JEE Main 2021TrigonometryTrigonometric Ratio and IdentitesMediumQuestion2sin(π22)sin(3π22)sin(5π22)sin(7π22)sin(9π22)2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)2sin(22π)sin(223π)sin(225π)sin(227π)sin(229π) is equal to :OptionsA316\frac{3}{16}163B116\frac{1}{16}161C132\frac{1}{32}321D932\frac{9}{32}329Check AnswerHide SolutionSolution2sinπ22sin3π22sin5π22sin7π22sin9π222\sin {\pi \over {22}}\sin {{3\pi } \over {22}}\sin {{5\pi } \over {22}}\sin {{7\pi } \over {22}}\sin {{9\pi } \over {22}}2sin22πsin223πsin225πsin227πsin229π =2sin(11π−10π22)sin(11π−8π22)sin(11π−6π22)sin(11π−4π22)sin(11π−2π22) = 2\sin \left( {{{11\pi - 10\pi } \over {22}}} \right)\sin \left( {{{11\pi - 8\pi } \over {22}}} \right)\sin \left( {{{11\pi - 6\pi } \over {22}}} \right)\sin \left( {{{11\pi - 4\pi } \over {22}}} \right)\sin \left( {{{11\pi - 2\pi } \over {22}}} \right)=2sin(2211π−10π)sin(2211π−8π)sin(2211π−6π)sin(2211π−4π)sin(2211π−2π) =2cosπ11cos2π11cos3π11cos4π11cos5π11 = 2\cos {\pi \over {11}}\cos {{2\pi } \over {11}}\cos {{3\pi } \over {11}}\cos {{4\pi } \over {11}}\cos {{5\pi } \over {11}}=2cos11πcos112πcos113πcos114πcos115π =2sin32π1125sinπ11 = {{2\sin {{32\pi } \over {11}}} \over {{2^5}\sin {\pi \over {11}}}}=25sin11π2sin1132π =116 = {1 \over {16}}=161