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Question: Answered & Verified by Expert
Let fx=sinx+3sin3x+5sin5x+3sin7xsin2x+2sin4x+3sin6x, wherever defined. If x1+x2=π2, where fx is defined at x1 and x2, then f2x1+f2x2 is
MathematicsTrigonometric Ratios & IdentitiesJEE Main
Options:
  • A cos2x
  • B sin2x
  • C 4
  • D 1
Solution:
2433 Upvotes Verified Answer
The correct answer is: 4

Numerator of fx=sinx+sin3x+2sin3x+sin5x+3sin5x+sin7x
=2cosxsin2x+2sin4x+3sin6x
Hence, fx=2cosx, if fx1=2cosx1 & fx2=2sinx1

i.e. f2x1+f2x2=4

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