A Viéte-like infinite product for $\sin (\pi/7)$
Let:
$$
\begin{align}
\LARGE{\rho}\normalsize &= \left(\small \sqrt{2} \,\sqrt{2-\sqrt{2}} \,\sqrt{2+\sqrt{2-\sqrt{2}}} \normalsize\right) \times \\[2mm]
& \space \left(\small \sqrt{2\color{red}{-}\sqrt{2+\sqrt{2-\sqrt{2}}}} \,\sqrt{2\color{red}{-}\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2}}}}} \,\color{red}{\sqrt{2+ \color{black}{\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2}}}}}}}} \normalsize\right) \times \cdots \\[4mm]
\LARGE{\sigma}\normalsize &= \color{red}{\sqrt{2- \color{black}{\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{\cdots}}}}}}}}}} \space\space\small\{\text{Same as last item of}\,\rho\,\text{with minus sign}\}
\end{align}
$$
Fundamentally, the two expressions are the same.
And to transpose between them, we use the property:
$$
\boxed{ \quad \color{red}{\Large{\rho \cdot \sigma = 2}} \quad } \\[4mm]
\frac{1}{\rho} = \frac{1}{\rho} \color{red}{\cdot \frac{\sigma}{\sigma}} = \frac{\sigma}{\rho\cdot\sigma} = \frac{\sigma}{2} = \quad\sin\left(\frac{\pi}{7}\right)\quad = \frac{\sigma}{2} = \frac{\sigma}{2} \color{red}{\cdot \frac{\rho}{\rho}} = \frac{\rho\cdot\sigma/2}{\rho} = \frac{1}{\rho}
$$
In order to trace the simplifying process, let us consider the definition of both $\left(\large{\rho}\right)$ and $\left(\large{\sigma}\right)$ upto $n$,
Where $\left(\large{\rho}\right)$ definition is consist of three square-roots at a time, as illustrated above:
$$
\begin{align}
\LARGE{\rho}\normalsize &= \left(\large{\rho}_{\normalsize 0}\,\large{\rho}_{\normalsize 1}\,\large{\rho}_{\normalsize 2}\right) \left(\large{\rho}_{\normalsize 3}\,\large{\rho}_{\normalsize 4}\,\large{\rho}_{\normalsize 5}\right) \cdots \left(\large{\rho}_{\normalsize 3n-3}\,\large{\rho}_{\normalsize 3n-2}\,\large{\rho}_{\normalsize 3n-1}\right) \left(\large{\rho}_{\normalsize 3n}\,\large{\rho}_{\normalsize 3n+1}\,\large{\rho}_{\normalsize 3n+2}\right) \\[2mm]
&= \left(\large{\rho}_{\normalsize 0}\,\large{\rho}_{\normalsize 1}\,\large{\rho}_{\normalsize 2}\right) \left(\large{\rho}_{\normalsize 3}\,\large{\rho}_{\normalsize 4}\,\large{\rho}_{\normalsize 5}\right) \cdots \left(\large{\rho}_{\normalsize 3n-3}\,\large{\rho}_{\normalsize 3n-2}\,\large{\rho}_{\normalsize 3n-1}\right) \left(\sqrt{2\color{red}{-}\rho_{\small 3n-1}}\,\sqrt{2\color{red}{-}\rho_{\small 3n}}\,\sqrt{2\color{red}{+}\rho_{\small 3n+1}}\right)
\end{align}
$$
And $\left(\large{\sigma}\right)$ equals same as last item of $\large{\rho}\,$ with minus sign, as mentioned above:
