Approximation near singularity of $1/\sin$

Question

I'm looking for an approximation $f(x)$ of $\frac{1}{\sin(x)}$ near the singularity at $x=0$.

Can you propose a function or literature or a key word, which leads me to $f(x)$? $f(x)$ must not have a singularity at $x=0$ and needs to be continous.

Answer 1

We have that by Taylor's expansion

$$\sin x = x-\frac16x^3+O(x^5)$$

therefore

$$\frac1{\sin x}=\frac1{x-\frac16x^3+\frac1{120}x^5+O(x^7)}=\frac1x\frac1{1-\frac16x^2+\frac1{120}x^4+O(x^6)}=\frac1x\left(1+\frac16x^2+\frac7{360}x^4+O(x^6)\right)=$$

$$=\frac1x+\frac16x+\frac7{360}x^3+O(x^5)$$

then we can eliminate the singularity by a suitable factor for the $\frac1x$ term that is for example

$$\frac{1-e^{1000x^2}}x+\frac16x+\frac7{360}x^3+O(x^5)$$

Answer 2

This may not the best approximation:

$f(x) = k\left(\sin\left(sign(x) \cdot \left|x\right|^c \cdot \frac{3\pi}{2}\right) + x\right)$

The solution for $\frac{1}{\sin(x)}=f(x)$ and $\left(\frac{1}{\sin(x)}\right)'=f'(x)$ for a fixed $k$ and a variable $c$ is hard to find.

Better proposals are welcome.

Answer 3

Multiply by anything that is close to $1$ far from $x=0$ and has a minimum at $(0,0)$.

Like

$$\frac{x^2}{(x^2+\epsilon)\sin(x)}.$$