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The value of $$\cos \left( {{{2\pi } \over 7}} \right) + \cos \left( {{{4\pi } \over 7}} \right) + \cos \left( {{{6\pi } \over 7}} \right)$$ is equal to :

JEE · Math · previous-year question

  1. A.$$-$$1
  2. B.$$-$$$${1 \over 2}$$correct
  3. C.$$-$$$${1 \over 3}$$
  4. D.$$-$$$${1 \over 4}$$

Answer

B. $$-$$$${1 \over 2}$$

Explanation

$$\cos {{2\pi } \over 7} + \cos {{4\pi } \over 7} + \cos {{6\pi } \over 7} = {{\sin 3\left( {{\pi \over 7}} \right)} \over {\sin {\pi \over 7}}}\cos {{\left( {{{2\pi } \over 7} + {{6\pi } \over 7}} \right)} \over 2}$$ $$ = {{\sin \left( {{{3\pi } \over 7}} \right)\,.\,\cos \left( {{{4\pi } \over 7}} \right)} \over {\sin \left( {{\pi \over 7}} \right)}}$$ $$ = {{2\sin {{4\pi } \over 7}\cos {{4\pi } \over 7}} \over {2\sin {\pi \over 7}}}$$ $$ = {{\sin \left( {{{8\pi } \over 7}} \right)} \over {2\sin {\pi \over 7}}} = {{ - \sin {\pi \over 7}} \over {2\sin {\pi \over 7}}} = {{ - 1} \over 2}$$

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