Determine the range of $f(x) = \log\frac{1+\sin(x)}{1-\sin(x)}$

Question

Determine the range of $f(x) = \log\frac{1+\sin(x)}{1-\sin(x)}$ and the preimage $f^{-1}([0, \log3>)$.

For the Range, I divided the function into a composition of 3 functions where $$f_1 = \sin(x)$$ $$f_2 = \frac{1+x}{1-x} $$ $$f_3 = \log(x) $$

So, firstly $\sin(x)$ "throws" the whole $\mathbb{R}$ to $[-1, 1]$.

Then we have $\frac{1+x}{1-x}$ which can be written as $$-\frac{2}{x-1} - 1$$ to make it easier to draw a sketch and then I concluded that the second function "throws" the segment $[-1, 1]$ to $[0, +\infty>$ and since $<0, +\infty> \subseteq [0, +\infty>$ the $\log$ function will "throw" that segment i.e. interval into $\mathbb{R}$ so I conclude that the range or image is the whole $\mathbb{R}$ Is that correct?

preimage $f^{-1}([0, \log3>)$

So we have $$ 0 \le\log\frac{1+\sin(x)}{1-\sin(x)} < \log3$$ So, $\log$ is an increasing function but I am not sure because of the $\sin(x)$ which is periodical if I may "act" with the $10^{a}$ function as follows:

$$ 10^{0} \le 10^{\log(\frac{1+\sin(x)}{1-\sin(x)})} < 10^{\log3} \\ 1 \le \frac{1+\sin(x)}{1-\sin(x)} <3 \implies 0 \le \sin(x) < \frac{1}{2}$$

And since $\arcsin$ is defined within this interval:

$$ \arcsin(0) \le \arcsin(\sin(x)) < \arcsin(\frac{1}{2}) \implies x\in[0, \frac{\pi}{6}>$$

i.e. $x\in[2k\pi, \frac{\pi}{6} + 2k\pi>$

So may I do it like this or is there a "problem" because of the periodical $\sin$ function?

Answer 1

The number $f(x)$ exists if the fraction in the log is well defined and positive. Since every sine is between $-1$ and $1$, the only problem is when the sine evaluates to $\pm1$. Thus, the domain of $f$ is $D=\{x\in\mathbb R\mid|\sin(x)|\ne1\}=\mathbb R\setminus\{\frac\pi2+n\pi\mid n\in\mathbb Z\}$.

For every $x$ in $D$, $0\leqslant f(x)\lt\log3$ is equivalent to $1\leqslant\frac{1+\sin x}{1-\sin x}\lt3$, which is equivalent to $1-\sin x\leqslant1+\sin x\lt3(1-\sin x)$, which is equivalent to $0\leqslant\sin x\lt\frac12$. Thus, $$f^{-1}([0,\log3))=\bigcup_{n\in\mathbb Z}[2n\pi,2n\pi+\tfrac\pi6)\cup(2n\pi+\tfrac{5\pi}6,(2n+1)\pi]. $$