The given proof of De Moivre's theorem is related to the operations of
complex numbers.
The Correct Responses;
- 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;
[tex]\mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1}}[/tex]
- Step A: By laws of indices, we have;
[tex]\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]}[/tex]
[tex]\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]}[/tex]
- Step B: By expanding, we have;
[tex]\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][/tex]
- 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;
[tex]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)}[/tex]
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