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Question

Find the value 6k=1(sin2πk7cos2πk7)

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Solution

6k=1(sin2πk7)6k=1(cos2πk7)
=6k=0sin2πk76k=0cos2πk7+1
=6k=0(Sum of imaginary part of seven seventh roots of unity)-6k=0(Sum of real part of seven seventh roots of unity)+1
=00+1=1

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