\( \sin \theta=-\frac{7}{25}, \frac{3 \pi}{2}<\theta<2 \pi \) Find \( \sin \theta \)

To find \( \sin \theta \) when \( \sin \theta = -\frac{7}{25} \) and \( \frac{3 \pi}{2}<\theta<2 \pi \), we already know the sine value given. Since \(\theta\) is in the fourth quadrant, where sine is negative, \( \sin \theta = -\frac{7}{25} \) is valid as is. Now, to solve for the other trigonometric functions: 1. We use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos \theta \). \[ \left(-\frac{7}{25}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{49}{625} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{49}{625} = \frac{576}{625} \] \[ \cos \theta = \pm\sqrt{\frac{576}{625}} = \pm\frac{24}{25} \] Since cosine is positive in the fourth quadrant, we have \( \cos \theta = \frac{24}{25} \). Thus, the functions are: \[ \sin \theta = -\frac{7}{25} \quad \text{and} \quad \cos \theta = \frac{24}{25}. \]