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Q.

The value of 1+sin2π9+icos2π91+sin2π9icos2π93 is

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a

12(1i3)

b

12(3+i)

c

12(3i)

d

12(1i3)

answer is A.

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Detailed Solution

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E=(1+sin2π9+icos2π91+sin2π9icos2π9)3       =[1+cos(π22π9)+isin(π22π9)cos(π4π9)1+cos(π42π9)+i2sin(π4π9)cos(π4π9)]3         =[2cos(π4π9){cos(π4π9)+isin(π4π9)}2cos(π4π9){cos(π4π9)isin(π4π9)}]           =[cis{3(3π4π9)}{3(π4π9)}]         =cis5π6+isin5π6       =12(3i)

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