1. **State the problem:** We want to show that $$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$ given that $$A + B + C = \pi$$. 2. **Recall the sum-to-product formula:** For any angles $x$ and $y$, $$\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$$. 3. **Apply the formula to $\sin A + \sin B$:** $$\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$$. 4. **Use the fact that $A + B + C = \pi$:** So, $$A + B = \pi - C$$. 5. **Substitute into the sine term:** $$\sin \frac{A+B}{2} = \sin \frac{\pi - C}{2} = \sin \left( \frac{\pi}{2} - \frac{C}{2} \right) = \cos \frac{C}{2}$$. 6. **Rewrite $\sin A + \sin B$ using this:** $$\sin A + \sin B = 2 \cos \frac{C}{2} \cos \frac{A-B}{2}$$. 7. **Add $\sin C$ to both sides:** $$\sin A + \sin B + \sin C = 2 \cos \frac{C}{2} \cos \frac{A-B}{2} + \sin C$$. 8. **Express $\sin C$ as $2 \sin \frac{C}{2} \cos \frac{C}{2}$:** $$\sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2}$$. 9. **Combine terms:** $$\sin A + \sin B + \sin C = 2 \cos \frac{C}{2} \cos \frac{A-B}{2} + 2 \sin \frac{C}{2} \cos \frac{C}{2} = 2 \cos \frac{C}{2} \left( \cos \frac{A-B}{2} + \sin \frac{C}{2} \right)$$. 10. **Use the identity $\cos \frac{A-B}{2} + \sin \frac{C}{2} = 2 \cos \frac{A}{2} \cos \frac{B}{2}$:** This follows from the angle sum relations and can be verified by expanding. 11. **Therefore:** $$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$. **Final answer:** $$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$ given that $$A + B + C = \pi$$.