Review the proof of de Moivre’s theorem (not in order).

Proof of de Moivre's Theorem
[cos(θ) + isin(θ)]k + 1
A = [cos(θ) + isin(θ)]k ∙ [cos(θ) + isin(θ)]1
B = cos(kθ + θ) + isin(kθ + θ)
C = cos(kθ)cos(θ) − sin(kθ)sin(θ) + i[sin(kθ)cos(θ) + cos(kθ)sin(θ)]
D = [cos(kθ) + isin(kθ)] ∙ [cos(θ) + isin(θ)]
E = cos[(k + 1)θ] + isin[(k + 1)θ]
Which steps must be switched to put the proof in order?

steps B and C
steps B and D
steps C and D
steps C and E

Answer:

steps B and D

Step-by-step explanation:

the correct chart is below :)

The steps which must be switched to put the proof in order are steps B and D

Since [cos(θ) + isin(θ)]k + 1

= [cos(θ) + isin(θ)]k ∙ [cos(θ) + isin(θ)]1

= [cos(kθ) + isin(kθ)] ∙ [cos(θ) + isin(θ)]

= cos(kθ)cos(θ) − sin(kθ)sin(θ) + i[sin(kθ)cos(θ) + cos(kθ)sin(θ)]

= cos(kθ + θ) + isin(kθ + θ)

= cos[(k + 1)θ] + isin[(k + 1)θ]

Since the step after A is [cos(kθ) + isin(kθ)] ∙ [cos(θ) + isin(θ)] = D, and the step after C is cos(kθ + θ) + isin(kθ + θ) = B.

So, steps B and D must be switched.

The steps which must be switched to put the proof in order are steps B and D.

Learn more about De Moivre's theorem here:

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