A) 1
B) \[-1\]
C) \[i\]
D) \[-i\]
Correct Answer: C
Solution :
\[{{\left[ \frac{\sin \frac{\pi }{6}+i\left( 1-\cos \frac{\pi }{6} \right)}{\sin \frac{\pi }{6}-i\left( 1-\cos \frac{\pi }{6} \right)} \right]}^{3}}\] \[=\,\,{{\left[ \frac{2\sin \,\frac{\pi }{12}\cos \frac{\pi }{12}+i\left( 2{{\sin }^{2}}\frac{\pi }{12} \right)}{2\sin \,\frac{\pi }{12}\cos \frac{\pi }{12}-i\left( 2{{\sin }^{2}}\,\frac{\pi }{12} \right)} \right]}^{3}}\] \[=\,\,{{\left[ \frac{\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}}{\cos \frac{\pi }{12}-i\sin \frac{\pi }{12}} \right]}^{3}}={{\left( \frac{{{e}^{i\frac{\pi }{12}}}}{{{e}^{-i\frac{\pi }{12}}}} \right)}^{3}}\] \[=\,\,{{\left( {{e}^{i\frac{\pi }{6}}} \right)}^{3}}={{e}^{i\,\times \,3\,\times \,\frac{\pi }{6}}}={{e}^{i\,\frac{\pi }{2}}}\] \[=\,\,\,\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i\]
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