To solve this problem, we are given that:

$\sin(\theta) = -\frac{2}{5}$ and that $\theta$ is in Quadrant III. We need to find $\sin(2\theta)$, $\cos(2\theta)$, and $\tan(2\theta)$.

Step 1: Determine $\cos(\theta)$

Since $\sin^2(\theta) + \cos^2(\theta) = 1$, we can find $\cos(\theta)$ as follows:

$\sin^2(\theta) = \left(-\frac{2}{5}\right)^2 = \frac{4}{25}$

Substitute into the Pythagorean identity:

$\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \frac{4}{25} = \frac{21}{25}$

Thus,

$\cos(\theta) = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5}$

We take the negative square root because $\theta$ is in Quadrant III, where both sine and cosine are negative.

Step 2: Find $\sin(2\theta)$

Using the double-angle identity for sine:

$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$

Substitute $\sin(\theta) = -\frac{2}{5}$ and $\cos(\theta) = -\frac{\sqrt{21}}{5}$:

$\sin(2\theta) = 2 \cdot \left(-\frac{2}{5}\right) \cdot \left(-\frac{\sqrt{21}}{5}\right) = 2 \cdot \frac{2\sqrt{21}}{25} = \frac{4\sqrt{21}}{25}$

Step 3: Find $\cos(2\theta)$

Using the double-angle identity for cosine:

$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$

Substitute $\cos^2(\theta) = \frac{21}{25}$ and $\sin^2(\theta) = \frac{4}{25}$:

$\cos(2\theta) = \frac{21}{25} - \frac{4}{25} = \frac{17}{25}$

Step 4: Find $\tan(2\theta)$

Using the double-angle identity for tangent:

$\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}$

Substitute $\sin(2\theta) = \frac{4\sqrt{21}}{25}$ and $\cos(2\theta) = \frac{17}{25}$:

$\tan(2\theta) = \frac{\frac{4\sqrt{21}}{25}}{\frac{17}{25}} = \frac{4\sqrt{21}}{17}$

Summary of Results

[ \sin(2\theta