Simplify the expression: sin(2cos^(-1)6x) Thoughts? =)

Let cos^-1(6x) = arccos
6x

sin(2cos^(-1)(6x)) = sin(2arccos
(6x))

We'll use the following
identity:

sin (2alpha) = 2sin
(alpha)*cos(alpha)

We'll put alpha = arccos
6x

sin(2arccos (6x)) = 2sin (arccos 6x)*cos(arccos
6x)

But sin(arccos alpha) = sqrt(1 - alpha^2) and
cos(arccos alpha) = alpha

sin (arccos 6x) = sqrt(1 -
36x^2)

cos (arccos 6x) =
6x

sin(2arccos (6x)) = 2*6x*sqrt(1 -
36x^2)

sin(2arccos (6x)) = 12xsqrt(1 -
36x^2)

Therefore, the requested value is
sin(2arccos (6x)) = 12xsqrt(1 - 36x^2).