Prove: (sin…

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

Step 1 · Locate the zeros

$\dfrac{\sin n\theta}{\sin\theta}=0$ exactly when $\sin n\theta=0$ and $\sin\theta\neq0$, i.e. $n\theta=k\pi$ with $\theta\neq0,\pi$:

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

giving $n-1$ distinct zeros of $\cos\theta$ at $x_k=\cos\frac{k\pi}n$.

Step 2 · $\sin n\theta/\sin\theta$ is a degree-$(n-1)$ polynomial in $\cos\theta$

This is the classical Chebyshev identity $\dfrac{\sin n\theta}{\sin\theta}=U_{n-1}(\cos\theta)$, where $U_{n-1}$ is a monic-up-to-scalar polynomial of degree $n-1$. Since it has exactly $n-1$ distinct roots $x_k=\cos\frac{k\pi}n$, it factors completely as:

$$ U_{n-1}(x)=A\prod_{k=1}^{n-1}\left(x-\cos\frac{k\pi}n\right) $$

for some constant $A$.

Step 3 · Determine the leading coefficient $A=2^{n-1}$

Write $\cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}2$ and $\sin n\theta=\dfrac{e^{in\theta}-e^{-in\theta}}{2i}$. As $\theta\to0$ (equivalently $x=\cos\theta\to1$), examine the leading behavior in $e^{i\theta}$: the highest power of $x$ in $U_{n-1}(x)$ arises from the term $e^{i(n-1)\theta}$ in the expansion of $\dfrac{\sin n\theta}{\sin\theta}$ written via $\dfrac{e^{in\theta}-e^{-in\theta}}{e^{i\theta}-e^{-i\theta}}=e^{i(n-1)\theta}+e^{i(n-3)\theta}+\cdots+e^{-i(n-1)\theta}$ (a geometric sum), whose leading term $e^{i(n-1)\theta}=(2x)^{n-1}$-order contributes coefficient $2^{n-1}$ to $x^{n-1}$. Hence $A=2^{n-1}$.

Step 4 · Conclusion

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

Verification

The identity was also checked symbolically for $n=2,3,4,5$.