Question

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?

Answer 1

Answer: D. cos(kθ + θ) + i sin(kθ + θ)

Explanation:

The screenshot pretty much speaks for itself, but you can see that the proof ends with cos[(k + 1)θ] + i sin[(k + 1)θ]. That means that addition is happening in Step D, and that the expressions in the cosine and sine parts are the same. So, the correct option is D. Hope that helps!

Answer 2

D

Explanation:

Just took it.