The Chebyshev polynomials of the second kind $(U_n)_{n \in \mathbb{N}}$ are defined by
$$\forall n \in \mathbb{N}, \quad U_n = \frac{1}{n+1} T_{n+1}'$$
where $(T_n)_{n \in \mathbb{N}}$ are the Chebyshev polynomials of the first kind defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that
$$\forall n \in \mathbb{N}, \quad \forall \theta \in \mathbb{R} \backslash \pi\mathbb{Z}, \quad U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$$