How to simplify $\arcsin(\sin x)$ for $π/2 ≤ x ≤ 3π/2$? I understand why it is necessary to find the equation within this domain as this is the range of the equation $\arcsin(x).$ However, I am confused by this method that was used to find the equation of the line in this range:
$$\arcsin(\sin x) = \arcsin(\sin(π+(x-π))) = -\arcsin(\sin(x-π))= π-x.$$
 A: First of all, let us clarify what $\arcsin$ means here. Start with regular $\sin\colon \mathbb R \to [-1,1]$, which is surjective, but not injective function. Non-injectivity prevents us from inverting $\sin$, so we want to restrict its domain so it becomes injective. This can be done in infinite number of ways, but the usual choice is to restrict the domain to $[-\pi/2,\pi/2]$ so the restricted version of our sine is now both injective and surjective, so it has inverse $\arcsin\colon [-1,1]\to [-\pi/2,\pi/2]$.
What we get is that $\sin(\arcsin x) = x$, for all $x\in\mathbb [-1,1]$, but $\arcsin(\sin x) = x$ only for $x\in [-\pi/2,\pi/2]$. For other $x$'s, the last formula doesn't make sense since the RHS of it is not in the range of $\arcsin$. Therefore, for $x$'s not in $[-\pi/2,\pi/2]$, we first want to transform it to that segment, i.e. we want to transform it to the 1st or 4th quadrant. In your example, you have $x$ in 2nd or 3rd quadrant and if you draw trigonometric circle the way to transform an angle from 2nd or 3rd quadrant to 1st or 4th such that sines remain the same is to simply reflect the angle with respect to the $y$-axis. Algebraically, this is done by $x\mapsto \pi -x$ (note that the formula works since your $x$ is in the segment $[\pi/2,3\pi/2]$, otherwise you would have to add $2k\pi$ for appropriate integer $k$ first, but more on it later). Indeed, $\sin(\pi - x) = \sin x$ and if $x\in[\pi/2,3\pi/2]$, then $\pi - x \in [-\pi/2,\pi/2]$. This ultimately gives us $$\arcsin(\sin x) = \arcsin(\sin(\pi - x)) = \pi - x,\quad x\in [\pi/2,3\pi/2]$$ where the 2nd equality is true since $\arcsin$ and $\sin$ do cancel  simply for angles in $[-\pi/2,\pi/2]$, as we mentioned above.
The above reasoning is very visual and should be able to explain to you what is going on. It also provides a way to think about $\arcsin(\sin x)$: it won't necessarily give you the starting angle $x$, but what it will give you is the unique angle $x' \in [-\pi/2,\pi/2]$ such that $\sin x = \sin x'$. So, $\arcsin\sin$ transforms any angle to the 1st or 4th quadrant (more precisely $[-\pi/2,\pi/2]$) such that sines remain the same. As an exercise, you can think of what $\arccos\cos$ does, but you have to think about different quadrants.
We could have done the above work purely algebraically without any care whatsoever about the trigonometric circle and quadrants. We can start with inequality $\pi/2 \leq x \leq 3\pi/2$ and realize that we can get to the desired segment $[-\pi/2,\pi/2]$ by subtracting $\pi$, i.e. $-\pi/2 \leq x - \pi \leq \pi/2$. Let $y = x - \pi$. Now we have
\begin{align}
\arcsin(\sin x) &= \arcsin(\sin(y + \pi)) \\
&= \arcsin(\sin y \cos \pi + \sin \pi \cos y) \\
&= \arcsin(-\sin y)\\
&= - \arcsin(\sin y) = -y = \pi - x
\end{align}
where in the last line we used that $\arcsin$ is an odd function and $\arcsin(\sin y) = y$ since $y\in[-\pi/2,\pi/2]$.
Just for completeness, we can also cover the general case when $x\in\mathbb R$. First determine integer $k$ such that $-\pi/2 \leq x - k\pi \leq \pi/2$ and set $y = x - k\pi$. Now we have
\begin{align}
\arcsin(\sin x) &= \arcsin(\sin(y + k\pi)) \\
&= \arcsin(\sin y \cos (k\pi) + \sin (k\pi) \cos y) \\
&= \arcsin((-1)^k\sin y)\\
&= (-1)^k \arcsin(\sin y) \\
&= (-1)^k y = (-1)^k(x - k\pi),\quad x\in[ -\pi/2 + k\pi , \pi/2 + k\pi]
\end{align}
where in the last line we used that $\arcsin$ is an odd function and $\arcsin(\sin y) = y$ since $y\in[-\pi/2,\pi/2]$.
A: Elaborating on the given working:

$$\arcsin(\sin x) 
\\= \arcsin(\sin(π+(x-π)))$$

$$=\arcsin(\sin\pi\cos(x-\pi)+\cos\pi\sin(x-\pi))\\
=\arcsin(0-\sin(x-\pi))$$

$$= -\arcsin(\sin(x-π))$$

The justification for this step is that $\arcsin$ is an odd function, i.e., $\arcsin(-x)=-\arcsin(x).$

$$= π-x$$

This step is justified by the fact that $\arcsin(\sin(x-π))=(x-π),$ because $(x-\pi)$ lies in $\arcsin$'s principal range $\left[-\fracπ2,\fracπ2\right],$ because $x$ lies in $\left[\fracπ2,\frac{3π}2\right].$
A: The range of the $\arcsin$ function is $[-\frac{\pi} 2,\frac{\pi} 2]$. Since the sines of the angular arguments in your given domain are all positive, you need to have outputs in the restricted range $[0, \frac{\pi} 2]$. Basically you want a first quadrant reference angle corresponding to the third quadrant angle of your argument $x$, and you get that by taking pi minus the angle, hence $\pi - x$.
A: A visual aid might clarify what is going on with this exercise.
On the unit circle below is labled

*

*an angle $x$ for which $\frac{\pi}{2}\le x\le \frac{3\pi}{2}$ It lies on the left half of the circle.


*the corresponding angle $\arcsin(\sin x)$ lying on the right half of the circle.

Notice that the two angles add to $\pi$.
If we place $x$ in quadrant III rather than in quadrant II, the corresponding angle $\arcsin(\sin x)$ will be a negative angle in quadrant IV, so $x$ and $\arcsin(sin x)$ will still sum to $\pi$.
Thus we have $x+\arcsin(\sin x)=\pi$ so
$$ \arcsin(\sin x)=\pi-x \quad \text{ for } \frac{\pi}{2}\le x\le \frac{3\pi}{2}$$
