1. The problem is to express $\cos 5x$ in terms of powers of $\cos x$ using De Moivre's theorem. 2. De Moivre's theorem states that $(\cos x + i \sin x)^n = \cos nx + i \sin nx$. For $n=5$, we have: $$\cos 5x + i \sin 5x = (\cos x + i \sin x)^5$$ 3. Expanding the right-hand side using the binomial theorem: $$(\cos x + i \sin x)^5 = \sum_{k=0}^5 \binom{5}{k} (\cos x)^{5-k} (i \sin x)^k$$ 4. Simplify powers of $i$ and separate real and imaginary parts. The real part gives $\cos 5x$: $$\cos 5x = \binom{5}{0} \cos^5 x + \binom{5}{2} \cos^3 x (i^2 \sin^2 x) + \binom{5}{4} \cos x (i^4 \sin^4 x)$$ 5. Recall that $i^2 = -1$ and $i^4 = 1$, so: $$\cos 5x = \cos^5 x - 10 \cos^3 x \sin^2 x + 5 \cos x \sin^4 x$$ 6. Use the Pythagorean identity $\sin^2 x = 1 - \cos^2 x$ to write $\cos 5x$ as a polynomial in $\cos x$ only: $$\cos 5x = \cos^5 x - 10 \cos^3 x (1 - \cos^2 x) + 5 \cos x (1 - \cos^2 x)^2$$ 7. Expand terms: $$\cos 5x = \cos^5 x - 10 \cos^3 x + 10 \cos^5 x + 5 \cos x (1 - 2 \cos^2 x + \cos^4 x)$$ 8. Further expand: $$\cos 5x = \cos^5 x - 10 \cos^3 x + 10 \cos^5 x + 5 \cos x - 10 \cos^3 x + 5 \cos^5 x$$ 9. Combine like terms: $$\cos 5x = (1 + 10 + 5) \cos^5 x - (10 + 10) \cos^3 x + 5 \cos x = 16 \cos^5 x - 20 \cos^3 x + 5 \cos x$$ 10. Final expression using De Moivre's theorem: $$\boxed{\cos 5x = 16 \cos^5 x - 20 \cos^3 x + 5 \cos x}$$