Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To determine which expression is equivalent to
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5, \][/tex]
we will apply De Moivre's Theorem, which states that for a complex number in the form [tex]\( r(\cos \theta + i \sin \theta) \)[/tex], its [tex]\( n \)[/tex]-th power is given by
[tex]\[ [r(\cos \theta + i \sin \theta)]^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right). \][/tex]
Given the expression, we set:
[tex]\[ z = \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \][/tex]
We want to find [tex]\( z^5 \)[/tex]:
[tex]\[ z^5 = \left( \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \right)^5 \][/tex]
According to De Moivre's Theorem:
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos \left( 5 \cdot \frac{\pi}{5} \right) + i \sin \left( 5 \cdot \frac{\pi}{5} \right) \right) \][/tex]
Simplifying the terms inside the cosine and sine functions:
[tex]\[ 5 \cdot \frac{\pi}{5} = \pi \][/tex]
Thus,
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Next, we compute [tex]\( \left( \frac{1}{2} \right)^5 \)[/tex]:
[tex]\[ \left( \frac{1}{2} \right)^5 = \frac{1}{32} \][/tex]
Now, we substitute back into the expression:
[tex]\[ z^5 = \frac{1}{32} \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Since [tex]\( \cos(\pi) = -1 \)[/tex] and [tex]\( \sin(\pi) \approx 0 \)[/tex], we have:
[tex]\[ \cos(\pi) = -1 \][/tex]
[tex]\[ \sin(\pi) = 0 \][/tex]
Thus, the expression simplifies to:
[tex]\[ z^5 = \frac{1}{32} \left( -1 + 0i \right) = -\frac{1}{32} \][/tex]
Therefore, the expression equivalent to
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5 \][/tex]
is:
[tex]\[ \frac{1}{32} [\cos (\pi) + i \sin (\pi)] \][/tex]
So, the answer is:
[tex]\[ \boxed{\frac{1}{32} [\cos (\pi) + i \sin (\pi)]} \][/tex]
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5, \][/tex]
we will apply De Moivre's Theorem, which states that for a complex number in the form [tex]\( r(\cos \theta + i \sin \theta) \)[/tex], its [tex]\( n \)[/tex]-th power is given by
[tex]\[ [r(\cos \theta + i \sin \theta)]^n = r^n \left( \cos(n \theta) + i \sin(n \theta) \right). \][/tex]
Given the expression, we set:
[tex]\[ z = \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \][/tex]
We want to find [tex]\( z^5 \)[/tex]:
[tex]\[ z^5 = \left( \frac{1}{2} \left( \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right) \right) \right)^5 \][/tex]
According to De Moivre's Theorem:
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos \left( 5 \cdot \frac{\pi}{5} \right) + i \sin \left( 5 \cdot \frac{\pi}{5} \right) \right) \][/tex]
Simplifying the terms inside the cosine and sine functions:
[tex]\[ 5 \cdot \frac{\pi}{5} = \pi \][/tex]
Thus,
[tex]\[ z^5 = \left( \frac{1}{2} \right)^5 \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Next, we compute [tex]\( \left( \frac{1}{2} \right)^5 \)[/tex]:
[tex]\[ \left( \frac{1}{2} \right)^5 = \frac{1}{32} \][/tex]
Now, we substitute back into the expression:
[tex]\[ z^5 = \frac{1}{32} \left( \cos(\pi) + i \sin(\pi) \right) \][/tex]
Since [tex]\( \cos(\pi) = -1 \)[/tex] and [tex]\( \sin(\pi) \approx 0 \)[/tex], we have:
[tex]\[ \cos(\pi) = -1 \][/tex]
[tex]\[ \sin(\pi) = 0 \][/tex]
Thus, the expression simplifies to:
[tex]\[ z^5 = \frac{1}{32} \left( -1 + 0i \right) = -\frac{1}{32} \][/tex]
Therefore, the expression equivalent to
[tex]\[ \left(\frac{1}{2}\left[\cos \left(\frac{\pi}{5}\right)+i \sin \left(\frac{\pi}{5}\right)\right]\right)^5 \][/tex]
is:
[tex]\[ \frac{1}{32} [\cos (\pi) + i \sin (\pi)] \][/tex]
So, the answer is:
[tex]\[ \boxed{\frac{1}{32} [\cos (\pi) + i \sin (\pi)]} \][/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.