Given the discrete signal $h(n)=r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}$ if $n \geq 0$ and $h(n)=0$ otherwise, find the $Z$ transform of $h(n)$.
What I did:
We know that $H(z)=\sum_{n=0}^{\infty}h(n)z^{-n}=\sum_{n=0}^{\infty}r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}z^{-n}$ which is equal to
\begin{equation*} \frac{1}{\sin \theta}\sum_{n=0}^{\infty}r^n[\sin{(n\theta)}\cos \theta+\cos{(n\theta)}\sin \theta]z^{-n}=\cot \theta\sum_{n=0}^{\infty}r^n \sin(n\theta)z^{-n}+\sum_{n=0}^{\infty}r^n\cos(n\theta)z^{-n} \end{equation*}
and now I am stuck and don't know how to continue. We wish it was some sort of geometric series in the sum, so we will know a closed form for it, but sadly the $\sin(n\theta)$ and $\cos(n\theta)$ ruin it for us.
How do we continue?
The fastest way to find the Z-transform of $h(n)=r^n\frac{\sin{[(n+1)\theta]}}{\sin{\theta}}$ is to utilize the Chebyshev polynomials of the second kind. These polynomials are given as \begin{align} U_{n}(\cos \theta) = \frac{\sin(n+1)\theta}{\sin \theta} \end{align} with the generating function \begin{align} \sum_{n=0}^{\infty} U_{n}(\cos \theta) \, t^{n} = \frac{1}{1 - 2 \cos\theta \, t + t^{2}}. \end{align} From these components the desired Z-transform is given by \begin{align} \mathcal{Z}\{h(n)\} &= \sum_{n=0}^{\infty} \frac{\sin(n+1)\theta}{\sin\theta} \, \left(\frac{r}{z}\right)^{n} = \frac{z^{2}}{z^{2} - 2 \cos\theta \, r z + r^{2}}. \end{align}
As to the question of the problem proposed consider the series $$S_{1} = \sum_{n=0}^{\infty}r^n\cos(n\theta)z^{-n}.$$ The evaluation is as follows. \begin{align} S_{1} &= \sum_{n=0}^{\infty}r^n\cos(n\theta)z^{-n} \\ &= \frac{1}{2} \, \sum_{n=0}^{\infty} \left( e^{i n \theta} + e^{- i n \theta} \right) \, \left(\frac{r}{z}\right)^{n} \\\ &= \frac{1}{2} \, \left[ \sum_{n=0}^{\infty} \left(\frac{e^{i\theta} r}{z}\right)^{n} + \sum_{n=0}^{\infty} \left(\frac{e^{-i\theta} r}{z}\right)^{n} \right] \\ &= \frac{1}{2} \left[ \frac{z}{z - e^{i\theta} r} + \frac{z}{z - e^{-i\theta} r} \right] \\ &= \frac{z^{2} - r z \, \cos\theta}{z^{2} - 2 r z \, \cos\theta + r^{2}}. \end{align} The sine series follows the same pattern and the resulting Z-transform is identical to the one presented above.