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

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