To determine which steps must be switched to put the proof of de Moivre's Theorem in order, let's review the steps carefully:
1. We start with the expression:
[tex]\[
[\cos(\theta) + i \sin(\theta)]^{k+1}
\][/tex]
2. Using the properties of exponents, we can break this down as follows:
[tex]\[
[\cos(\theta) + i \sin(\theta)]^{k+1} = [\cos(\theta) + i \sin(\theta)]^k \cdot [\cos(\theta) + i \sin(\theta)]
\][/tex]
This is Step A.
3. Next, we apply the distributive property (expanding the product of two complex numbers):
[tex]\[
[\cos(\theta) + i \sin(\theta)]^k \cdot [\cos(\theta) + i \sin(\theta)] = [\cos(k\theta) + i \sin(k\theta)] \cdot [\cos(\theta) + i \sin(\theta)]
\][/tex]
This is Step D.
4. We then express it using the angle addition formulas for cosine and sine:
[tex]\[
[\cos(k\theta) + i \sin(k\theta)] \cdot [\cos(\theta) + i \sin(\theta)] = \cos(k\theta) \cos(\theta) - \sin(k\theta) \sin(\theta) + i[\sin(k\theta) \cos(\theta) + \cos(k\theta) \sin(\theta)]
\][/tex]
This is Step C.
5. Combining terms, we get:
[tex]\[
\cos((k+1)\theta) + i \sin((k+1)\theta)
\][/tex]
This is Step E.
6. Since we know the result of angle addition directly:
[tex]\[
= \cos(k\theta + \theta) + i \sin(k\theta + \theta)
\][/tex]
This is Step B.
So, in order for the sequence to be correct, we need to switch Step B and Step D. Specifically:
- Step A remains the same.
- Step D should come right after Step A.
- Step C follows Step D.
- Step B should come before Step C.
- Step E is the final step.
Therefore, the correct answer is:
Steps B and D.
