Question

Which of the following is NOT
\[\sqrt{8(\cos \frac{ 11\pi }{ 6 } + i \sin \frac{ 11\pi }{ 6 })}\]

a)
\[2^{3/2}(\cos \frac{ 11\pi }{ 12 } + \sin \frac{ 11\pi }{ 12 })\]
b)
\[2^{3/2}(-\cos \frac{ \pi }{ 12 } +i \sin \frac{ \pi }{ 12 })\]
c)
\[8^{1/2}(\cos \frac{ 23\pi }{ 12 } +i \sin \frac{ 23\pi }{ 12 })\]
d)
\[8^{1/2}(\cos \frac{ \pi }{ 12 } -i \sin \frac{ \pi }{ 12 })\]
e)
\[2^{3/2}(\cos \frac{23 \pi }{ 12 } -i \sin \frac{ 23\pi }{ 12 })\]

Answer 1

r u doin a test?

Answer 2

Exam prep.

Answer 3

We can't help u sry...

Answer 4

No problem.

Answer 5

Have a great day sir

Answer 6

Thank you, you 2!

Answer 7

use DeMoivre's theorem
\[ \large [\rho(\cos\phi+i\sin\phi)]^n=\rho^n(\cos(n\phi)+i\sin(n\phi)) \]

Answer 8

But what is the n value in this case? since it is under a square root?

Answer 9

n=1/2

Answer 10

May I ask how you find this?

Answer 11

\[ \large \sqrt{8(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})}=
\left(8(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})\right)^{1/2} \]
\[ \large =8^{1/2}(\cos\frac{1}{2}\frac{11\pi}{6}+i\sin\frac{1}{2}\frac{11\pi}{6}) \]

Answer 12

Thus option e is the correct one. (The one that is not equal to the above). Since it doesn't give -2?

Answer 13

\[ \large =(2^3)^{1/2}(\cos\frac{11\pi}{12}+i\sin\frac{11\pi}{12}) \]

Answer 14

@ helder, I really don't catch on as to what you are trying to tell me now. Don't you agree with my answer?

Answer 15

sorry. this site aparently crashed. and i had to go to work.

the correct answer is A.