$\begingroup$

I was reading a maths book which stated something like:

It is true, that: $\csc x = {1\over{\sin x}}$

But, it is false, that: $\sin x = {1\over{\csc x}}$

But how is it possible, it is somehow related to their domain and range, but how and why? Are they not same?

$\endgroup$ 9

2 Answers

$\begingroup$

Let's say I would have written a different statement if I had been the author of the book. But the statements are both true.

---

The function $\csc x$ is defined to be $\csc x := \frac 1{\sin x}$, and thus $\csc x$ makes sense for $x\neq 2k\pi$, $k\in \mathbb{Z}$. So the first sentence of your book is true since it is simply the definition of the cosecant function.

The second sentence of your book is true, that is, the equality there is false since the RHS is defined only for $x\neq 2k\pi$, $k\in \mathbb{Z}$ (since it contains the cosecant function, that is never $0$ by the way) and thus it cannot be equal to the LHS for all $x\in \mathbb{R}$, that is the domain of the sine function. The second sentence simply means that the two expressions are equal just for $x\neq 2k\pi$, $k\in \mathbb{Z}$.

$\endgroup$ 1 $\begingroup$

The expression $1/\sin x$ makes sense only when $\sin x \neq 0$, whereas $\sin x$ is defined for every real $x$. So the second equality holds only when both $\sin$ and $\csc$ are defined, whereas the first one makes sense when $\sin x \neq 0$.

$\endgroup$ 3

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy