\( \sin \theta=\frac{3}{5}, \frac{\pi}{2}<\theta<2 \pi \) Find \( \tan (2 \theta) \)

To find \( \tan(2\theta) \), we can use the double angle formula: \[ \tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} \] First, we need to find \( \tan(\theta) \). Since \( \sin(\theta) = \frac{3}{5} \) and \( \theta \) is in the third quadrant (where both sine and cosine are negative), we can find \( \cos(\theta) \). Using the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] we can find \( \cos(\theta) \): \[ \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 \] \[ \frac{9}{25} + \cos^2(\theta) = 1 \] \[ \cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} \] So, \( \cos(\theta) = -\frac{4}{5} \) (negative in the third quadrant). Now, we can find \( \tan(\theta) \): \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \] Next, we substitute \( \tan(\theta) \) into the double angle formula: \[ \tan(2\theta) = \frac{2 \left(-\frac{3}{4}\right)}{1 - \left(-\frac{3}{4}\right)^2} \] Calculating the denominator: \[ 1 - \left(-\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \] Now substituting back, we get: \[ \tan(2\theta) = \frac{-\frac{6}{4}}{\frac{7}{16}} = \frac{-\frac{6}{4} \cdot 16}{7} = \frac{-24}{7} \] Thus, the value of \( \tan(2\theta) \) is \[ \boxed{-\frac{24}{7}} \]