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]