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(1x^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(1x^2)*x
siny = 2x sqrt(1x^2)
so sin(2arccos(x) = sin(y) = 2xsqrt(1x^2)

by the double angle identity
this is 2sin(arccos(x))cos(arccos(x))
=2 * â(1x^2) *x
=2xâ(1x^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