# Points of discontinuity disappearing?

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• September 27th 2012, 07:05 AM
dblaisewatson
Points of discontinuity disappearing?
Hey all, so I had a question where it seems that points of discontinuity are just disappearing. Take a look:

$cot(\theta)=\frac{1}{tan(\theta)}=\frac{1}{\frac{s in(\theta)}{cos(\theta)}}$ $=\frac{cos(\theta)}{sin(\theta)}$

So, when I have the $\frac{1}{tan(\theta)}$, the points of discontinuity are wherever $tan(\theta)$ is 0 or discontinuous, which occurs whenever $sin(\theta)=0$ or $cos(\theta)=0$, but when I have $\frac{cos(\theta)}{sin(\theta)}$, the only points of discontinuity are when $sin(\theta)=0$. Am I crazy or did a point of discontinuity just disappear through a simple rearrangement? What is going on here?
• September 27th 2012, 07:23 AM
johnsomeone
Re: Points of discontinuity disappearing?
If we *define* cot as 1/tan, then technically you're right. The odd integer multiples of pi/2 will be missing from its domain.

However, they'll be removeable discontinuities that, when the function is defined to be 0 there, will even make the function smooth there. Thus no one would let that situation endure, any more than people would talk about the function exp(x) as having, for no good reason, the point x=5 removed from its domain. Yet, to be thoroughly technically correct, if we had first defined defined cot as 1/tan, then that would be its definition, and this new function with those missing points filled in would have to be given a different name.

If we *define* cot as cos/sin, then there is no confusion or technicality to fret over. So, why don't we just agree that cot is defined as cos/sin? It comes down to hair-splitting on the precise definition of the cot function, so let's choose the best definition - the one we're going to work with regardless.

So, no, you're not crazy. It's a truly irrelevant point, but it's a valid observation. You've a sharp eye.
• September 27th 2012, 10:38 AM
HallsofIvy
Re: Points of discontinuity disappearing?
A little more specifically, we define $tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$ with the stipulation that it is undefined when $cos(\theta)= 0$ and we define $cot(\theta)= \frac{cos(\theta)}{sin(\theta)}$ with the stipulation when it is undefined when $sin(\theta)= 0$. We can then show that $cot(\theta)= \frac{1}{tan(\theta)}$ provided both sides are defined.