1. **State the problem:** We need to find the value of $\sin\left(\arccos\left(-\frac{1}{2}\right) + \arctan(1)\right)$.\n\n2. **Recall the formula:** For angles $A$ and $B$, $\sin(A+B) = \sin A \cos B + \cos A \sin B$.\n\n3. **Identify the angles:** Let $A = \arccos\left(-\frac{1}{2}\right)$ and $B = \arctan(1)$.\n\n4. **Evaluate $\sin A$ and $\cos A$:** Since $\cos A = -\frac{1}{2}$, use $\sin^2 A + \cos^2 A = 1$ to find $\sin A$.\n$$\sin A = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.$$\nBecause $A = \arccos(-\frac{1}{2})$ is in the second quadrant, $\sin A$ is positive.\n\n5. **Evaluate $\sin B$ and $\cos B$:** Since $B = \arctan(1)$, $\tan B = 1$, so $B = \frac{\pi}{4}$.\nTherefore, $\sin B = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $\cos B = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.\n\n6. **Apply the sine addition formula:**\n$$\sin(A+B) = \sin A \cos B + \cos A \sin B = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \left(-\frac{1}{2}\right) \cdot \frac{\sqrt{2}}{2}.$$\n\n7. **Simplify:**\n$$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.$$\n\n**Final answer:**\n$$\sin\left(\arccos\left(-\frac{1}{2}\right) + \arctan(1)\right) = \frac{\sqrt{6} - \sqrt{2}}{4}.$$