Prove…

Prove: $\ \dfrac{\sin(2n+1)\theta}{\sin\theta}=(2n+1)\displaystyle\prod_{k=1}^n\left[1-\dfrac{\sin^2\theta}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right]$.

Step 1 · Locate the zeros

$\dfrac{\sin(2n+1)\theta}{\sin\theta}=0$ exactly when $\sin(2n+1)\theta=0$ but $\sin\theta\neq0$, i.e. $(2n+1)\theta=k\pi$ with $\theta\neq0$:

$$ \theta_k=\frac{k\pi}{2n+1}, \qquad k=1,2,\ldots,2n $$

Step 2 · Reduce to $n$ distinct values of $\sin^2\theta$

Since $\sin\left(\frac{(2n+1-k)\pi}{2n+1}\right)=\sin\left(\pi-\frac{k\pi}{2n+1}\right)=\sin\frac{k\pi}{2n+1}$, the $2n$ zeros pair up as $\theta_k$ and $\theta_{2n+1-k}$ sharing the same value of $\sin^2\theta$. This leaves exactly $n$ distinct values:

$$ s_k=\sin^2\left(\frac{k\pi}{2n+1}\right), \qquad k=1,\ldots,n $$

Step 3 · The ratio as a polynomial in $\sin^2\theta$

Because $\dfrac{\sin(2n+1)\theta}{\sin\theta}$ can be written entirely in terms of $\sin^2\theta$ (an even, degree-$2n$ polynomial in $\sin\theta$), and it vanishes exactly at $\sin^2\theta=s_k$ for $k=1,\ldots,n$, it factors as:

$$ \frac{\sin(2n+1)\theta}{\sin\theta}=B\prod_{k=1}^n\left(1-\frac{\sin^2\theta}{s_k}\right) $$

where the constant is normalized so each factor equals $1$ at $\theta=0$.

Step 4 · Determine $B$ from $\theta\to0$

As $\theta\to0$, $\ \dfrac{\sin(2n+1)\theta}{\sin\theta}\to2n+1$ (by L'Hôpital, or the standard limit $\frac{\sin m\theta}{\sin\theta}\to m$), while every factor on the right tends to $1$. Hence:

$$ B=2n+1 $$

Step 5 · Conclusion

$$ \boxed{\frac{\sin(2n+1)\theta}{\sin\theta}=(2n+1)\prod_{k=1}^n\left[1-\frac{\sin^2\theta}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right]} $$

Verification

The general identity was also checked numerically for $n=1,2,3,4$ across multiple values of $\theta$.