$$ \LARGE{\sigma}\normalsize = \sqrt{2\color{red}{-}\rho_{\small 3n+1}} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$
So:
$$
\begin{align}
& \left(\large{\rho}_{\normalsize 3n}\,\large{\rho}_{\normalsize 3n+1}\,\large{\rho}_{\normalsize 3n+2}\right) \,\sqrt{2-\rho_{\small 3n+1}} \\[2mm]
& \qquad = \sqrt{2-\rho_{\small 3n-1}}\,\sqrt{2-\rho_{\small 3n}}\,\sqrt{2+\rho_{\small 3n+1}}\,\sqrt{2-\rho_{\small 3n+1}} = \sqrt{2-\rho_{\small 3n-1}}\,\sqrt{2-\rho_{\small 3n}}\,\sqrt{4-\rho_{\small 3n+1}^2} \\[2mm]
& \qquad = \sqrt{2-\rho_{\small 3n-1}}\,\sqrt{2-\rho_{\small 3n}}\,\sqrt{4-\left(2-\rho_{\small 3n}\right)} = \sqrt{2-\rho_{\small 3n-1}}\,\sqrt{4-\rho_{\small 3n}^2} \\[2mm]
& \qquad = \sqrt{2-\rho_{\small 3n-1}}\,\sqrt{4-\left(2-\rho_{\small 3n-1}\right)} = \sqrt{4-\rho_{\small 3n-1}^2} = \sqrt{4-\left(2+\rho_{\small 3n-2}\right)} = \color{red}{\sqrt{2-\rho_{\small 3n-2}}}
\end{align}
$$
Thus:
$$
\begin{align}
\color{red}{\Large{\rho \cdot \sigma}}\normalsize &= \left[\left(\large{\rho}_{\normalsize 0}\,\large{\rho}_{\normalsize 1}\,\large{\rho}_{\normalsize 2}\right) \left(\large{\rho}_{\normalsize 3}\,\large{\rho}_{\normalsize 4}\,\large{\rho}_{\normalsize 5}\right) \cdots \left(\large{\rho}_{\normalsize 3n-3}\,\large{\rho}_{\normalsize 3n-2}\,\large{\rho}_{\normalsize 3n-1}\right) \left(\large{\rho}_{\normalsize 3n}\,\large{\rho}_{\normalsize 3n+1}\,\large{\rho}_{\normalsize 3n+2}\right)\right]\,\sqrt{2-\large{\rho}_{\normalsize 3n+1}} \\[2mm]
&= \left[\left(\large{\rho}_{\normalsize 0}\,\large{\rho}_{\normalsize 1}\,\large{\rho}_{\normalsize 2}\right) \left(\large{\rho}_{\normalsize 3}\,\large{\rho}_{\normalsize 4}\,\large{\rho}_{\normalsize 5}\right) \cdots \left(\large{\rho}_{\normalsize 3n-3}\,\large{\rho}_{\normalsize 3n-2}\,\large{\rho}_{\normalsize 3n-1}\right)\right]\,\sqrt{2-\large{\rho}_{\normalsize 3n-2}} \\[2mm]
&= \left[\left(\large{\rho}_{\normalsize 0}\,\large{\rho}_{\normalsize 1}\,\large{\rho}_{\normalsize 2}\right) \left(\large{\rho}_{\normalsize 3}\,\large{\rho}_{\normalsize 4}\,\large{\rho}_{\normalsize 5}\right) \cdots \right]\,\sqrt{2-\large{\rho}_{\normalsize 3n-5}} \\[2mm]
&= \quad \cdots \quad \cdots \quad \cdots \\[2mm]
&= \left[\left(\large{\rho}_{\normalsize 0}\,\large{\rho}_{\normalsize 1}\,\large{\rho}_{\normalsize 2}\right)\right]\,\sqrt{2-\large{\rho}_{\normalsize 1}} = \large{\rho}_{\normalsize 0}\normalsize\,\sqrt{4-\rho_{\small 0}^2} \normalsize = \sqrt{2}\,\sqrt{4-2} = \color{red}{2 \quad \forall\space n \ge 1}
\end{align}
$$
Hence:
