1. **Problem Statement:** Given that $\pi < \theta < \frac{3\pi}{2}$ and $\tan(\theta) = 4$, find $\cos(\theta)$ and $\sin(\theta)$. 2. **Recall the definition of tangent:** $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$ Given $\tan(\theta) = 4$, we have $$\sin(\theta) = 4 \cos(\theta)$$ 3. **Use the Pythagorean identity:** $$\cos^2(\theta) + \sin^2(\theta) = 1$$ Substitute $\sin(\theta) = 4 \cos(\theta)$ into the identity: $$\cos^2(\theta) + (4 \cos(\theta))^2 = 1$$ $$\cos^2(\theta) + 16 \cos^2(\theta) = 1$$ $$17 \cos^2(\theta) = 1$$ 4. **Solve for $\cos(\theta)$:** $$\cos^2(\theta) = \frac{1}{17}$$ $$\cos(\theta) = \pm \sqrt{\frac{1}{17}} = \pm \frac{1}{\sqrt{17}}$$ 5. **Determine the sign of $\cos(\theta)$ and $\sin(\theta)$:** Since $\pi < \theta < \frac{3\pi}{2}$, $\theta$ is in the third quadrant where both sine and cosine are negative. Therefore, $$\cos(\theta) = - \frac{1}{\sqrt{17}}$$ 6. **Find $\sin(\theta)$:** Recall $\sin(\theta) = 4 \cos(\theta)$, so $$\sin(\theta) = 4 \times \left(- \frac{1}{\sqrt{17}}\right) = - \frac{4}{\sqrt{17}}$$ **Final answers:** $$\cos(\theta) = - \frac{1}{\sqrt{17}}, \quad \sin(\theta) = - \frac{4}{\sqrt{17}}$$