Using De Moivre's Theorem, evaluate (cos(pi/6) + i sin(pi/6))^6.
Answer and explanation
Correct answer: B. -1
- Step 1: State De Moivre's Theorem: [cos(theta) + i sin(theta)]^n = cos(n theta) + i sin(n theta).
- Step 2: Here theta = pi/6 and n = 6.
- Step 3: Compute argument: 6 * (pi / 6) = pi.
- Step 4: Evaluate: cos(pi) + i sin(pi) = -1 + i(0) = -1.
Answer
-1.
Why not the others: '1' assumes rotation completes a full 2pi cycle; 'i' calculates rotation as pi/2; '-i' calculates 3pi/2.
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