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Question: Answered & Verified by Expert

If L=sin2π16-sin2π8 and M=cos2π16-sin2π8

MathematicsTrigonometric Ratios & IdentitiesJEE MainJEE Main 2020 (05 Sep Shift 2)
Options:
  • A L=-122+12cosπ8
  • B L=142-14cosπ8
  • C M=142+14cosπ8
  • D M=122+12cosπ8
Solution:
2248 Upvotes Verified Answer
The correct answer is: M=122+12cosπ8

L=sinπ16+π8sinπ16-π8
=sin3π16·sin-π16

=12cos3π16+π16-cos3π16-π16=1212-cosπ8
M=cosπ16+π8cosπ16-π8

=cos3π16·cos-π16
=12cos3π16+π16+cos3π16-π16=1212+cosπ8

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