Evaluate $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$ How do you evaluate $$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$$ using the identity $$\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)?$$
I assume I'll have to express $1+\frac{1}{n^2}+\frac{1}{n^4}$ as $\left(1+\frac{a}{n^2}\right)\left(1+\frac{b}{n^2}\right)$ for some $a,b\in\mathbb{C}$ so that I could use the identity given, but I can't seem to factor the above appropriately.
 A: Observing that
$$
\left(1+\frac1{n^2}+\frac1{n^4}\right)
=\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right)
$$
where
$$
a^2+b^2=-1\\
a^2b^2=1
$$
find $a$ and $b$ by a direct computation.
We get
\begin{align*}
\prod_{n\ge1}\left(1+\frac1{n^2}+\frac1{n^4}\right)
=&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right)\\
=&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right)
\prod_{n\ge1}\left(1-\frac{b^2}{n^2}\right)\\
=&\frac{\sin(\pi a)}{\pi a}\frac{\sin(\pi b)}{\pi b}
\end{align*}
A: We have $1+z+z^2=\Phi_3(z)=(z-\omega)(z-\bar{\omega})=(1-\omega z)(1-\bar{\omega}z)$, with $\omega=\exp{\frac{2\pi i}{3}}$.
By considering $z=\frac{1}{n^2}$ and $\eta=\exp\frac{\pi i}{3}$ we have that
$$ \frac{\sin(\pi w)}{\pi w}=\prod_{n\geq 1}\left(1-\frac{w^2}{n^2}\right)\tag{1} $$
implies:
$$ \prod_{n\geq 1}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right) = \frac{\sin(\pi\eta)\sin(\pi\bar{\eta})}{\pi^2}=\frac{\cos(\pi(\eta-\bar{\eta}))-\cos(\pi(\eta+\bar{\eta}))}{2\pi^2}\tag{2}$$
so that:
$$ \prod_{n\geq 1}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right) = \color{red}{\frac{1+\cosh(\pi\sqrt{3})}{2\pi^2}}=\left(\frac{1}{\pi}\,\cosh\frac{\pi\sqrt{3}}{2}\right)^2.\tag{3}$$
