Using de Moivre's theorem, evaluate (cos(pi/6) + i*sin(pi/6))^6.
Answer and explanation
Correct answer: D. -1, multiplying argument by 6 to yield cos(pi) + i*sin(pi) = -1 + 0i
- Step 1: By de Moivre's theorem, [cos(theta) + isin(theta)]^n = cos(ntheta) + isin(ntheta). Step 2. Here n = 6 and theta = pi/6: 6(pi/6) = pi. Thus cos(pi) + isin(pi) = -1 + 0 = -1. *Answer: -1, multiplying argument by 6 to yield cos(pi) + isin(pi) = -1 + 0i. Why not the others:** The angle rotates to pi on the real axis, which equals -1.
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