Question 108 of 455MediumMCQ
The value of (1+sin92π−icos92π1+sin92π+icos92π)3 is:
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Official solution
\[{\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}\]
= \[{\left( {{{1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) + i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)} \over {1 + \sin \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right) - i\cos \left( {{\pi \over 2} - {{5\pi } \over {18}}} \right)}}} \right)^3}\]
= \[{\left( {{{1 + \cos \left( {{{5\pi } \over {18}}} \right) + i\sin \left( {{{5\pi } \over {18}}} \right)} \over {1 + \cos \left( {{{5\pi } \over {18}}} \right) - i\sin \left( {{{5\pi } \over {18}}} \right)}}} \right)^3}\]
= \[{\left( {{{2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) + 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)} \over {2{{\cos }^2}\left( {{{5\pi } \over {36}}} \right) - 2i\sin \left( {{{5\pi } \over {36}}} \right)\cos \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}\]
= \[{\left( {{{\cos \left( {{{5\pi } \over {36}}} \right) + i\sin \left( {{{5\pi } \over {36}}} \right)} \over {\cos \left( {{{5\pi } \over {36}}} \right) - i\sin \left( {{{5\pi } \over {36}}} \right)}}} \right)^3}\]
= \[{\left( {{{{e^{i{{5\pi } \over {36}}}}} \over {{e^{ - i{{5\pi } \over {36}}}}}}} \right)^3}\]
= \[{\left( {{e^{i{{5\pi } \over {18}}}}} \right)^3}\]
= \[{{e^{i{{5\pi } \over {18}} \times 3}}}\] = \[{{e^{i{{5\pi } \over 6}}}}\]
= \[\cos {{5\pi } \over 6} + i\sin {{5\pi } \over 6}\]
= \[ - {{\sqrt 3 } \over 2} + {i \over 2}\]
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