Question: Use double angle formulas to show that: $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$$
1. **State the problem:** We want to prove the identity $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$$ using double angle formulas.
2. **Recall the double angle formula for sine:**
$$\sin 2\theta = 2 \sin \theta \cos \theta$$
3. **Express $$\sin 3\theta$$ as $$\sin(2\theta + \theta)$$:**
$$\sin 3\theta = \sin(2\theta + \theta)$$
4. **Use the sine addition formula:**
$$\sin(a + b) = \sin a \cos b + \cos a \sin b$$
So,
$$\sin 3\theta = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta$$
5. **Substitute the double angle formulas:**
$$\sin 2\theta = 2 \sin \theta \cos \theta$$
$$\cos 2\theta = 1 - 2 \sin^2 \theta$$
6. **Replace in the expression:**
$$\sin 3\theta = (2 \sin \theta \cos \theta) \cos \theta + (1 - 2 \sin^2 \theta) \sin \theta$$
7. **Simplify:**
$$\sin 3\theta = 2 \sin \theta \cancel{\cos \theta \cos \theta}^{\cos^2 \theta} + \sin \theta - 2 \sin^3 \theta$$
8. **Rewrite:**
$$\sin 3\theta = 2 \sin \theta \cos^2 \theta + \sin \theta - 2 \sin^3 \theta$$
9. **Use the Pythagorean identity:**
$$\cos^2 \theta = 1 - \sin^2 \theta$$
10. **Substitute:**
$$\sin 3\theta = 2 \sin \theta (1 - \sin^2 \theta) + \sin \theta - 2 \sin^3 \theta$$
11. **Distribute:**
$$\sin 3\theta = 2 \sin \theta - 2 \sin^3 \theta + \sin \theta - 2 \sin^3 \theta$$
12. **Combine like terms:**
$$\sin 3\theta = (2 \sin \theta + \sin \theta) - (2 \sin^3 \theta + 2 \sin^3 \theta)$$
$$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$$
**Final answer:**
$$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$$
This completes the proof using double angle formulas.