Write sin(2 sin^−1 x) as an algebraic expression in x.
If y = arcsin(x) then x = sin(y)
sin(2 sin^-1 x) = sin(2y) = 2 siny cosy = 2x √(1-x^2)
let sin^−1 x = θ, such that sinθ = x
sin θ = x/1
construct a right-angled triangle with base angle θ, opposite as x, and
hypotenuse as 1. Of course you should see that sinθ = x
by Pythagoras , base^2 + x^2 = 1^2
base = √(1 - x^2)
cos θ = √(1-x^2)
then 2 sin^−1 x = 2θ
and we want sin 2θ
recall sin 2θ = 2sinθ cosθ
= 2x√(1-x^2)
sin(2 sin^−1 x) = 2x√(1-x^2)