Answer to Question #50654 in Trigonometry for labu

Question

2Sin^(-1) x = Sin^(-1)⁡ {2x*√(1-x^2)}⁡ it's okay. but i think it's proof is if we assume x= sin z?
2Sin^(-1) x = Sin^(-1)⁡ {2x*√(1-x^2)}⁡ it's okay. but i think
it's proof is
if we assume x= sin z

sin⁻¹(2sin z*sqrt(1-sin^2 z))
=sin⁻¹(2sin z cos z )
=sin⁻¹(sin 2z)
=2z
=2 sin⁻¹x
2cos^(-1) x = Sin^(-1)⁡ {2x*√(1-x^2)}⁡ it's also okay

it's my thinking
if we assume x= cos z

sin⁻¹(2cos z*sqrt(1-cos^2 z))
=sin⁻¹(2cos z sin z )
=sin⁻¹(sin 2z)
=2z
=2 cos⁻¹x

so now my question is are they both correct?
i want to differentiate this Sin^(-1)⁡ {2x*√(1-x^2)}
so if we use 2 sin^-1 x , then the answer will be 2/√(1-x^2)

and if we use 2 cos^-1 x , then the answer will be -2/√(1-x^2). there are two different answer after differentiate. please let me know.which one is correct or both correct??