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JEE Main 2018
Trigonometry
Trigonometric Ratio and Identites
Easy

Question

The 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) is :

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Solution

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) 2sin2π8sin22π8sin23π82{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\sin ^2}{{3\pi } \over 8} sin2π8sin23π8{\sin ^2}{\pi \over 8}{\sin ^2}{{3\pi } \over 8} sin2π8cos2π8{\sin ^2}{\pi \over 8}{\cos ^2}{\pi \over 8} 14sin2(π4)=18{1 \over 4}{\sin ^2}\left( {{\pi \over 4}} \right) = {1 \over 8}

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