Using De Moivre's theorem, $(\cos\theta + i\sin\theta)^4$ equals:
A$\cos\theta + i\sin\theta$, the same unchanged power here
B$4(\cos\theta + i\sin\theta)$, scaled linearly without the angle factor
C$\cos(4 + \theta) + i\sin(4 + \theta)$, adding 4 to the angle
D$\cos 4\theta + i\sin 4\theta$, multiplying the angle by 4
Answer & Solution
Correct answer: D. $\cos 4\theta + i\sin 4\theta$, multiplying the angle by 4
De Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$.
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