1. **State the problem:** We want to show that $\sin x + \sin y$ is equivalent to $2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$. 2. **Formula used:** This is a standard trigonometric identity known as the sum-to-product formula for sine: $$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$ 3. **Explanation:** The formula transforms the sum of two sine functions into a product of sine and cosine functions, which can simplify many trigonometric expressions. 4. **Derivation:** Start with the right-hand side (RHS): $$2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$$ Use the angle addition formulas: $$\sin P \cos Q = \frac{1}{2} [\sin(P+Q) + \sin(P-Q)]$$ So, $$2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) = \sin \left( \frac{x+y}{2} + \frac{x-y}{2} \right) + \sin \left( \frac{x+y}{2} - \frac{x-y}{2} \right)$$ Simplify inside the sine functions: $$\frac{x+y}{2} + \frac{x-y}{2} = \frac{(x+y)+(x-y)}{2} = \frac{2x}{2} = x$$ $$\frac{x+y}{2} - \frac{x-y}{2} = \frac{(x+y)-(x-y)}{2} = \frac{2y}{2} = y$$ Therefore, $$\sin x + \sin y$$ which matches the left-hand side (LHS). 5. **Conclusion:** We have shown step-by-step that $$\sin x + \sin y = 2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$$ This confirms the identity.