Question 770262
using double angle formulas we write:

{{{cos(2theta)=1-2sin^2(theta)

}}}

with 
{{{theta=arcsin(x)}}}


then we notice:

{{{sin(theta)=sin(arcsin(x))=x}}}


and therefore

{{{cos(2arcsinx)=cos(2theta)=1-2sin^2(theta)=1-2x^2

}}}



:)