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Question

The value of 4k=1 (sin2π k7icos2π k7) is

A
-1
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B
0
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C
-i
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D
i
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Solution

The correct option is D i

Let z=cos2π7+isin2π7, then by De Moiver's theorem Zk=cos2π k7+isin2π k7
Now the given sum

s=6k=1(sin2π k7icos2π k7)

=6k=1[(i)(cos2π k7+isin2π k7)]

=(i)6k=1(cos2π k7+isin2π k7)=(i)6k=1zk

which is a G.P. of which the first term is z, number of the terms is 6 and the common

ratio is z=cos2π7+isin2π71 So summing up the G.P., we have

S=(i)z(1z6)1z=(i)zz71z=(i)z11z=i

[z7=(cos2π7+isin2π7)7=cos2π+isin2π=1


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