We are given the function $f(x) = \log \sin x$. The function is defined for $x \in (0, \pi)$ because $\sin x > 0$ in this interval. To determine whether $f$ is increasing or decreasing, we find its derivative.

Differentiating $f(x)$ with respect to $x$, we get:

$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$

Now consider the interval $(0, \pi/2)$. For any $x \in (0, \pi/2)$, we have $\sin x > 0$ and $\cos x > 0$. Therefore,

$\cot x = \frac{\cos x}{\sin x} > 0$

Hence $f'(x) > 0$ on $(0, \pi/2)$. This implies that $f$ is increasing on $(0, \pi/2)$.

Next, consider the interval $(\pi/2, \pi)$. For any $x \in (\pi/2, \pi)$, $\sin x > 0$ but $\cos x < 0$. Therefore,

$\cot x = \frac{\cos x}{\sin x} < 0$

Hence $f'(x) < 0$ on $(\pi/2, \pi)$. This implies that $f$ is decreasing on $(\pi/2, \pi)$.

Thus, we have proved that $f(x) = \log \sin x$ is increasing on $(0, \pi/2)$ and decreasing on $(\pi/2, \pi)$.