Question

Find left-bracket 2 left-parenthesis cosine 240 degree-sign plus i sine 240 degree-sign right-parenthesis right-bracket superscript 4 baseline.
(1 point)
Responses

8 left-parenthesis cosine 240 degree-sign plus i sine 240 degree-sign right-parenthesis
Image with alt text: 8 left-parenthesis cosine 240 degree-sign plus i sine 240 degree-sign right-parenthesis

16 left-parenthesis cosine 240 degree-sign plus i sine 240 degree-sign right-parenthesis
Image with alt text: 16 left-parenthesis cosine 240 degree-sign plus i sine 240 degree-sign right-parenthesis

8 left-parenthesis cosine 60 degree-sign plus i sine 60 degree-sign right-parenthesis
Image with alt text: 8 left-parenthesis cosine 60 degree-sign plus i sine 60 degree-sign right-parenthesis

16 left-parenthesis cosine 60 degree-sign plus i sine 60 degree-sign right-parenthesis

Answer 1

To solve the expression \( [2(\cos 240^\circ + i \sin 240^\circ)]^4 \), we can start by using De Moivre's Theorem, which states that:

\[
[r(\cos \theta + i \sin \theta)]^n = r^n (\cos(n\theta) + i \sin(n\theta))
\]

Here, \( r = 2 \), \( \theta = 240^\circ \), and \( n = 4 \).

1. **Calculate \( r^n \)**:
\[
r^4 = 2^4 = 16
\]

2. **Calculate \( n\theta \)**:
\[
n\theta = 4 \times 240^\circ = 960^\circ
\]
Now we need to find an equivalent angle for \( 960^\circ \) within the range of \( [0^\circ, 360^\circ] \).

To find this, we can subtract \( 360^\circ \) until we are within that range:
\[
960^\circ - 2 \times 360^\circ = 960^\circ - 720^\circ = 240^\circ
\]

3. **Using De Moivre's Theorem**:
\[
[2(\cos 240^\circ + i \sin 240^\circ)]^4 = 16 (\cos 960^\circ + i \sin 960^\circ) = 16 (\cos 240^\circ + i \sin 240^\circ)
\]

Thus, we find:
\[
[2(\cos 240^\circ + i \sin 240^\circ)]^4 = 16(\cos 240^\circ + i \sin 240^\circ)
\]

The correct response is:
\[
\boxed{16(\cos 240^\circ + i \sin 240^\circ)}
\]