Review the proof of de Moivre’s theorem. 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) + i sin(kθ)] ∙ [cos(θ) + i sin(θ)]C= cos(kθ)cos(θ) − sin(kθ)sin(θ) + i [sin(kθ)cos(θ) + cos(kθ)sin(θ)]D= ?E= cos[(k + 1)θ] + i sin[(k + 1)θ]
Answer
A
cos(kθ – θ) + i sin(kθ – θ)
B
cos(kθ – θ) + i sin(kθ + θ)
C
cos(kθ + θ) + i sin(kθ – θ)
D
cos(kθ + θ) + i sin(kθ + θ)
Explanation
Apply the angle-addition identities to line C: cos(a)cos(b) − sin(a)sin(b) = cos(a + b), and sin(a)cos(b) + cos(a)sin(b) = sin(a + b). With a = kθ and b = θ, line C becomes cos(kθ + θ) + i sin(kθ + θ). This is choice D and matches line E because kθ + θ = (k + 1)θ.
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