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JEE Main 2019
Trigonometry
Trigonometric Ratio and Identites
Easy

Question

The value of cos(2π7)+cos(4π7)+cos(6π7)\cos \left( {{{2\pi } \over 7}} \right) + \cos \left( {{{4\pi } \over 7}} \right) + \cos \left( {{{6\pi } \over 7}} \right) is equal to :

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Solution

cos2π7+cos4π7+cos6π7=sin3(π7)sinπ7cos(2π7+6π7)2\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(3π7).cos(4π7)sin(π7) = {{\sin \left( {{{3\pi } \over 7}} \right)\,.\,\cos \left( {{{4\pi } \over 7}} \right)} \over {\sin \left( {{\pi \over 7}} \right)}} =2sin4π7cos4π72sinπ7 = {{2\sin {{4\pi } \over 7}\cos {{4\pi } \over 7}} \over {2\sin {\pi \over 7}}} =sin(8π7)2sinπ7=sinπ72sinπ7=12 = {{\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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