How to express $\sqrt{2-\sqrt{2}}$ in terms of the basis for $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ I've already shown that $\sqrt{2-\sqrt{2}}\in\mathbb{Q}(\sqrt{2+\sqrt{2}})$ but I want to write $\sqrt{2-\sqrt{2}}=a+b\alpha+c\alpha^2+d\alpha^3$ where $\alpha=\sqrt{2+\sqrt{2}}$. By squaring $\sqrt{2-\sqrt{2}}$ I get a system of four equations with four variables, but everything I try (Sage, for instance) takes too long to solve it. Is there a simpler way to write it in terms of the basis?
 A: First observe that
$$ \sqrt{2-\sqrt{2}}=\frac{\sqrt{2}}{\alpha}=\frac{\alpha^2-2}{\alpha}=\alpha-\frac{2}{\alpha}$$
Next, the minimal polynomial for $\alpha$ is $f(x)=x^4-4x^2+2$, hence
$$ -\frac{2}{\alpha}=\alpha^3-4\alpha$$
and therefore
$$ \sqrt{2-\sqrt{2}}=\alpha+(\alpha^3-4\alpha)=\alpha^3-3\alpha$$
A: There's a trigonometry trick for this particular case.
Let $\alpha=\sqrt{2+\sqrt{2}}$; then
$$
\alpha=2\sqrt{\frac{1+\sqrt{2}/2}{2}}=2\cos\frac{\pi}{8}
$$
whereas
$$
\beta=\sqrt{2-\sqrt{2}}=2\sqrt{\frac{1-\sqrt{2}/2}{2}}=2\sin\frac{\pi}{8}=
2\cos\left(\frac{\pi}{2}-\frac{\pi}{8}\right)=
2\cos\frac{3\pi}{8}
$$
Now
$$
\cos3t=4\cos^3t-3\cos t
$$
so
$$
\beta=2\left(4\frac{\alpha^3}{8}-3\frac{\alpha}{2}\right)=
\alpha^3-3\alpha
$$
A: $$\begin{align} \dfrac{2-\sqrt2}{2+\sqrt2}\, &=\, 3-2\sqrt 2\,=\, (\sqrt 2 - 1)^{\large 2}\\[.5em]
\Rightarrow\ \sqrt{2-\sqrt2} 
\,&=\, \sqrt{2+\sqrt2}\,\ \Big(\sqrt 2\, -\, 1\Big)\\
&=\,\sqrt{2+\sqrt2}\left(\sqrt{2+\sqrt 2}^{\large \,2}-3\right)\\[.1em]
&=\qquad\qquad\! \alpha\ (\alpha^{\large 2}\ -\ 3)
\end{align}$$
