Verified Solution • 2025–26 Session moderate

QJEEM21Maa24b: JEE Main 2021 Mathematics - Class 11 JEEM21Maa24b Solution

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 :

Multiple Choice Options

  • A 142{1 \over {4\sqrt 2 }}
  • B 14{1 \over 4}
  • C 18{1 \over 8}
  • D 182{1 \over {8\sqrt 2 }}

Step-by-Step Verified 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)

2sin⁡2π8sin⁡22π8sin⁡23π82{\sin ^2}{\pi \over 8}{\sin ^2}{{2\pi } \over 8}{\sin ^2}{{3\pi } \over 8}

sin⁡2π8sin⁡23π8{\sin ^2}{\pi \over 8}{\sin ^2}{{3\pi } \over 8}

sin⁡2π8cos⁡2π8{\sin ^2}{\pi \over 8}{\cos ^2}{\pi \over 8}

14sin⁡2(π4)=18{1 \over 4}{\sin ^2}\left( {{\pi \over 4}} \right) = {1 \over 8}

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