Step 1: Derive $\cos 5\theta$ step by step using De Moivre's theorem.
By De Moivre's theorem, $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$. Expand the left side using the binomial theorem.
Step 2: Expand $(\cos\theta + i\sin\theta)^5$.
Using $C = \cos\theta$, $S = \sin\theta$: \[ (C+iS)^5 = C^5 + 5C^4(iS) + 10C^3(iS)^2 + 10C^2(iS)^3 + 5C(iS)^4 + (iS)^5 \] Since $i^2=-1, i^3=-i, i^4=1, i^5=i$: \[ = C^5 + 5iC^4S - 10C^3S^2 - 10iC^2S^3 + 5CS^4 + iS^5 \]
Step 3: Separate real and imaginary parts.
Real part: $C^5 - 10C^3S^2 + 5CS^4 = \cos 5\theta$.
Step 4: Convert to powers of $\cos\theta$ only.
Replace $S^2 = 1 - C^2$, $S^4 = (1-C^2)^2 = 1-2C^2+C^4$: \[ \cos 5\theta = C^5 - 10C^3(1-C^2) + 5C(1-2C^2+C^4) \] \[ = C^5 - 10C^3 + 10C^5 + 5C - 10C^3 + 5C^5 \] \[ = 16C^5 - 20C^3 + 5C \]
Step 5: Match with the options.
We found $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$. This is option 4.
Step 6: State the answer.
\[ \boxed{\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta} \]