1. **Problem statement:** Prove the trigonometric identity $$\sin 4x \equiv 4 \sin x (2 \cos^3 x - \cos x)$$. 2. **Recall the double-angle formulas:** - $$\sin 2x = 2 \sin x \cos x$$ - $$\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$$ 3. **Express $$\sin 4x$$ using double-angle formula:** $$\sin 4x = \sin (2 \cdot 2x) = 2 \sin 2x \cos 2x$$ 4. **Substitute the formulas for $$\sin 2x$$ and $$\cos 2x$$:** $$\sin 4x = 2 (2 \sin x \cos x)(2 \cos^2 x - 1) = 4 \sin x \cos x (2 \cos^2 x - 1)$$ 5. **Rewrite the right side of the identity:** $$4 \sin x (2 \cos^3 x - \cos x) = 4 \sin x \cos x (2 \cos^2 x - 1)$$ 6. **Compare both sides:** Both sides simplify to $$4 \sin x \cos x (2 \cos^2 x - 1)$$, so the identity holds true. **Final answer:** The identity $$\sin 4x \equiv 4 \sin x (2 \cos^3 x - \cos x)$$ is proven.