How do I write tan(2 sin^-1(x)) as an algebraic expression?

Answer 1

Consider a right-angled triangle with hypotenuse = 1, and opposite = x

sinθ = opposite/hypotenuse = x/1
then by Pythagoras the adjacent side = √(1-x^2)
and cosθ = √(1-x^2)

sin^-1(x) is the angle θ so that sinθ = x
then tan(2 sin^-1 (x) )
= tan (2θ)
= sin 2θ / cos 2θ
= 2sinθcosθ/(cos^2 θ - sin^2 θ)
= 2x√(1-x^2) / (1-x^2 - x^2)
= 2x√(1-x^2) / (1- 2x^2)