1. **Problem statement:** Prove that $$\cos^4 A = \frac{3}{8} + \frac{1}{2} \cos 2A + \frac{1}{8} \cos 4A.$$\n\n2. **Formula and identities used:** We use the double-angle identity for cosine: $$\cos 2A = 2\cos^2 A - 1,$$ and the power-reduction formula: $$\cos^2 A = \frac{1 + \cos 2A}{2}.$$\n\n3. **Step-by-step proof:**\n- Start with $$\cos^4 A = (\cos^2 A)^2.$$\n- Substitute the power-reduction formula: $$\cos^4 A = \left( \frac{1 + \cos 2A}{2} \right)^2 = \frac{(1 + \cos 2A)^2}{4}.$$\n- Expand the numerator: $$ (1 + \cos 2A)^2 = 1 + 2\cos 2A + \cos^2 2A.$$\n- So, $$\cos^4 A = \frac{1 + 2\cos 2A + \cos^2 2A}{4}.$$\n- Apply the power-reduction formula again to $$\cos^2 2A$$: $$\cos^2 2A = \frac{1 + \cos 4A}{2}.$$\n- Substitute back: $$\cos^4 A = \frac{1 + 2\cos 2A + \frac{1 + \cos 4A}{2}}{4} = \frac{1 + 2\cos 2A}{4} + \frac{1 + \cos 4A}{8}.$$\n- Combine terms: $$\cos^4 A = \frac{1}{4} + \frac{2\cos 2A}{4} + \frac{1}{8} + \frac{\cos 4A}{8} = \frac{1}{4} + \frac{1}{2} \cos 2A + \frac{1}{8} + \frac{1}{8} \cos 4A.$$\n- Add the constants: $$\frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}.$$\n- Final expression: $$\cos^4 A = \frac{3}{8} + \frac{1}{2} \cos 2A + \frac{1}{8} \cos 4A,$$ which is what we wanted to prove.\n\n4. **Explanation:** We used the power-reduction formula twice to express $$\cos^4 A$$ in terms of cosines of multiple angles. This method simplifies powers of cosine into sums of cosines with different arguments, making it easier to work with in integrals or series expansions.