what will be sin^2(sin^-1 x)

Sorry Zegroes but you messed up!

$$sin^2(sin^{-1}x)\\ =[sin(sin^{-1}x)]^2\\ =x^2$$

This is because sine and inverse sine cancel each cancel each other out.

sin(2u) = 2sin(u)cos(u)

sin(2sin^-1(x)) = 2sin(sin^-1(x))cos(sin^-1(x)) = 2xcos(sin^-1(x))

let y = sin^-1 x
x =siny

sin^2(y) = 1- cos^2(y)

cos^2(y) = 1 - sin^2(y)
cos^2(y) = 1 -x^2 >= 0 ( since anything^2 is always positive)

+/-cos(y) = sqrt(1-x^2)



sin(2sin^-1(x)) = 2sin(sin^-1(x))cos(sin^-1(x)) = 2xcos(sin^-1(x)) =+/- 2xsqrt(1-x^2)



Hopefully I didn't mess up somewhere.

yeah did way to much......