We are given that $\cos(\pi + \alpha) = \frac{3}{5}$ and $\alpha \in (\pi, 2\pi)$. We want to find the value of $\sin \alpha$.

1. Problem Description

We are given that $\cos(\pi + \alpha) = \frac{3}{5}$ and $\alpha \in (\pi, 2\pi)$. We want to find the value of $\sin \alpha$.

2. Solution Steps

We know that

$$\cos(\pi + \alpha) = \cos \pi \cos \alpha - \sin \pi \sin \alpha$$

Since $\cos \pi = -1$ and $\sin \pi = 0$, we have

$$\cos(\pi + \alpha) = -\cos \alpha$$

We are given that $\cos(\pi + \alpha) = \frac{3}{5}$, so

$$-\cos \alpha = \frac{3}{5}$$

$$\cos \alpha = -\frac{3}{5}$$

We also know the trigonometric identity:

$$\sin^2 \alpha + \cos^2 \alpha = 1$$

Substituting $\cos \alpha = -\frac{3}{5}$, we get

$$\sin^2 \alpha + \left(-\frac{3}{5}\right)^2 = 1$$

$$\sin^2 \alpha + \frac{9}{25} = 1$$

$$\sin^2 \alpha = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$$

$$\sin \alpha = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$$

Since $\alpha \in (\pi, 2\pi)$, $\alpha$ is in the third or fourth quadrant. In the third quadrant, $\sin \alpha < 0$ and in the fourth quadrant $\sin \alpha < 0$. Therefore, $\sin \alpha$ must be negative.

$$\sin \alpha = -\frac{4}{5}$$

3. Final Answer

$-\frac{4}{5}$