Formula for $\pi$ by a $15$-gon Is it possible to generate a formula for $\pi$ using a $15$-gon as first inscribed polygon followed by a viete-like formula?

This question concerns application of the identity$$\operatorname{sinc}x=\prod_{k\ge1}\cos\frac{x}{2^k}.$$The factors are all knowable if $\cos\frac{x}{2}$ is known; given $\cos y$ with $y$ acute, one can calculate $\cos\frac{y}{2}$ as $\sqrt{\frac{1+\cos y}{2}}$.
The usual Viète's formula is the special case $x=\frac{\pi}{2}$, the internal angle of a square. If we take the $n$-gon as our starting point, the internal angle is $\pi-\frac{2\pi}{n}$, so we need to start with $\cos\left(\frac{\pi}{2}-\frac{\pi}{n}\right)=\sin\frac{\pi}{n}$. So in your example, we need an expression for $\sin\frac{\pi}{15}$. Take your pick from this expression or this equivalent one. I've seen neither proven, but such exercises in multiple-angle formulae are familiar.
But now for the tough part: what is our product of cosines? It's$$\operatorname{sinc}\frac{14\pi}{15}=\frac{15}{14\pi}\cos\frac{\pi}{15}.$$You'll find the cosine in the same sources.
