Missing delay in heavyside step function
We know that $\displaystyle
\frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ has value $0$ for $n &lt; -1$ since $u[n+1]=0$ for $n &lt; -1$. At $n=-1$, $u[n+1]$ jumps to value $1$, but $\sin(\theta (n+1))\bigr|_{n=-1} = \sin(\theta (-1+1))$ has value $0$, and so $\displaystyle
\frac{\sin(\theta (n+1))}{\sin(\theta)} u[n+1]$ just happens to equal $\displaystyle
\frac{\sin(\theta (n+1))}{\sin(\theta)} u[n]$. 

No need to convolve with $\delta[n+1]$ (or subtract $\delta[n+1]$ as Ch. Siva Ram Kishore suggests, why I don't know).