What is the justification for step B?
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 = cos(kθ + θ) + i sin(kθ + θ)
E = cos[(k + 1)θ] + i sin[(k + 1)θ]
A) distributive property
B) factoring
C) multiplication rule
D) assumption (for n = k step)
Answers
Given : 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 = cos(kθ + θ) + i sin(kθ + θ)
E = cos[(k + 1)θ] + i sin[(k + 1)θ]
To Find : justification for step B
A) distributive property
B) factoring
C) multiplication rule
D) assumption (for n = k step)
Solution:
A = [cos(θ) + i sin(θ)]^k ∙ [cos(θ) + i sin(θ)]^1
here law of exponent used
aⁿ⁺¹ = aⁿ.a¹
B = [cos(kθ) + i sin(kθ)] ∙ [cos(θ) + i sin(θ)]
Here [cos(θ) + i sin(θ)]^k = [cos(kθ) + i sin(kθ)]
is assumption for n=k step
[cos(θ) + i sin(θ)]^1 = [cos(θ) + i sin(θ)] is property a¹ = a
justification for step B is assumption for n=k step
as Mathematical Induction is being used
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Answer:
D. assumption (for n = k step)
Step-by-step explanation:
edg 22
