The value of {\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3} is :

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Answer: B. -{1 \over 2}\left( {\sqrt 3 - i} ight) {\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} ight)^3} = {\left( {{{1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} ight) + i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} ight)} \over {1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} ight) - i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} ight)}}} ight)^3} = {\left( {{{1 + \cos \left( {{{5\pi } \over {18}}} ight) + i\sin \left( {{{5\pi } \over {18}}} ight)} \over {1 + \cos \left( {{{5\pi } \over {18}}} ight) - i\sin \left( {{{5\pi } \over {18}}} ight)}}} ight)^3} = {\left( {{{2{{\cos }^2}\left( {{{5\pi } \over {36}}} ight) + 2i\sin \left( {{{5\pi } \over {36}}} ight)\cos \left( {{{5\pi } \over {36}}} ight)} \over {2{{\cos }^2}\left( {{{5\pi } \over {36}}} ight) - 2i\sin \left( {{{5\pi } \over {36}}} ight)\cos \left( {{{5\pi } \over {36}}} ight)}}} ight)^3} = {\left( {{{\cos \left( {{{5\pi } \over {36}}} ight) + i\sin \left( {{{5\pi } \over {36}}} ight)} \over {\cos \left( {{{5\pi } \over {36}}} ight) - i\sin \left( {{{5\pi } \over {36}}} ight)}}} ight)^3} = {\left( {{{{e^{i{{5\pi } \over {36}}}}} \over {{e^{ - i{{5\pi } \over {36}}}}}}} ight)^3} = {\left( {{e^{i{{5\pi } \over {18}}}}} ight)^3} = {{e^{i{{5\pi } \over {18}} 	imes 3}}} = {{e^{i{{5\pi } \over 6}}}} = \cos {{5\pi } \over 6} + i\sin {{5\pi } \over 6} = - {{\sqrt 3 } \over 2} + {i \over 2}

The value of $${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$$ is :

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Explanation

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