Q57 / 90 · JEE Main 2019 (10 Jan Shift 2) · Mathematics

The value of cos⁡π22⋅cos⁡π23⋅…⋅cos⁡π210⋅sin⁡π210 is:

  • A
    11024
  • Correct answer
    B
    1512
  • C
    12
  • D
    1256

Solution

Given cosπ22·cosπ23·...·cosπ210·sinπ210

=sin29.π21029.sinπ210.sinπ210

Using  cosA.cos2A.cos22A.cos23A.cos24A.....cos2n-1A=sin2nA2nsinA

=129

=1512

Hence, the correct answer is option: 'B'.

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