How do you simplify “sin (2 arccos (x) )” further?

NetherCraft 0

Thanks, guys.

3 Answers

  • let 2 arccosx = y

    arccosx = y/2

    by definition

    cos(y/2) = x

    1 – cos^2(y/2) = 1 – x^2

    sin^2(y/2) = (1 – x^2)

    sin(y/2) = sqrt(1-x^2)

    now siny = sin(2y/2)

    => 2sin(y/2)cos(y/2)

    substituting sin(y/2) and cos(y/2) values in terms of x

    siny = 2sqrt(1-x^2)*x

    siny = 2x sqrt(1-x^2)

    so sin(2arccos(x) = sin(y) = 2xsqrt(1-x^2)

  • by the double angle identity

    this is 2sin(arccos(x))cos(arccos(x))

    =2 * √(1-x^2) *x

    =2x√(1-x^2)

  • let cos(y) = x

    arccos(x) = y

    sin(2 y) = 2 sin y cos y = 2sqrt(1 – (cos y)^2) * cos y

    sin (2 arccos(x)) = 2 * sqrt(1 – x^2) * x

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