# Thread: How do I simplify this

1. ## How do I simplify this

$arcsin(sin(x+\pi))$ for $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$

I can see that $arcsin(sin(x+\pi)) = x+\pi$

How does $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$ apply to my answer?

(I have an answer $x-2\pi$ and didn't understand the logic)

Thanks.

2. ## Re: How do I simplify this

Originally Posted by terrorsquid
$arcsin(sin(x+\pi))$ for $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$
I can see that $arcsin(sin(x+\pi)) = x+\pi$
How does $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$ apply to my answer?
(I have an answer $x-2\pi$ and didn't understand the logic
First, take note that $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$
is the same domain as $x\in \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$.
That happens to be the range of $\arcsin(x).$
Observe that $\frac{3\pi}{2}-2\pi=\frac{-\pi}{2}$

3. ## Re: How do I simplify this

I can see it's domain, but I am still confused as to how I can derive an answer from that information. There is a disconnect between $x+\pi$ and the domain for me. How do I combine them to form an answer?

4. ## Re: How do I simplify this

Originally Posted by terrorsquid
I can see it's domain, but I am still confused as to how I can derive an answer from that information. There is a disconnect between $x+\pi$ and the domain for me. How do I combine them to form an answer?
To transform $x\in \left[\frac{3\pi}{2},\frac{5\pi}{2}\right]$ to $x\in \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ we use $x-2\pi$.

I don't know any other way to explain it.

5. ## Re: How do I simplify this

Hmm ok.What was the point of the original equation and finding $arcsin(sin(x+ \pi)) = x+ \pi$ then if $x-2\pi$ is found just by looking at the domain?

6. ## Re: How do I simplify this

Originally Posted by terrorsquid
Hmm ok.What was the point of the original equation and finding $arcsin(sin(x+ \pi)) = x+ \pi$ then if $x-2\pi$ is found just by looking at the domain?
Frankly, I do not understand the wording of the original question.
I was trying to give you a reason for the proposed answer.

I can see that $\arcsin(\sin(x+\pi))=\arcsin(-\sin(x))$ that is a simplification.