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TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Section: Mathematics

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Question : 9 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)

=

$i$
$-i$
$2i$
$-2i$

Solution:

\sin \frac{2\pi k}{7} \; -i\cos \frac{2\pi k}{7}

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

= -i e^{i \frac{2\pi k}{7}}

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

\because \quad Z^{7} - 1 = 0

has roots

Z = e^{i \frac{2\pi k}{7}}, k = 0, 1, 2, 3, 4, 5, 6

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

\Rightarrow \quad

[Sum of roots of unity is 0

]

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

\;

t

e^{6} = -1 \, \frac{2\pi k}{7}

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

= -i e^{i \frac{2\pi k}{7}}

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

\because \quad Z^{7} - 1 = 0

\text{ has roots } Z = e^{i \frac{2\pi k}{7}}, k = 0, 1, 2, 3, 4, 5, 6

\therefore \quad \sum\limits_{k=0}^{6} e^{i \frac{2\pi k}{7}} = 0 \quad \left[ \text{ Sum of roots of unity is } 0 \right]

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

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

From Eq. (i),

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

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