I managed to prove the formula
\[
\sqrt{2} \sin \left( \frac{\pi (z + 1)}{4} \right) = \left( 1 + \frac{z}{1} \right)\left( 1-\frac{z}{3} \right)\left( 1 + \frac{z}{5} \right)\left( 1-\frac{z}{7} \right) \cdots
\] that appear in the post with the help of the generative AI. I will be sharing the proof.
Proof
Substituting \(w = \frac{z+1}{4}\) into the formula
\begin{equation}
\sin \pi w = \pi w \prod_{n = 1}^{\infty} \left( 1-\frac{w^2}{n^2} \right)
\end{equation} yields the formula
\[
\sin \frac{\pi (z + 1)}{4} = \frac{\pi (z + 1)}{4} \prod_{n = 1}^{\infty} \left( 1-\frac{(z+1)^2}{16n^2} \right). \tag{*} \label{sin-prod}
\] Since
\begin{align}
1-\frac{(z+1)^2}{16n^2} &= \frac{16n^2-(z+1)^2}{16n^2} \\
&= \frac{(4n-z-1)(4n+z+1)}{16n^2} \\
&= \frac{(4n-1-z)(4n+1+z)}{16n^2} \\
&= \frac{4n-1-z}{4n-1} \cdot \frac{4n+1+z}{4n+1} \cdot \frac{(4n-1)(4n+1)}{16n^2} \\
&= \left( 1-\frac{z}{4n-1} \right) \left( 1+\frac{z}{4n+1} \right) \cdot \frac{(4n-1)(4n+1)}{16n^2},
\end{align} it holds that
\begin{align}
\sin \frac{\pi (z + 1)}{4} &= \frac{\pi}{4} \cdot (1+z) \cdot \prod_{n=1}^{\infty} \left( 1-\frac{(z+1)^2}{16n^2} \right) \\
&= \frac{\pi}{4} \cdot P(z) \cdot \prod_{n=1}^{\infty} \frac{(4n-1)(4n+1)}{16n^2},
\end{align} where
\[
P(z) = \left( 1+\frac{z}{1} \right) \left( 1-\frac{z}{3} \right) \left( 1+\frac{z}{5} \right) \left( 1-\frac{z}{7} \right) \cdots.
\] Let
\[
C = \prod_{n=1}^{\infty} \frac{(4n-1)(4n+1)}{16n^2}.
\] Substituting \(z = 0\) into the formula \eqref{sin-prod} yields
\[
\frac{1}{\sqrt{2}} = \frac{\pi}{4} \cdot C.
\] Therefore, it holds that
\[
\sin \frac{\pi (z + 1)}{4} = \frac{1}{\sqrt{2}} P(z)
\] i.e.
\[
\sqrt{2} \sin \left( \frac{\pi (z + 1)}{4} \right) = \left( 1 + \frac{z}{1} \right)\left( 1-\frac{z}{3} \right)\left( 1 + \frac{z}{5} \right)\left( 1-\frac{z}{7} \right) \cdots.
\] This completes the proof.
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