I miss?
A related question was asked here; but the answer omitted the details I need. Any help would be appreciated.