<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] Interesting product. Let us define the Kepler-Bouwkamp constant as $c$. Then
$$c=\prod_{k=1}^\infty \cos\left(\frac{\pi}{k+2}\right)$$
Let $\text{Si}(x)=\frac{\sin(x)}{x}$. Then, $\cos(x)=\frac{\text{Si}(2x)}{\text{Si}(x)}$. Plugging this in, we get
$$c=\prod_{k=1}^\infty\frac{\text{Si}\left(\frac{2\pi}{k+2}\right)}{\text{Si}\left(\frac{\pi}{k+2}\right)}$$
Now, observe that every term in the denominator cancels with a term in the numerator because $\frac{\pi}{k+2}=\frac{2\pi}{2(k+1)}$. All that remains is the second term in the numerator and the odd terms of the numerator, viz
$$c=\prod_{k=1}^\infty \text{Si}\left(\frac{2\pi}{2+2}\right)\text{Si}\left(\frac{2\pi}{2k+1}\right)=\text{Si}\left(\frac{\pi}{2}\right)\prod_{k=1}^\infty \text{Si}\left(\frac{2\pi}{2k+1}\right)$$
Simplifying, and plugging in the actual product expression for $c$, we have
$$\prod_{k=1}^\infty \cos\left(\frac{\pi}{k+2}\right)=\frac{2}{\pi}\prod_{k=1}^\infty\frac{2k+1}{2\pi}\sin\left(\frac{2\pi}{2k+1}\right)$$
Rearranging gives us that $$\frac{\pi}{2}=\prod_{k=1}^\infty\frac{2k+1}{2\pi}\sin\left(\frac{2\pi}{2k+1}\right)\sec\left(\frac{\pi}{k+2}\right)$$
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