1. **State the problem:** Prove the trigonometric identity $$\cos 4x = 8\sin^4 x - 8\sin^2 x + 1$$. 2. **Recall the double-angle and power-reduction formulas:** - $$\cos 2x = 1 - 2\sin^2 x$$ - $$\cos 4x = 2\cos^2 2x - 1$$ 3. **Express $$\cos 4x$$ in terms of $$\sin x$$:** Start with $$\cos 4x = 2\cos^2 2x - 1$$. 4. Substitute $$\cos 2x = 1 - 2\sin^2 x$$: $$\cos 4x = 2(1 - 2\sin^2 x)^2 - 1$$. 5. Expand the square: $$\cos 4x = 2(1 - 4\sin^2 x + 4\sin^4 x) - 1$$. 6. Distribute the 2: $$\cos 4x = 2 - 8\sin^2 x + 8\sin^4 x - 1$$. 7. Simplify constants: $$\cos 4x = 1 - 8\sin^2 x + 8\sin^4 x$$. 8. Rearrange terms to match the right side of the identity: $$\cos 4x = 8\sin^4 x - 8\sin^2 x + 1$$. **Final answer:** The identity is proven as $$\cos 4x = 8\sin^4 x - 8\sin^2 x + 1$$.