1. **State the problem:** We need to prove the trigonometric identity $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$. 2. **Recall the sine difference formula:** The sine of a difference of two angles is given by $$\sin(A - B) = \sin A \cos B - \cos A \sin B$$. 3. **Apply the formula:** Letting $A = \alpha$ and $B = \beta$, we substitute directly: $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$. 4. **Explanation:** This formula comes from the unit circle definitions and the angle addition formulas. It expresses the sine of a difference in terms of sines and cosines of the individual angles. 5. **Conclusion:** The identity is shown by direct application of the sine difference formula, so it holds true. **Final answer:** $$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$.