sin^-1(2x√(1 - x^2)) for -1 ≤ x ≤ 1 is equal to

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sin^-1(2x√(1 – x^2)) for -1 ≤ x ≤ 1 is equal to (A) π – 2sin^-1(x) 1/√2 ≤ x ≤ 1

1 Answer

  1. Answer:

    To prove sin−1(2x1x2)=2sin−1x

    Let’s take derivative of LHS w.r.t x i.e.
    dxdsin−1(2x1x2)
     =1(2x1x2)21×d(2x1x2)
     =14x2(1x2)1×(21x22x(−2x)+1x2(2))
     =14x2+4x41×(1x2−2x2+22x2)
     =(12x2)1x224x2
     
     =(12x2)1x22(12x2)
     =1x22
    dxdsin−1(2x1x2)=1x22
    dxdsin−1(2x1x2)=1x22
    d(sin−1(2x1x2))=1x22dx
    sin−1(2x1x2)=2sin−1x  for 2−1x21

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