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Mathematics, 18.03.2021 01:20 nique0808

Review the proof of de Moivre’s theorem. Proof of de Moivre's Theorem
[cos(θ) + i sin(θ)]k + 1
A = [cos(θ) + i sin(θ)]k ∙ [cos(θ) + i sin(θ)]1
B = [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]

Which expression will complete the proof?

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Answers: 3

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Review the proof of de Moivre’s theorem. Proof of de Moivre's Theorem
[cos(θ) + i sin(θ)]k +...
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