I have tried to draw the graph of the $f(\theta)$ function.
Firstly the domain of the function is,
$$D(\sin\theta): -\infty<\sin\theta<\infty$$
$$D(\cos\theta+1): -\infty<\cos\theta<\infty \Rightarrow -\infty<\cos\theta + 1<\infty$$
The intersection of these domains is $-\infty<f(\theta)<\infty$.
I have stuck on the finding asymptots of the function phase.
$C$ is a constant that varies in $0<C<1$. For the vertical asymptotes;
$$\lim_{\cos\theta\to -C^{+}}(\frac{\sin\theta}{\cos\theta+C})=+\infty$$
But I'm not sure about it...
No, you're definitely confusing domain and codomain. This is not good. The domain of the numerator is all real numbers, so you should write $$-\infty<\theta<\infty,$$ not what you've written. The domain of the denominator is, likewise, all real numbers, but to consider the find the domain of $f(\theta)$, you must eliminate all the points where the denominator becomes $0$. So we need to remove $\theta = \arccos(-C)$ (which, by the way, falls between $\pi/2$ and $\pi$) and all the other values of $\theta$ for which $\cos\theta = -C$. This will be all the values $$\theta_* = 2\pi n\pm\arccos(-C) \quad\text{for all integers }n.$$
As we discussed yesterday, you will have vertical asymptotes in both directions because $\cos\theta+C$ will be positive on one side of each $\theta_*$ I've listed and negative on the other side (and $\sin\theta_*\ne 0$).