Review the proof of de Moivre’s theorem (not in order). Proof of de Moivre's Theorem [cos(θ) + i sin(θ)]k + 1A= [cos(θ) + i sin(θ)]k ∙ [cos(θ) + i sin(θ)]1B= cos(kθ + θ) + i sin(kθ + θ)C= cos(kθ)cos(θ) − sin(kθ)sin(θ) + i [sin(kθ)cos(θ) + cos(kθ)sin(θ)]D= [cos(kθ) + i sin(kθ)] ∙ [cos(θ) + i sin(θ)]E= cos[(k + 1)θ] + i sin[(k + 1)θ]

The proof should begin by rewriting the power as a product (step A). Then use the inductive hypothesis to replace [cos(θ) + i sin(θ)]^k with cos(kθ) + i sin(kθ), which is step D. Expanding gives step C, applying the angle-sum identities gives step B, and simplifying kθ + θ to (k + 1)θ gives step E. Thus steps B and D need to switch places; choice B is “steps B and D.”

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