Question

Use DeMoivre’s Theorem to find left parenthesis 4 cis start fraction pi over 18 end fraction right parenthesis cubed.
(1 point)
Responses

32 plus 32 start root 3 end root lower i
Image with alt text: 32 plus 32 start root 3 end root lower i

32 start root 3 end root plus 32 lower i
Image with alt text: 32 start root 3 end root plus 32 lower i

6 plus 6 start root 3 end root lower i
Image with alt text: 6 plus 6 start root 3 end root lower i

6 start root 3 end root plus 6 lower i
Image with alt text: 6 start root 3 end root plus 6 lower i

Answer 1

To solve \( \left( 4 \text{cis} \frac{\pi}{18} \right)^3 \) using DeMoivre's Theorem, we first note that DeMoivre's theorem states:

\[
(r \text{cis} \theta)^n = r^n \text{cis}(n\theta)
\]

In this case:

- \( r = 4 \)
- \( \theta = \frac{\pi}{18} \)
- \( n = 3 \)

Now we apply the theorem:

1. Calculate \( r^n \):
\[
r^n = 4^3 = 64
\]

2. Calculate \( n \theta \):
\[
n\theta = 3 \times \frac{\pi}{18} = \frac{\pi}{6}
\]

Putting it all together, we have:
\[
\left( 4 \text{cis} \frac{\pi}{18} \right)^3 = 64 \text{cis} \frac{\pi}{6}
\]

Now we need to express \( 64 \text{cis} \frac{\pi}{6} \) in rectangular form. Recall that \( \text{cis} \theta = \cos \theta + i \sin \theta \). Therefore:

\[
\text{cis} \frac{\pi}{6} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}
\]

We know:
- \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
- \( \sin \frac{\pi}{6} = \frac{1}{2} \)

Thus:
\[
\text{cis} \frac{\pi}{6} = \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}
\]

Now plug this into our expression:
\[
64 \text{cis} \frac{\pi}{6} = 64 \left( \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \right) = 64 \cdot \frac{\sqrt{3}}{2} + 64 \cdot i \cdot \frac{1}{2}
\]
\[
= 32\sqrt{3} + 32i
\]

So, the final answer is:
\[
32\sqrt{3} + 32i
\]

Therefore, the correct response is:

**32 start root 3 end root plus 32 lower i**.