1. **Problem:** Prove the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$. 2. **Formula and rules:** The triple-angle formula for sine is: $$\sin 3x = 3 \sin x - 4 \sin^3 x$$ This formula can be derived using the angle addition formula: $$\sin(a+b) = \sin a \cos b + \cos a \sin b$$ 3. **Derivation:** Start with: $$\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x$$ 4. Use double-angle formulas: $$\sin 2x = 2 \sin x \cos x$$ $$\cos 2x = 1 - 2 \sin^2 x$$ 5. Substitute these into the expression: $$\sin 3x = (2 \sin x \cos x) \cos x + (1 - 2 \sin^2 x) \sin x$$ 6. Simplify: $$\sin 3x = 2 \sin x \cos^2 x + \sin x - 2 \sin^3 x$$ 7. Use the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$: $$\sin 3x = 2 \sin x (1 - \sin^2 x) + \sin x - 2 \sin^3 x$$ 8. Expand: $$\sin 3x = 2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x$$ 9. Combine like terms: $$\sin 3x = (2 \sin x + \sin x) - (2 \sin^3 x + 2 \sin^3 x) = 3 \sin x - 4 \sin^3 x$$ 10. **Conclusion:** The identity is proven: $$\boxed{\sin 3x = 3 \sin x - 4 \sin^3 x}$$ This completes the solution for the first question.