JEE Main 2018TrigonometryTrigonometric Ratio and IdentitesEasyQuestionThe value of 2sin(π8)sin(2π8)sin(3π8)sin(5π8)sin(6π8)sin(7π8)2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)2sin(8π)sin(82π)sin(83π)sin(85π)sin(86π)sin(87π) is :OptionsA142{1 \over {4\sqrt 2 }}421B14{1 \over 4}41C18{1 \over 8}81D182{1 \over {8\sqrt 2 }}821Check AnswerHide SolutionSolution2sin(π8)sin(2π8)sin(3π8)sin(5π8)sin(6π8)sin(7π8)2\sin \left( {{\pi \over 8}} \right)\sin \left( {{{2\pi } \over 8}} \right)\sin \left( {{{3\pi } \over 8}} \right)\sin \left( {{{5\pi } \over 8}} \right)\sin \left( {{{6\pi } \over 8}} \right)\sin \left( {{{7\pi } \over 8}} \right)2sin(8π)sin(82π)sin(83π)sin(85π)sin(86π)sin(87π) 2sin2π8sin22π8sin23π82{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\sin ^2}{{3\pi } \over 8}2sin28πsin282πsin283π sin2π8sin23π8{\sin ^2}{\pi \over 8}{\sin ^2}{{3\pi } \over 8}sin28πsin283π sin2π8cos2π8{\sin ^2}{\pi \over 8}{\cos ^2}{\pi \over 8}sin28πcos28π 14sin2(π4)=18{1 \over 4}{\sin ^2}\left( {{\pi \over 4}} \right) = {1 \over 8}41sin2(4π)=81