Should cosecant be defined as $\csc \theta = \frac{1}{\sin \theta}$, specifying the constraint: $\sin \theta \neq 0$?

Question

I'm studying trigonometry on my own, and I keep noticing that the trigonometric functions are never defined with constraints to deal with divide-by-zero issues. As an example, I've seen cosecant defined like this: $\csc \theta = \frac{1}{\sin \theta}$

I've encountered this definition in the book I'm currently working through, and online resources ^{1 2 3}, but as $\sin \theta$ has the range [-1, 1] shouldn't cosecant really be defined as: $\csc \theta = \frac{1}{\sin \theta}, \sin \theta \neq 0$?

Am I missing something?

It is so obvious that no one writes it. – Kartik Aug 23 '15 at 10:46 — Kartik, Aug 23 '15 at 10:46

Deusovi · Accepted Answer · 2015-08-25T15:10:19.127

2

Those are the same thing. Either it's undefined as part of the definition (i.e. part of $\mathbb{R}$ is not in the domain) or it's undefined as a consequence of $\frac{1}{0}$ being undefined.

It's just like how $f(x)=\frac{1}{2-x}$ is completely clear without a note saying "but x can't be 2!" If a function is undefined, then the values for which it is undefined are not part of its domain by definition.

edited Aug 25 '15 at 15:10

answered Aug 23 '15 at 10:43

Deusovi

2,792
1
20
23

1

@JohnH No problem, glad I could help. :) – Deusovi Aug 23 '15 at 12:49

Should cosecant be defined as $\csc \theta = \frac{1}{\sin \theta}$, specifying the constraint: $\sin \theta \neq 0$?

1 Answers1