Let $ n\geq 2 $, and $ x\in\left]0,\pi\right[ $, prove the following formulas :
$$\begin{align}
\prod_{k=1}^{n-1}{\left(1-\frac{\sin^{2}{\left(\frac{x}{2n}\right)}}{\sin^{2}{\left(\frac{k\pi}{2n}\right)}}\right)}&=\frac{\sin{x}}{n\sin{\left(\frac{x}{n}\right)}} \\[8pt]
\prod_{k=1}^{n-1}{\left(1-\frac{\tan^{2}{\left(\frac{x}{2n}\right)}}{\tan^{2}{\left(\frac{k\pi}{2n}\right)}}\right)}&=\frac{\sin{x}}{n\sin{\left(\frac{x}{n}\right)}\cos^{2n-2}{\left(\frac{x}{2n}\right)}}
\end{align}$$
These beautiful formulas have served me to build a rigourous proof (Using squeezing theorem and the fact that if $ 0< x\leq y<\frac{\pi}{2} $ then $ \frac{\tan{x}}{\tan{y}}\leq\frac{x}{y}\leq\frac{\sin{x}}{\sin{y}} $) for Euler's well-known formula : $$ \left(\forall x\in\left]-\pi,\pi\right[\right),\ \sin{x}=x\prod_{n=1}^{+\infty}{\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)} $$
Best Answer
Both can be deduced from the multiple-angle formula for $\sin 2n\theta$.
For the first one, fix $n$ and write $\sin 2n\theta=P(\sin^2\theta)\sin 2\theta$, where $P$ is some polynomial of degree $n-1$ (we don't even care what it is exactly). Now if $\theta_k=\frac{k\pi}{2n}$ for $0<k<n$, then $\sin 2n\theta_k=0$, but $\sin 2\theta_k\neq 0$, which implies that $\sin^2\theta_k$ are roots of $P$, hence $P(x)=A\prod_{k=1}^{n-1}(x-\sin^2\theta_k)$ or $$\frac{\sin 2n\theta}{\sin 2\theta}=B\prod_{k=1}^{n-1}\left(1-\frac{\sin^2\theta}{\sin^2\theta_k}\right),\qquad(\sin 2\theta\neq 0)$$ where $A$ and $B$ are constants (not depending on $x$ and $\theta$, respectively). But then $B=\lim\limits_{\theta\to 0}\frac{\sin 2n\theta}{\sin 2\theta}=n$.
Similarly, for the second one, we can write $\sin 2n\theta=Q(\tan^2\theta)\cos^{2n-2}\theta\sin 2\theta$, where $Q$ is again a polynomial of degree $n-1$, and go literally through the same sequence of steps.