Question

What is the correct justification for the indicated steps?

Answer 1

Answer:

Step A: Product of Powers

Step C: Distributive Property

Step D: Trigonometric Sum Identity

Step-by-step explanation:

Correct on Edge :)

Answer 2

The given proof of De Moivre's theorem is related to the operations of

complex numbers.

The Correct Responses;

• Step A: Laws of indices

• Step C: Expanding and collecting like terms

• Step D: Trigonometric formula for the cosine and sine of the sum of two numbers

Reasons that make the above selection correct;

The given proof is presented as follows;

$\mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1}}$

• Step A: By laws of indices, we have;

$\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1} = \mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}$

$\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = \mathbf{\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}$

• Step B: By expanding, we have;

$\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right]$

• Step D: From trigonometric addition formula, we have;

cos(A + B) = cos(A)·cos(B) - sin(A)·sin(B)

sin(A + B) = sin(A)·cos(B) + sin(B)·cos(A)

Therefore;

$cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right] = \mathbf{ cos(k \cdot \theta + \theta) + i \cdot sin(k \cdot \theta + \theta)}$

Learn more about  complex numbers here:

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