1. **State the problem:** Given that $\cos \theta = \frac{7}{25}$ and $\theta$ is in the interval $\left(\frac{3\pi}{2}, 2\pi\right)$, find $\sin 2\theta$. 2. **Recall the formula:** The double-angle identity for sine is $$\sin 2\theta = 2 \sin \theta \cos \theta$$ 3. **Determine the sign of $\sin \theta$:** Since $\theta$ is in the fourth quadrant ($\frac{3\pi}{2}$ to $2\pi$), $\sin \theta$ is negative. 4. **Find $\sin \theta$ using Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ $$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{625}{625} - \frac{49}{625} = \frac{576}{625}$$ $$\sin \theta = -\sqrt{\frac{576}{625}} = -\frac{24}{25}$$ 5. **Calculate $\sin 2\theta$:** $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \left(-\frac{24}{25}\right) \times \frac{7}{25} = 2 \times \left(-\frac{168}{625}\right) = -\frac{336}{625}$$ **Final answer:** $$\sin 2\theta = -\frac{336}{625}$$