JEE Main 2019TrigonometryTrigonometric Ratio and IdentitesEasyQuestionThe 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)cos(72π)+cos(74π)+cos(76π) is equal to :OptionsA−-−1B-$$$${1 \over 2}C-$$$${1 \over 3}D-$$$${1 \over 4}Check AnswerHide SolutionSolutioncos2π7+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}cos72π+cos74π+cos76π=sin7πsin3(7π)cos2(72π+76π) =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)}}=sin(7π)sin(73π).cos(74π) =2sin4π7cos4π72sinπ7 = {{2\sin {{4\pi } \over 7}\cos {{4\pi } \over 7}} \over {2\sin {\pi \over 7}}}=2sin7π2sin74πcos74π =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}=2sin7πsin(78π)=2sin7π−sin7π=2−1