$$ \lim_{n\rightarrow\infty}\frac{1}{\rho} = \lim_{n\rightarrow\infty}\frac{\sigma}{2} = \sin\left(\frac{\pi}{7}\right) \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$
In order use this technique in general, for a period of length $\color{red}{m}$, we have:
$$
\begin{align}
\color{red}{{\large\rho}_{-1}=0},\,\, {\large\rho}_{0} &= \sqrt{2\pm{\rho}_{-1}},\,\, {\large\rho}_{1}=\sqrt{2\pm{\rho}_{0}},\,\, {\large\rho}_{2}=\sqrt{2\pm{\rho}_{1}},\,\, \cdots ,\,\, {\large\rho}_{m-1}=\sqrt{2\pm{\rho}_{m-2}},\,\, \cdots \\[2mm]
{\Large\rho} &= \prod_{k=0}^{n}\left( {\large\rho}_{mk}\, {\large\rho}_{mk+1}\, \cdots\, {\large\rho}_{mk+m-2}\, {\large\rho}_{mk+m-1} \right) \qquad \\[2mm]
&= \prod_{k=0}^{n}\left( \sqrt{2\pm{\large\rho}_{mk-1}}\, \sqrt{2\pm{\large\rho}_{mk}}\, \cdots\, \sqrt{2\pm{\large\rho}_{mk+m-3}}\, \sqrt{2\color{red}{\pm}{\large\rho}_{mk+m-2}} \right) \\[2mm]
{\Large\sigma} &= \sqrt{2\color{red}{\mp}{\large\rho}_{mn+m-2}}
\end{align}
$$
Hence, $\,\left({\large\sigma}\right)\,$ must start with a different sign from the actual period!
Nevertheless, if $\,{\large\sigma}\,$ is not comply with this condition, we still have options to write the product form of related/equivalent function by applying simple manipulations to the original $\,{\large\sigma}\,$.
Consider the following example $\left\{\,\text{period}\,\left(+,+,-,-,-\right)\,\right\}$:
$$
\begin{align}
\cos\left(\frac{\pi}{11}\right) &= \frac12\,\sqrt{2\color{red}{+}\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2\color{blue}{+}\sqrt{\cdots}}}}}}} \\[2mm]
\implies \qquad &\space \underbrace{\color{red}{+}{\large/}+,-,-,-,} \underbrace{\color{blue}{+}{\large/}+,-,-,-,} \underbrace{\color{blue}{+}{\large/}+,-,-,-,} \color{blue}{+}{\large/}\cdots
\end{align}
$$
And as it appears, after isolating the first sign, the pattern is not obeying the condition.
We have two options to convert it into applicable format:
$$
\color{blue}{\underline{\left({\bf1}\right) - \,\text{Convert to}\, \sin(\pi/11)}} \space\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\[2mm]
\begin{align}
\sin\left(\frac{\pi}{11}\right) &= \sqrt{1-\cos^2\left(\frac{\pi}{11}\right)} = \frac12\,\sqrt{4-\left[2\cos\left(\frac{\pi}{11}\right)\right]^2} \Rightarrow \\[2mm]
\sin\left(\frac{\pi}{11}\right) &= \frac12\,\sqrt{2\color{red}{-}\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2\color{blue}{+}\sqrt{\cdots}}}}}}} \\[2mm]
\implies \qquad &\space \underbrace{\color{red}{-}{\large/}+,-,-,-,} \underbrace{\color{blue}{+}{\large/}+,-,-,-,} \underbrace{\color{blue}{+}{\large/}+,-,-,-,} \color{blue}{+}{\large/}\cdots \\[2mm]
\end{align} \\[2mm]
