I found the following task that was inspired by an example in the book A. V. Oppenheim and R. W. Schafer, "Discrete-Time Signal Processing", 3rd Edition, 2014.
Task:
Consider the 2nd-order IIR filter $$H(z) = \frac{1}{1 - 2\cos(\theta)z^{-1} + z^{-2}}\,.$$
Show that the impulse response $$h[n] = \frac{\sin(\theta(n+1))}{\sin(\theta)} u[n] \, ,$$ with $u[n]$ the heavyside step function.
My solution:
I started from $$ \sin(\theta n) u[n] \leftrightarrow \frac{\sin(\theta)z^{-1}}{1-2\cos(\theta) z^{-1} + z^{-2}} \, , $$ which I looked up in a table and divided both sides by $\sin(\theta)$, i.e. $$ \frac{\sin(\theta n)}{\sin(\theta)} u[n] \leftrightarrow \frac{z^{-1}}{1-2\cos(\theta) z^{-1} + z^{-2}} \, . $$ Now I just had to get rid of $z^{-1}$, hence I multiplied with $z$ in z-domain, which corresponds to a convolution with $\delta[n+1]$ in time-domain, i.e. $$ \frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1] \leftrightarrow \frac{1}{1-2\cos(\theta) z^{-1} + z^{-2}} \, . $$
Comparing my solution with $h[n]$ in the task above, the only difference is the delayed function $u[n]$.
Does anybody see what I might did wrong or what properties can be exploited here?