Step 1: Use the half-angle formula instead of a sum or difference formula.
Since \(\frac{\pi}{12}\) is half of \(\frac{\pi}{6}\), use \(\cos^2\theta=\frac{1+\cos2\theta}{2}\) with \(\theta=\frac{\pi}{12}\), so \(2\theta=\frac{\pi}{6}\).
Step 2: Substitute the known value of cos(pi/6).
\[ \cos^2\frac{\pi}{12}=\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4} \]
Step 3: Take the square root and simplify the nested surd.
Since \(\frac{\pi}{12}\) is in the first quadrant, the cosine is positive: \(\cos\frac{\pi}{12}=\sqrt{\frac{2+\sqrt3}{4}}=\frac{\sqrt{2+\sqrt3}}{2}\).
Noting \(2+\sqrt3=\frac{4+2\sqrt3}{2}=\frac{(\sqrt3+1)^2}{2}\), we get \(\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}\).
Step 4: Rationalize to match the answer choices.
\[ \cos\frac{\pi}{12}=\frac{\sqrt3+1}{2\sqrt2}=\frac{(\sqrt3+1)\sqrt2}{4}=\frac{\sqrt6+\sqrt2}{4} \]
\[ \boxed{\frac{\sqrt2+\sqrt6}{4}} \]