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

Option: 1

-1


Option: 2

-\frac{1}{2}


Option: 3

-\frac{1}{3}


Option: 4

-\frac{1}{4}


Answers (1)

best_answer

Using Summation of cosine series

\mathrm{\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=\frac{\sin 3 \times \frac{2 \pi}{2 \times 7}}{\sin \frac{2 \pi}{2 \times 7}} \times \cos \left(\frac{\frac{2 \pi}{7}+\frac{6 \pi}{7}}{2}\right)}

                                                  \mathrm{=\frac{\sin \frac{3 \pi}{7}}{\sin \frac{\pi}{7}} \times \cos \frac{4 \pi}{7}} \\

                                                  \mathrm{=\frac{\sin \left(\pi-\frac{4 \pi}{7}\right) \cos \frac{4 \pi}{7}}{\sin \frac{\pi}{7}}}

                                                  \mathrm{=\frac{2 \sin \frac{4 \pi}{7} \cos \frac{4 \pi}{7}}{2 \sin \frac{\pi}{7}}} \\

                                                  \mathrm{=\frac{\sin \frac{8 \pi}{7}}{2 \sin \frac{\pi}{7}}=-\frac{1}{2} }

Hence the correct answer is option 2

Posted by

Sanket Gandhi

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