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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...

Kenta S
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M. Çağlar TUFAN
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  • Your way to describe the domain of a function looks like you mix it up with the range. – Arnaud Mortier Dec 10 '19 at 21:14
  • @ArnaudMortier Yes, I do it in that way. I learned it by this way. It's a rational function and the way to find a rational function's domain, you basically find the domains of the nominator and the denominator functions. Then you take the intersection of these ranges and you finally find the domain of the main function. Am I wrong? – M. Çağlar TUFAN Dec 10 '19 at 21:26
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    If we forget about the way it's written, then you're still wrong in the sense that you're dividing by zero plenty of times here. You seem to know that there are vertical asymptotes, so you should also know that the domain can't be $\Bbb R$. – Arnaud Mortier Dec 10 '19 at 22:01

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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 $\sin\theta_*\ne 0$).

Ted Shifrin
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