Field Extensions over the Rationals

Let $p$ be an odd prime. For which values of $p$ is the following true: $$\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right)\right)=\mathbb{Q}\left(\sin\left(2\pi/p\right)\right),$$ where those are field extensions over $\mathbb{Q}$.

From $$p-1=[\mathbb{Q}(e^{2\pi/p}):\mathbb{Q}]=[\mathbb{Q}(e^{2\pi/p}):\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right)\right)]\cdot[\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right):\mathbb{Q}\right)]$$

I get $$[\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right)\right):\mathbb{Q}]=\frac{p-1}{2}.$$

But then $$[\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right)\right):\mathbb{Q}]=[\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right):\mathbb{Q}\sin(2\pi/p)\right)]\cdot[\mathbb{Q}\left(\sin\left(2\pi/p\right)\right):\mathbb{Q}]$$ $$\Rightarrow \frac{p-1}{2}=[\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right):\mathbb{Q}\sin(2\pi/p)\right)].[\mathbb{Q}\left(\sin\left(2\pi/p\right)\right):\mathbb{Q}]$$

I came across the following article What is $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q(\sin(2\pi k/n))$? which states that (since clearly $p$ prime $\Rightarrow p$ is not $0\mod4$) $$[\mathbb{Q}(\sin(2π/p)):\mathbb{Q}]=ϕ(p)=p-1$$

which forces $$[\mathbb{Q}\left(\sin\left(2\pi/p\right),\cos\left(2\pi/p\right):\mathbb Q(\sin(2\pi/p)\right)]=\frac{1}{2}.$$

Clearly something went wrong but I'm not sure what - I have a feeling I'm making some elementary error.

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With $s=\sin(2\pi/p), c=\cos(2\pi/p)$, note that $\mathbb Q(s,c,i)=\mathbb Q(\zeta_{4p})$ (because $i(c+is)$ is a primitive $4p$th root of unity and on the other hand $i=\zeta_{4p}^p, c=\frac{\zeta_{4p}+\zeta_{4p}^{-1}}2,c=\frac{\zeta_{4p}-\zeta_{4p}^{-1}}{2i}$) and is of degree $\phi(4p)=2(p-1)$ over $\mathbb Q$ and clearly of degree 2 over $\mathbb Q(s,c)$. Hence I get $[\mathbb Q(s,c):\mathbb Q]=p-1$, not $\frac{p-1}2$.
From expanding $(c+is)^p=1$ and taking the real part, we find a polynomial for $c$ with coefficients in $\mathbb Q(s)$ and such that only odd powers of $c$ appear (because only even powers of $is$ appear in the real part). Since $c^{2k+1}=c\cdot (1-s^2)^k$ we obtain in fact a linear polynomial for $c$, hence see directly that $\mathbb Q(s,c)=\mathbb Q(s)$.
@QiaochuYuan That's why I say $[\mathbb Q(s,c,i):\mathbb Q(s,c)]=2$. Am I (still) missing something? –  Hagen von Eitzen Nov 14 '12 at 22:49
Isn't $\mathbb{Q}(s,c,i)$ the same as $\mathbb{Q}(e^{2\pi/p})$ though? –  Rcwt Nov 14 '12 at 22:53
No, $i\notin \mathbb Q(e^{2\pi/p})$. (Even if you mean $\mathbb Q(e^{i\cdot2\pi/p})$) –  Hagen von Eitzen Nov 14 '12 at 23:00
As it is shown here $\mathbb Q(s,c)\neq\mathbb Q(s)$. Can you reconciliate your answer with that one? –  user26857 Aug 15 '13 at 20:53