<s>[INST] Show that $\prod_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$ Show that
$$\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$$
Context:
Inspired by this question, I considered the product of the areas of every regular odd-gon inscribed in a circle of area $1$. This is $\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi} \sin{\frac{2\pi}{2k+1}}=0.18055...$ which I guess does not have a closed form.
On a whim, I divided this number by the Kepler-Bouwkamp constant, $\prod\limits_{k=1}^\infty \cos{\frac{\pi}{k+2}}=0.11494...$, and numerical calculation suggests that the result is $\pi/2$.
My attempt:
I used the double angle formula for sine to get $\prod\limits_{k=1}^\infty \frac{2k+1}{\pi}\sin{\left(\frac{\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{2k+2}\right)}$ but this seems just as intractable as the original form of the product.
 [/INST] Let
$$K = \prod_{k=3}^\infty \frac1{\cos\left(\frac\pi k\right)}$$
As Bouwkamp points out under First method in this paper,
$$\cos(x) = \frac{\sin(2x)}{2x} \cdot \frac x{\sin(x)} \tag{1}$$
Then
$$\begin{align*}
\frac1K &= \prod_{k=3}^\infty \cos\left(\frac\pi k\right) \\[1ex]
&= \prod_{k=3}^\infty \frac{\sin\left(\frac{2\pi}k\right)}{\frac{2\pi}k} \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{1}\\[1ex]
&= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}} \cdot \prod_{k=2}^\infty \frac{\sin\left(\frac{2\pi}{2k}\right)}{\frac{2\pi}{2k}}\right] \cdot \prod_{k=3}^\infty \frac{\frac\pi k}{\sin\left(\frac\pi k\right)} \tag{2}\\[1ex]
&= \left[\prod_{k=1}^\infty \frac{\sin\left(\frac{2\pi}{2k+1}\right)}{\frac{2\pi}{2k+1}}\right] \cdot \frac2\pi
\end{align*}$$
where in $(2)$ we split the product over even and odd indices.
</s>