{\Large\rho} = \prod_{k=0}^{n}\left( \sqrt{2-{\large\rho}_{5k-1}}\, \sqrt{2-{\large\rho}_{5k\space\space\space\space}}\, \sqrt{2-{\large\rho}_{5k+1}}\, \sqrt{2+{\large\rho}_{5k+2}}\, \sqrt{2\color{red}{+}{\large\rho}_{5k+3}} \right) \\[2mm]
{\Large\sigma} = \sqrt{2\color{red}{-}{\large\rho}_{5n+3}}\,,\quad {\Large\rho}\cdot{\Large\sigma}=2\,,\quad \sin\left(\frac{\pi}{11}\right)=\frac{{\large\sigma}}{2}=\frac{1}{{\large\rho}}\,,\quad \cos\left(\frac{\pi}{11}\right)=\sqrt{1-\left(1/{\large\rho}\right)^2} \\[4mm]
$$
$$
\color{blue}{\underline{\left({\bf2}\right) - \,\text{Convert to}\, \cos(\pi/22)}} \space\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\[2mm]
\begin{align}
\cos\left(\frac{\pi}{22}\right) &= \sqrt{\frac12+\frac12\,\cos\left(\frac{\pi}{11}\right)} = \frac12\,\sqrt{2+2\,\cos\left(\frac{\pi}{11}\right)} \Rightarrow \\[2mm]
\cos\left(\frac{\pi}{22}\right) &= \frac12\,\sqrt{2\color{red}{+}\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2\color{blue}{-}\sqrt{\cdots}}}}}}} \\[2mm]
\implies \qquad &\space \underbrace{\color{red}{+}{\large/}+,+,-,-,} \underbrace{\color{blue}{-}{\large/}+,+,-,-,} \underbrace{\color{blue}{-}{\large/}+,+,-,-,} \color{blue}{-}{\large/}\cdots \\[2mm]
\end{align} \\[2mm]
{\Large\rho} = \prod_{k=0}^{n}\left( \sqrt{2-{\large\rho}_{5k-1}}\, \sqrt{2-{\large\rho}_{5k\space\space\space\space}}\, \sqrt{2+{\large\rho}_{5k+1}}\, \sqrt{2+{\large\rho}_{5k+2}}\, \sqrt{2\color{red}{-}{\large\rho}_{5k+3}} \right) \\[2mm]
{\Large\sigma} = \sqrt{2\color{red}{+}{\large\rho}_{5n+3}}\,,\quad {\Large\rho}\cdot{\Large\sigma}=2\,,\quad \cos\left(\frac{\pi}{22}\right)=\frac{{\large\sigma}}{2}=\frac{1}{{\large\rho}}\,,\quad \cos\left(\frac{\pi}{11}\right)=\sqrt{\frac12+\frac12\,\left(1/{\large\rho}\right)} \\[2mm]
$$
And here are Mathematica codes for these two options:
n = 4;
r = 1; e = 0; For[k = 0, n >= k, k++, {a = Sqrt[2 - e]; b = Sqrt[2 - a]; c = Sqrt[2 - b]; d = Sqrt[2 + c]; e = Sqrt[2 + d]; r = r (a b c d e)}];
s = 1; x = Sqrt[2 + Sqrt[2 - Sqrt[2 - Sqrt[2]]]]; For[k = 0, (n - 1) >= k, k++, {x = Sqrt[2 + Sqrt[2 - Sqrt[2 - Sqrt[2 - Sqrt[2 + x]]]]]}]; s = Sqrt[2 - x];
Print[N[r s], " ", N[1/r], " ", N[s/2], " ", N[Sin[Pi/11]]]
n = 4;
r = 1; e = 0; For[k = 0, n >= k, k++, {a = Sqrt[2 - e]; b = Sqrt[2 - a]; c = Sqrt[2 + b]; d = Sqrt[2 + c]; e = Sqrt[2 - d]; r = r (a b c d e)}];
s = 1; x = Sqrt[2 + Sqrt[2 + Sqrt[2 - Sqrt[2]]]]; For[k = 0, (n - 1) >= k, k++, {x = Sqrt[2 + Sqrt[2 + Sqrt[2 - Sqrt[2 - Sqrt[2 - x]]]]]}]; s = Sqrt[2 + x];
Print[N[r s], " ", N[1/r], " ", N[s/2], " ", N[Cos[Pi/22]]]