The value of 1+sin2π9+icos2π91+sin2π9−icos2π93 is

E=(1+sin2π9+icos2π91+sin2π9−icos2π9)3 =[1+cos(π2−2π9)+isin(π2−2π9)cos(π4−π9)1+cos(π4−2π9)+i2sin(π4−π9)cos(π4−π9)]3 =[2cos(π4−π9){cos(π4−π9)+isin(π4−π9)}2cos(π4−π9){cos(π4−π9)−isin(π4−π9)}] =[cis{3(3π4−π9)}−{3(π4−π9)}] =cis5π6+isin5π6 =−12(3−i)

The value of 1+sin2π9+icos2π91+sin2π9−icos2π93 is

E=(1+sin2π9+icos2π91+sin2π9−icos2π9)3 =[1+cos(π2−2π9)+isin(π2−2π9)cos(π4−π9)1+cos(π4−2π9)+i2sin(π4−π9)cos(π4−π9)]3 =[2cos(π4−π9){cos(π4−π9)+isin(π4−π9)}2cos(π4−π9){cos(π4−π9)−isin(π4−π9)}] =[cis{3(3π4−π9)}−{3(π4−π9)}] =cis5π6+isin5π6 =−12(3−i)

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

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

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

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

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

answer is A.

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Detailed Solution

$E = {(\frac{1 + \sin \frac{2 π}{9} + i \cos \frac{2 π}{9}}{1 + \sin \frac{2 π}{9} - i \cos \frac{2 π}{9}})}^{3} = {[\frac{1 + \cos (\frac{π}{2} - \frac{2 π}{9}) + i \sin (\frac{π}{2} - \frac{2 π}{9}) \cos (\frac{π}{4} - \frac{π}{9})}{1 + \cos (\frac{π}{4} - \frac{2 π}{9}) + i 2 \sin (\frac{π}{4} - \frac{π}{9}) \cos (\frac{π}{4} - \frac{π}{9})}]}^{3} = [\frac{2 \cos (\frac{π}{4} - \frac{π}{9}) {\cos (\frac{π}{4} - \frac{π}{9}) + i \sin (\frac{π}{4} - \frac{π}{9})}}{2 \cos (\frac{π}{4} - \frac{π}{9}) {\cos (\frac{π}{4} - \frac{π}{9}) - i \sin (\frac{π}{4} - \frac{π}{9})}}] = [c i s {3 (\frac{3 π}{4} - \frac{π}{9})} - {3 (\frac{π}{4} - \frac{π}{9})}] = c i s \frac{5 π}{6} + i \sin \frac{5 π}{6} = - \frac{1}{2} (\sqrt{3} - i)$