The minimum value of the function $y = -\sin(x)$ is $-1$, occurring precisely when $\sin(x) = 1$.

To understand why this is true, let’s first review the properties of the sine function. The sine function, $\sin(x)$, oscillates between $-1$ and $1$ for all real values of $x$. Thus, the maximum value of $\sin(x)$ is $1$, and the minimum value is $-1$.

Next, consider the function $y = -\sin(x)$. This function is simply the sine function reflected about the horizontal axis. When we multiply $\sin(x)$ by $-1$, we invert its values. Consequently, the maximum value of $\sin(x)$ becomes the minimum value of $-\sin(x)$, and vice versa.

Since the maximum value of $\sin(x)$ is $1$, the minimum value of $-\sin(x)$ will be $-1$. This is because multiplying $1$ by $-1$ results in $-1$. Therefore, we conclude that the minimum value of $y = -\sin(x)$ is indeed $-1$.

To determine when this minimum value occurs, we need to identify the points where $\sin(x) = 1$. The sine function equals $1$ at specific angles, notably at $x = \frac{\pi}{2} + 2k\pi$, where $k$ is any integer. At these values of $x$, the function $y = -\sin(x)$ reaches its minimum of $-1$.

In summary, the minimum value of $y = -\sin(x)$ is $-1$, and this occurs whenever $\sin(x) = 1$, specifically at $x = \frac{\pi}{2} + 2k\pi$ for any integer $k$.