1. **State the problem:** Given $\cos \theta = \frac{7}{25}$ and $\theta$ is in the interval $\left(\frac{3\pi}{2}, 2\pi\right)$, find $\sin 2\theta$, $\cos 2\theta$, and $\tan 2\theta$. 2. **Identify the quadrant and sign of $\sin \theta$:** Since $\theta$ is in the fourth quadrant ($\frac{3\pi}{2}$ to $2\pi$), $\cos \theta$ is positive and $\sin \theta$ is negative. 3. **Find $\sin \theta$ using Pythagorean identity:** $$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625}$$ So, $$\sin \theta = -\sqrt{\frac{576}{625}} = -\frac{24}{25}$$ (negative because $\theta$ is in the fourth quadrant) 4. **Use double-angle formulas:** - $\sin 2\theta = 2 \sin \theta \cos \theta$ - $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ - $\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}$ 5. **Calculate $\sin 2\theta$:** $$\sin 2\theta = 2 \times \left(-\frac{24}{25}\right) \times \frac{7}{25} = 2 \times -\frac{168}{625} = -\frac{336}{625}$$ 6. **Calculate $\cos 2\theta$:** $$\cos 2\theta = \left(\frac{7}{25}\right)^2 - \left(-\frac{24}{25}\right)^2 = \frac{49}{625} - \frac{576}{625} = -\frac{527}{625}$$ 7. **Calculate $\tan 2\theta$:** $$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{336}{625}}{-\frac{527}{625}} = \frac{336}{527}$$ **Final answers:** $$\sin 2\theta = -\frac{336}{625}, \quad \cos 2\theta = -\frac{527}{625}, \quad \tan 2\theta = \frac{336}{527}$$ This matches **Option 4**.