The value of 1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π93is

E=1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π93=1+cos⁡π2−2π9+isin⁡π2−2π9cos⁡π4−π91+cos⁡π4−2π9+i2sin⁡π4−π9cos⁡π4−π93=2cos⁡π4−π9cos⁡π4−π9+isin⁡π4−π92cos⁡π4−π9cos⁡π4−π9−isin⁡π4−π9=cis⁡33π4−π9−3π4−π9=cis⁡5π6+isin⁡5π6=−12(3−i)

The value of 1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π93is

E=1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π93=1+cos⁡π2−2π9+isin⁡π2−2π9cos⁡π4−π91+cos⁡π4−2π9+i2sin⁡π4−π9cos⁡π4−π93=2cos⁡π4−π9cos⁡π4−π9+isin⁡π4−π92cos⁡π4−π9cos⁡π4−π9−isin⁡π4−π9=cis⁡33π4−π9−3π4−π9=cis⁡5π6+isin⁡5π6=−12(3−i)

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The value of ${(\frac{1 + \sin \frac{2 π}{9} + icos \frac{2 π}{9}}{1 + \sin \frac{2 π}{9} - icos \frac{2 π}{9}})}^{3}$ is

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$- \frac{1}{2} (\sqrt{3} - i)$

$\frac{1}{2} (\sqrt{3} + i)$

$\frac{1}{2} (1 - i \sqrt{3})$

$- \frac{1}{2} (1 - i \sqrt{3})$

answer is A.

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Detailed Solution

$E = {(\frac{1 + \sin \frac{2 π}{9} + icos \frac{2 π}{9}}{1 + \sin \frac{2 π}{9} - icos \frac{2 π}{9}})}^{3}$
$= {[\frac{1 + \cos (\frac{π}{2} - \frac{2 π}{9}) + i \sin (\frac{π}{2} - \frac{2 π}{9}) \cos (\frac{π}{4} - \frac{π}{9})}{1 + \cos (\frac{π}{4} - \frac{2 π}{9}) + i 2 \sin (\frac{π}{4} - \frac{π}{9}) \cos (\frac{π}{4} - \frac{π}{9})}]}^{3}$
$\begin{array}{l} = [\frac{2 \cos (\frac{π}{4} - \frac{π}{9}) \{\cos (\frac{π}{4} - \frac{π}{9}) + isin (\frac{π}{4} - \frac{π}{9})\}}{2 \cos (\frac{π}{4} - \frac{π}{9}) \{\cos (\frac{π}{4} - \frac{π}{9}) - isin (\frac{π}{4} - \frac{π}{9})\}}] \\ = [cis \{3 (\frac{3 π}{4} - \frac{π}{9})\} - \{3 (\frac{π}{4} - \frac{π}{9})\}] \\ = cis \frac{5 π}{6} + isin \frac{5 π}{6} \\ = - \frac{1}{2} (\sqrt{3} - i) \end{array}$