Answer
De Moivre's Theorem states that for any real number \( \theta \) and any integer \( n \):
\[
(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)
\]
**Proof:**
1. **Base Case (n = 1):**
\[
(\cos \theta + i \sin \theta)^1 = \cos \theta + i \sin \theta
\]
This is true.
2. **Inductive Step:**
Assume the theorem holds for \( n \):
\[
(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)
\]
For \( n + 1 \):
\[
(\cos \theta + i \sin \theta)^{n+1} = (\cos(n\theta) + i \sin(n\theta))(\cos \theta + i \sin \theta)
\]
Expanding:
\[
= \cos(n\theta)\cos \theta - \sin(n\theta)\sin \theta + i(\sin(n\theta)\cos \theta + \cos(n\theta)\sin \theta)
\]
Using angle addition formulas:
\[
= \cos((n+1)\theta) + i \sin((n+1)\theta)
\]
3. **Conclusion:**
By induction, De Moivre's Theorem is true for all integers \( n \).
This completes the proof.
Solution
To prove De Moivre's Theorem, we start with the statement of the theorem itself:
**De Moivre's Theorem** states that for any real number \( \theta \) and any integer \( n \):
\[
(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)
\]
### Proof:
1. **Base Case (n = 1)**:
For \( n = 1 \):
\[
(\cos \theta + i \sin \theta)^1 = \cos \theta + i \sin \theta
\]
This is trivially true.
2. **Inductive Step**:
Assume the theorem holds for some integer \( n \), i.e.,
\[
(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)
\]
We need to show it holds for \( n + 1 \):
\[
(\cos \theta + i \sin \theta)^{n+1} = (\cos \theta + i \sin \theta)^n (\cos \theta + i \sin \theta)
\]
By the inductive hypothesis, we can substitute:
\[
= (\cos(n\theta) + i \sin(n\theta))(\cos \theta + i \sin \theta)
\]
3. **Expanding the Right Side**:
Using the distributive property (FOIL method):
\[
= \cos(n\theta)\cos \theta + i \cos(n\theta)\sin \theta + i \sin(n\theta)\cos \theta - \sin(n\theta)\sin \theta
\]
Rearranging gives:
\[
= (\cos(n\theta)\cos \theta - \sin(n\theta)\sin \theta) + i(\sin(n\theta)\cos \theta + \cos(n\theta)\sin \theta)
\]
4. **Using Angle Addition Formulas**:
We recognize the expressions:
- The real part:
\[
\cos(n\theta + \theta) = \cos(n\theta)\cos \theta - \sin(n\theta)\sin \theta
\]
- The imaginary part:
\[
\sin(n\theta + \theta) = \sin(n\theta)\cos \theta + \cos(n\theta)\sin \theta
\]
5. **Combining Results**:
Thus, we have:
\[
(\cos \theta + i \sin \theta)^{n+1} = \cos((n+1)\theta) + i \sin((n+1)\theta)
\]
6. **Conclusion**:
By the principle of mathematical induction, De Moivre's Theorem holds for all integers \( n \).
This completes the proof of De Moivre's Theorem.
Reviewed and approved by the UpStudy tutoring team
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