1. **Problem:** Verify the trigonometric identity $$\sin 3x = 3 \sin x - 4 \sin^3 x$$. 2. **Formula and rules:** Use the triple-angle formula for sine: $$\sin 3x = 3 \sin x - 4 \sin^3 x$$ is a known identity derived from angle addition formulas. 3. **Verification:** Start from the left side: $$\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x$$ 4. Substitute double-angle formulas: $$\sin 2x = 2 \sin x \cos x$$ $$\cos 2x = 1 - 2 \sin^2 x$$ 5. So, $$\sin 3x = (2 \sin x \cos x) \cos x + (1 - 2 \sin^2 x) \sin x = 2 \sin x \cos^2 x + \sin x - 2 \sin^3 x$$ 6. Factor out $\sin x$: $$\sin 3x = \sin x (2 \cos^2 x + 1 - 2 \sin^2 x)$$ 7. Use $\cos^2 x = 1 - \sin^2 x$: $$2 (1 - \sin^2 x) + 1 - 2 \sin^2 x = 2 - 2 \sin^2 x + 1 - 2 \sin^2 x = 3 - 4 \sin^2 x$$ 8. Therefore, $$\sin 3x = \sin x (3 - 4 \sin^2 x) = 3 \sin x - 4 \sin^3 x$$ 9. This matches the right side, so the identity is verified. --- 1. **Problem:** Verify the trigonometric identity $$\cos 4x = 8 \cos^4 x - 8 \cos^2 x + 1$$. 2. **Formula and rules:** Use the double-angle formula repeatedly: $$\cos 2x = 2 \cos^2 x - 1$$ 3. Express $\cos 4x$ as $\cos 2(2x)$: $$\cos 4x = 2 \cos^2 2x - 1$$ 4. Substitute $\cos 2x$: $$\cos 4x = 2 (2 \cos^2 x - 1)^2 - 1$$ 5. Expand the square: $$(2 \cos^2 x - 1)^2 = 4 \cos^4 x - 4 \cos^2 x + 1$$ 6. Substitute back: $$\cos 4x = 2 (4 \cos^4 x - 4 \cos^2 x + 1) - 1 = 8 \cos^4 x - 8 \cos^2 x + 2 - 1$$ 7. Simplify: $$\cos 4x = 8 \cos^4 x - 8 \cos^2 x + 1$$ 8. This matches the right side, so the identity is verified. **Final answers:** Both identities are true.