Question
Use DeMoivre’s Theorem to evaluate the expression left bracket start root 3 end root left parenthesis cosine start fraction 5 pi over 12 end fraction plus lower I sine start fraction 5 pi over 12 end fraction right parenthesis right bracket superscript 4 baseline. Write the answer in rectangular form.
(1 point)
Responses
negative start fraction 9 start root 3 end root over 2 end fraction plus start fraction 9 over 2 end fraction lower i
Image with alt text: negative start fraction 9 start root 3 end root over 2 end fraction plus start fraction 9 over 2 end fraction lower i
start fraction 9 over 2 end fraction minus start fraction 9 start root 3 end root over 2 end fraction lower i
Image with alt text: start fraction 9 over 2 end fraction minus start fraction 9 start root 3 end root over 2 end fraction lower i
start fraction 9 over 2 end fraction plus start fraction 9 start root 3 end root over 2 end fraction lower i
Image with alt text: start fraction 9 over 2 end fraction plus start fraction 9 start root 3 end root over 2 end fraction lower i
negative start fraction 9 start root 3 end root over 2 end fraction minus start fraction 9 over 2 end fraction lower i
(1 point)
Responses
negative start fraction 9 start root 3 end root over 2 end fraction plus start fraction 9 over 2 end fraction lower i
Image with alt text: negative start fraction 9 start root 3 end root over 2 end fraction plus start fraction 9 over 2 end fraction lower i
start fraction 9 over 2 end fraction minus start fraction 9 start root 3 end root over 2 end fraction lower i
Image with alt text: start fraction 9 over 2 end fraction minus start fraction 9 start root 3 end root over 2 end fraction lower i
start fraction 9 over 2 end fraction plus start fraction 9 start root 3 end root over 2 end fraction lower i
Image with alt text: start fraction 9 over 2 end fraction plus start fraction 9 start root 3 end root over 2 end fraction lower i
negative start fraction 9 start root 3 end root over 2 end fraction minus start fraction 9 over 2 end fraction lower i
Answers
GPT-4o mini
To evaluate the expression \(\left( \sqrt{3} \left( \cos\frac{5\pi}{12} + i \sin\frac{5\pi}{12} \right) \right)^4\) using DeMoivre's Theorem, we can start by applying the theorem which states that:
\[
(r(\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta))
\]
Here, \(r = \sqrt{3}\), \(\theta = \frac{5\pi}{12}\), and \(n = 4\).
1. Calculate \(r^n\):
\[
r^n = (\sqrt{3})^4 = 3^2 = 9
\]
2. Calculate \(n\theta\):
\[
n\theta = 4 \cdot \frac{5\pi}{12} = \frac{20\pi}{12} = \frac{5\pi}{3}
\]
3. Now, evaluate \(\cos\left(\frac{5\pi}{3}\right)\) and \(\sin\left(\frac{5\pi}{3}\right)\):
Since \(\frac{5\pi}{3}\) is in the fourth quadrant:
\[
\cos\left(\frac{5\pi}{3}\right) = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}
\]
\[
\sin\left(\frac{5\pi}{3}\right) = \sin\left(2\pi - \frac{\pi}{3}\right) = \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}
\]
4. Now we can combine everything:
\[
\left(\sqrt{3} \left( \cos\frac{5\pi}{12} + i \sin\frac{5\pi}{12} \right) \right)^4 = 9 \left( \cos\left(\frac{5\pi}{3}\right) + i \sin\left(\frac{5\pi}{3}\right) \right)
\]
\[
= 9 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 9 \left( \frac{1}{2} \right) + 9 \left( -i \frac{\sqrt{3}}{2} \right)
\]
\[
= \frac{9}{2} - \frac{9\sqrt{3}}{2}i
\]
Thus, the final result in rectangular form is:
\[
\frac{9}{2} - \frac{9\sqrt{3}}{2} i
\]
The correct response is:
**start fraction 9 over 2 end fraction minus start fraction 9 start root 3 end fraction over 2 end fraction lower i**
\[
(r(\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta))
\]
Here, \(r = \sqrt{3}\), \(\theta = \frac{5\pi}{12}\), and \(n = 4\).
1. Calculate \(r^n\):
\[
r^n = (\sqrt{3})^4 = 3^2 = 9
\]
2. Calculate \(n\theta\):
\[
n\theta = 4 \cdot \frac{5\pi}{12} = \frac{20\pi}{12} = \frac{5\pi}{3}
\]
3. Now, evaluate \(\cos\left(\frac{5\pi}{3}\right)\) and \(\sin\left(\frac{5\pi}{3}\right)\):
Since \(\frac{5\pi}{3}\) is in the fourth quadrant:
\[
\cos\left(\frac{5\pi}{3}\right) = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}
\]
\[
\sin\left(\frac{5\pi}{3}\right) = \sin\left(2\pi - \frac{\pi}{3}\right) = \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}
\]
4. Now we can combine everything:
\[
\left(\sqrt{3} \left( \cos\frac{5\pi}{12} + i \sin\frac{5\pi}{12} \right) \right)^4 = 9 \left( \cos\left(\frac{5\pi}{3}\right) + i \sin\left(\frac{5\pi}{3}\right) \right)
\]
\[
= 9 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 9 \left( \frac{1}{2} \right) + 9 \left( -i \frac{\sqrt{3}}{2} \right)
\]
\[
= \frac{9}{2} - \frac{9\sqrt{3}}{2}i
\]
Thus, the final result in rectangular form is:
\[
\frac{9}{2} - \frac{9\sqrt{3}}{2} i
\]
The correct response is:
**start fraction 9 over 2 end fraction minus start fraction 9 start root 3 end fraction over 2 end fraction lower i**