3
$\begingroup$

For example, I write a function $$y=\sin x$$ and set its domain to be $\frac{\pi}{2}<x<\pi$.

Now, suppose I want to express $x$ from it, so the result is $$x=\arcsin y$$ But the domain for $x$ was $\frac{\pi}{2}<x<\pi$, while $\arcsin y$ outputs values between $\left[-\frac{\pi}{2};\frac{\pi}{2}\right]$.

Does it mean that I just changed the range of $\arcsin y$ to be $\left(\frac{\pi}{2};\pi\right)$, since $x$ and $\arcsin y$ are equal to each other in this scenario? I thought $\arcsin$ or any other inverse trigonometric function has a fixed range, yet here it feels I am redefining its range by setting a specific domain for the initial trigonometric function, which feels wrong, but I cannot explain why.

Thank you in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

$$y=\sin x, \quad \frac{\pi}{2}<x<\pi$$

The function $f:(\frac{\pi}{2},\pi) \to (1,0)$ is one-one onto i.e. a bijection and hence it is invertable. The inverse of $y$ can be denoted as $\text{myarcsin}$.

In other words,$$\sin x=y\iff x= \text{myarcsin} \,y \text{ for all } x\in(\frac{\pi}{2},\pi) \text{ and } y\in(1,0)$$

So, clearly you do not modify the domain of $\arcsin$ function. You have created a new function i.e $\text{myarcsin}$.

The $\arcsin$ function is by definition inverse of $\sin x:[-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1]$. So, it cannot be changed.


"Now, suppose I want to express $x$ from it, so the result is $$x=\arcsin y$$"

It is not possible. As, $x\in(\frac{\pi}{2},\pi)$. But you can create a inverse function as I mention in the answer. You can see the graph.

enter image description here

$\endgroup$
2
  • $\begingroup$ I think I get it, thank you! Just to clarify, what is "a bijection"? $\endgroup$
    – Tom
    Commented Jan 5 at 14:57
  • $\begingroup$ @Tom Bijection is term related to functions. In simple words, for every element in domain if we get a unique element in co-domain and co-domain doesnot contain any other element than these, then It is bijection or bijective function and one can only take inverse of bijection function. $\endgroup$
    – O M
    Commented Jan 5 at 15:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .