1. ## inverse function

$\displaystyle \mathfrak{hello}$
what is the domain of this function :
$\displaystyle f(x)=Arcsin(sin(x))$

2. Originally Posted by Raoh
what is the domain of this function :
$\displaystyle f(x)=Arcsin(sin(x))$
domain is all real numbers

3. thanks
but we have $\displaystyle arcsin(x) : \left [\frac{-\pi}{2},\frac{\pi}{2} \right ]\mapsto \left [ -1,1 \right ]$
i don't get it,would you explain it to me,thanks a lot.

4. the domain of the arcsin function is [-1,1]

the range of the sine function is [-1,1]

the domain of the sine function is all reals.

look at the graph ...

5. Hi!

Domain is all real numbers.

This is because $\displaystyle arcsin(x) : [-1,1] \to [-\frac{\pi}{2},\frac{\pi}{2}]$ and we have $\displaystyle sin(x) : \mathbb{R} \to [-1,1]$

Therfore, the inner function, $\displaystyle sin(x)$ , will have codomain equal to the domain of the outer function.

6. thanks to both of you.
just to make sure,i'm going to write to you what is in my papers :
$\displaystyle \forall x\in \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ], Arcsin(sin(x)) = x$
that why i got confused

7. Originally Posted by Raoh
thanks to both of you.
just to make sure,i'm going to write to you what is in my papers :
$\displaystyle \forall x\in \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ], Arcsin(sin(x)) = x$
that why i got confused

Ok, this is because if you let $\displaystyle x=\frac{3\pi}{4}$ , then $\displaystyle sin(x)=\frac{1}{\sqrt{2}}$

But $\displaystyle arcsin(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$ , not $\displaystyle \frac{3\pi}{4}$

This is why it is only true when $\displaystyle x\in [-\frac{\pi}{2},\frac{\pi}{2}]$

8. Awesome,thank you