<單選題>
已知\(y = g(x) = 2\cos 2x\)向右平移\(\theta\)後，可和\(y = f(x) = \sin 2x + \sqrt{3} \cos 2x\)重合，其中\(\theta\)為何？
(1) 0
(2) \(\frac{\pi}{12}\)
(3) \(\frac{\pi}{6}\)
(4) \(\frac{\pi}{3}\)
(5) \(\frac{\pi}{2}\)
答案
(2)
先將\(f(x)\)化為單一三角函數：\(f(x) = 2\sin(2x + \frac{\pi}{3})\)。題意為將\(g(x) = 2\cos 2x = 2\sin(2x + \frac{\pi}{2})\)向右平移\(\theta\)後得到\(2\sin(2(x-\theta) + \frac{\pi}{2}) = 2\sin(2x - 2\theta + \frac{\pi}{2})\)。令此式等於\(2\sin(2x + \frac{\pi}{3})\)，比較相位：\(-2\theta + \frac{\pi}{2} = \frac{\pi}{3} + 2k\pi\)，取最小正角解得\(\theta = \frac{\pi}{12}\)。
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