Find period of $f(x)$

Question

Consider $f(x) = \ln(\sin x)$ . First of all , Is $f(x)$ a periodic function ? And if it is periodic what is the period ?

My try : For finding period , this equation is obvious : $\ln(\sin x) = \ln(\sin (x+T))$ but I can't solve it .

Answer 1

$\ln$ is a strictly increasing function thus injective where defined.

So $\ln(\sin x)$ has same period than $\sin$ which is $T=2\pi$.

The domain of definition is $D=\bigcup\limits_{k\in\mathbb Z}]0,\pi[+kT$

---

To answer your question in comment : when using $x$ we generally refer to the real valued function. Wolfram-alpha in this case draw the complex-valued one.

But even though, it has the same period assuming we choose a proper determination $\operatorname{Ln}$ of the logarithm that agrees with $x\in\mathbb R$ (a cut at $\Im(x)>0$ for instance).

If $x\in]0,\pi[$ then $\operatorname{Ln}(\sin(x))=\ln(\sin(x))$

If $x\in]\pi,2\pi[$ then $\operatorname{Ln}(\sin(x))=\operatorname{Ln}(-1)+\ln(|\sin(x)|)=i\pi+\ln(\sin(x-\pi))$

On each of the intervals, the function is $2\pi$ periodic, so overall it is too. Only the domain of definition changes to $D=\mathbb R\setminus\pi\mathbb Z$.

Answer 2

That function is periodic. In general, if you have a function $f(x)$ and $g(x)$ so that $g(x)$ is periodic with period $p$, the function $f(g(x))$ will also be periodic with period $p$, because the argument of $f(x)$ is $g(x)$, and since the argument is repeating itself, the output will repeat itself.