1. We are asked to verify the trigonometric identity: $$\sin 3A = \sin A (3 \cos^2 A - \sin^2 A)$$. 2. Start with the left-hand side (LHS): $$\sin 3A$$. 3. Use the triple-angle formula for sine: $$\sin 3A = 3 \sin A - 4 \sin^3 A$$. 4. Now, consider the right-hand side (RHS): $$\sin A (3 \cos^2 A - \sin^2 A)$$. 5. Recall that $$\cos^2 A = 1 - \sin^2 A$$, so substitute this into the RHS: $$\sin A (3 (1 - \sin^2 A) - \sin^2 A) = \sin A (3 - 3 \sin^2 A - \sin^2 A) = \sin A (3 - 4 \sin^2 A)$$. 6. Distribute $$\sin A$$: $$3 \sin A - 4 \sin^3 A$$. 7. This matches exactly the LHS expression from step 3. 8. Therefore, the identity is verified: $$\sin 3A = \sin A (3 \cos^2 A - \sin^2 A)$$. Final answer: The identity holds true.