Complex Numbers And Quadratic Equations question 706
Question: What is $ {{[ \frac{\sin \frac{\pi }{6}+i( 1-\cos \frac{\pi }{6} )}{\sin \frac{\pi }{6}-i( 1-\cos \frac{\pi }{6} )} ]}^{3}} $ where $ i=\sqrt{-1}, $ equal to?
Options:
A) 1
B) $ -1 $
C) $ i $
D) $ -i $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{[ \frac{\sin \frac{\pi }{6}+i( 1-\cos \frac{\pi }{6} )}{\sin \frac{\pi }{6}-i( 1-\cos \frac{\pi }{6} )} ]}^{3}} $ $ ={{[ \frac{2\sin \frac{\pi }{12}\cos \frac{\pi }{12}+i( 2{{\sin }^{2}}\frac{\pi }{12} )}{2\sin \frac{\pi }{12}\cos \frac{\pi }{12}-i( 2{{\sin }^{2}}\frac{\pi }{12} )} ]}^{3}} $ $ ={{[ \frac{\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}}{\cos \frac{\pi }{12}-i\sin \frac{\pi }{12}} ]}^{3}}={{( \frac{{e^{i\frac{\pi }{12}}}}{{e^{-i\frac{\pi }{12}}}} )}^{3}} $ $ ={{( {e^{i\frac{\pi }{6}}} )}^{3}}={e^{i\times 3\times \frac{\pi }{6}}}={e^{i\frac{\pi }{2}}} $ $ =\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i $
BETA
