The problem asks us to draw the graph of the function $y = \sin^{-1}(-\sin x)$.

1. Problem Description

The problem asks us to draw the graph of the function $y = \sin^{-1}(-\sin x)$.

2. Solution Steps

First, we use the property that $\sin(-x) = -\sin x$. Thus $-\sin x = \sin(-x)$. Therefore, the function becomes $y = \sin^{-1}(\sin(-x))$.

We know that $\sin^{-1}(\sin x) = x$ if $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$. However, since we have $\sin^{-1}(\sin(-x))$, we want to determine the values of $x$ for which $-\frac{\pi}{2} \le -x \le \frac{\pi}{2}$. Multiplying by $-1$, we get $\frac{\pi}{2} \ge x \ge -\frac{\pi}{2}$, which is the same as $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$.

Therefore, for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$, we have $\sin^{-1}(\sin(-x)) = -x$.

Now, let's consider the interval $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$. In this interval, we can write $x = \pi - x'$, so $x' = \pi - x$. Note that if $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$, then $-\frac{\pi}{2} \le x' \le \frac{\pi}{2}$ is false. For example, if $x = \pi$, then $x' = 0$, which is true, but if $x = \frac{3\pi}{2}$, then $x' = -\frac{\pi}{2}$ which is also true. If $x = \frac{\pi}{2}$, $x' = \frac{\pi}{2}$.

Since $\sin x = \sin(\pi - x)$, we have $\sin(-x) = \sin(\pi + x)$.

However, since the range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$, we want to express $-\sin x$ as $\sin y$ for some $y \in [-\pi/2, \pi/2]$.

We know $-\sin x = \sin(-x)$. If we are considering the interval $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$, then $-\frac{3\pi}{2} \le -x \le -\frac{\pi}{2}$.

Let $z = x-\pi$. Then $x = z+\pi$. If $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$, then $-\frac{\pi}{2} \le z \le \frac{\pi}{2}$.

We have $\sin(-x) = \sin(-(z+\pi)) = \sin(-z-\pi) = -\sin(z+\pi) = -(-\sin z) = \sin z$. Thus $\sin(-x) = \sin(x-\pi)$.

So, $y = \sin^{-1}(\sin(-x)) = \sin^{-1}(\sin(x-\pi))$.

Since $-\frac{\pi}{2} \le x-\pi \le \frac{\pi}{2}$ when $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$, we have $\sin^{-1}(\sin(x-\pi)) = x-\pi$.

So we have:

$y = -x$ for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$

$y = x-\pi$ for $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$

Notice that the function is periodic with period $2\pi$. Thus, we want to find the function in the interval $[-\pi/2, \frac{3\pi}{2}]$.

The function is defined as:

$$$$

y =

\begin{cases}

-x & -\frac{\pi}{2} \le x \le \frac{\pi}{2} \\

x - \pi & \frac{\pi}{2} \le x \le \frac{3\pi}{2}

\end{cases}

$$$$

The graph repeats this pattern every $2\pi$.

3. Final Answer

The graph of $y = \sin^{-1}(-\sin x)$ is a sawtooth wave that consists of line segments. The line segments are defined as follows:

$y = -x$ for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$

$y = x - \pi$ for $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$

and this pattern repeats every $2\pi$.

Final Answer: The final answer is $\sin^{-1}(-\sin x)$.