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

1 Answer
Jan 23, 2015

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

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: #\{ x \in \mathbb{R} | \sin(x)\ne0\}#

The sine function is defined as the projection of a point on the unit circle on the #y# axis, and thus #\sin(x)=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 #(1,0)# and #(-1,0)#, and they are given by a rotation of 0 and #\pi# radiants. Because of the periodicity of the sine function, 0 radiants is the same as #2k\pi# radians, and #pi# radians are the same as #(2k+1)\pi# radians, for each #k \in \mathbb{Z}#.

Finally, our answer is ready: we need to exclude from the domain all the points of the form #2k\pi#, for each #k \in \mathbb{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}#}