# What is the domain of f(x)=cscx?

Jan 23, 2015

The question gets much easier if we recall that, by definition, the function you're dealing with is $\setminus \frac{1}{\setminus \sin \left(x\right)}$.

Being a fraction, we have to make sure that the denominator is non-zero; and since there are no roots or logarithms, it is the only thing we have to study.

The domain of the function will thus be the following set: $\setminus \left\{x \setminus \in \setminus m a t h \boldsymbol{R} | \setminus \sin \left(x\right) \setminus \ne 0 \setminus\right\}$

The sine function is defined as the projection of a point on the unit circle on the $y$ axis, and thus $\setminus \sin \left(x\right) = 0$ if and only if the point belongs to the $x$ axis.

The only two points which are on both the unit circle and on the $x$ axis are $\left(1 , 0\right)$ and $\left(- 1 , 0\right)$, and they are given by a rotation of 0 and $\setminus \pi$ radiants. Because of the periodicity of the sine function, 0 radiants is the same as $2 k \setminus \pi$ radians, and $\pi$ radians are the same as $\left(2 k + 1\right) \setminus \pi$ radians, for each $k \setminus \in \setminus m a t h \boldsymbol{Z}$.

Finally, our answer is ready: we need to exclude from the domain all the points of the form $2 k \setminus \pi$, for each $k \setminus \in \setminus m a t h \boldsymbol{Z}$. Using a proper notation, the domain is the set D_f={x \in \mathbb{R} | x \ne 2k\pi, \forall k \in \mathbb{Z}}