Which is the correct expansion of (\cos 3x)?

Correct answer: (4\cos^3 x-3\cos x)

The direct answer is option C: \(\cos 3x=4\cos^3x-3\cos x\). Start with the addition formula: \(\cos(2x+x)=\cos2x\cos x-\sin2x\sin x\). Using \(\cos2x=2\cos^2x-1\) and \(\sin2x=2\sin x\cos x\), we get \(\cos3x=(2\cos^2x-1)\cos x-2\sin^2x\cos x\). Since \(\sin^2x=1-\\cos^2x\), simplification gives \(4\cos^3x-3\cos x\). Option C is therefore correct. Option A has the terms reversed and equals \(-\cos3x\), not \(\cos3x\). Option B is the sine triple-angle expression, because \(\sin3x=3\sin x-4\sin^3x\), so its sign/order is also not the required cosine formula. Option D has an incorrect plus sign; the cubic and linear terms must be subtracted. Memory cue: cosine triple angle is “four cube minus three,” while sine is “three sine minus four sine-cube.”