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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

\sum\limits_{k=1}^{6} \left( \sin \frac{2\pi k}{7} - i \cos \frac{2\pi k}{7} \right)

Solution:

The value of the expression is calculated as

\sum\limits_{k=1}^{6} \left( \sin \;\frac{2\pi k}{7} - i \cos \;\frac{2\pi k}{7} \right) \;\; = -i \sum\limits_{k=1}^{6} \left[ \cos \left( \;\frac{2\pi k}{7} \right) - \frac{1}{i} \sin \left( \;\frac{2\pi k}{7} \right) \right]

\;\; = -i \sum\limits_{k=1}^{6} \left[ \cos \left( \;\frac{2\pi k}{7} \right) + i \sin \left( \;\frac{2\pi k}{7} \right) \right]

\;\; = -i \left[ \sum\limits_{k=1}^{6} e^{\frac{i 2\pi k}{7}} \right] \ldots (1)

Now from the given condition.

1 + e^{\frac{i 2\pi}{7}} + e^{\frac{i 4\pi}{7}} + e^{\frac{i 6\pi}{7}} + e^{\frac{i 8\pi}{7}} + e^{\frac{i 10\pi}{7}} + e^{\frac{i 12\pi}{7}} = 0

1 + \sum\limits_{k=1}^{6} e^{i\frac{2\pi k}{7}} = 0

\sum\limits_{k=1}^{6} e^{i\frac{2\pi k}{7}} = -1

From equation (1) and (2)

\left( \sin \;\frac{2\pi k}{7} - i \cos \;\frac{2\pi k}{7} \right) = -i(-1) = i

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