# A Viéte-like infinite product for $\sin (\pi/7)$

In the article "The unruly $\sin (\pi/7)$ of Samuel Moreno, the following infinite product is given:

Also, the same article shows that $\sin (\pi/7)$ is equal to the infinite nested radical

$\sin (\pi/7) = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2+...}}}}}}}}}}$

The question is: are these two expressions fundamentally different for $\sin(\pi/7)$, or is it possible to rewritte the infinite nested radical into that infinite product by simple algebraic manipulations?

• Are you sure that it's $$\sin\dfrac {\pi}7=\dfrac 12\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\cdots}}}}}}}}}}\tag1$$And not$$\sin\dfrac {\pi}7=\dfrac 12\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\cdots}}}}}}}}}\tag2$$ – Frank Jan 8 '17 at 17:48
• No, the expression $(1)$ is correct. – C.C. Babock Jan 9 '17 at 17:25

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]]]


• @HazemOrahi +1 Very clever analysis. Is it possible to use the above procedure to transform this type of infinite radicals into infinite products? For example, $\cos \frac{\pi}{11} = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2 + ...}}}}}}$ with period $(+, +, -, -, -)$. Can this be easily transformed into an infinite product, let's say $1/\rho$? – C.C. Babock Jan 9 '17 at 17:23
• I want to say @HazemOrabi – C.C. Babock Jan 9 '17 at 18:16
• @C.C.Babock: Thanks a lot. I’ll check your proposition and get back to you soon. – Hazem Orabi Jan 9 '17 at 22:30
• OK. I'll wait . – C.C. Babock Jan 11 '17 at 18:03
• @C.C.Babock: I edited the answer to include a sufficient condition for the technique to work in the general case. Nevertheless, I tried my best to find a direct infinite product for $\,\sigma\,$ of $\,\cos(\pi/11)\,$ but, unfortunately, the condition of having a different sign to start $\,\sigma\,$ with seems to be almost forced! Thanks again & good luck. – Hazem Orabi Jan 11 '17 at 22:34