Show that the Chebyshev-Lobatto points $(\eta_k)^n = \cos\left(\dfrac{kpi}{n}\right)$ are the zeros of the polynomial $$(1-x^2)\dfrac{\sin(n\theta)}{\sin\theta}$$
where $x = \cos\theta$. The polynomils $U_{n-1}(x):= \dfrac{\sin(n\theta)}{n\sin\theta}$ (note the entra $n$ in the denominator) is called the Chebyshev polynomial of the second kind, and the $\eta_k$ are sometimes called the Chebyshev points of the second kind.
I have been going at this question all day and I am confused on where to begin. When I plug in $x=\cos\theta$ it just end up getting stuck with the original $U_{n-1}$ equation.
Obviously, $\theta_k=\frac{k\pi}n$ are roots of $\sin(n\theta)$.
$\frac{\sin(n\theta)}{\sin\theta}$ removes $\theta_0$ and $\theta_n$ as roots.
Why this fraction? Because you want a polynomial in $x=\cos\theta$, and the expansion of $\sin(n\theta)$ in frequency-one cosine and sine via $$ \sin(n\theta)=Im(e^{in\theta})=Im\Bigl((\cos\theta+i\sin\theta)^n\Bigr) $$ will contain terms with odd powers of the sine that can reduced to a single sine via Pythagoras $\sin^2\theta=1-\cos^2\theta$
So if you still want the $x$ values $\pm 1$ of $\theta_0$ and $\theta_n$ as roots, you need the extra factor $(1-x^2)$.