Arcsine in terms of Twice Arctangent

Theorem

$\arcsin x = 2 \map \arctan {\dfrac x {1 + \sqrt {1 - x^2} } }$

where $x$ is a real number with $-1 < x < 1$.

Let:

$\theta = \arcsin x$

Then by the definition of arcsine:

$x = \sin \theta$

and:

$-\dfrac \pi 2 < \theta < \dfrac \pi 2$

Then:

$\ds \tan \dfrac \theta 2$

$=$

$\ds \dfrac {\sin \theta} {1 + \cos \theta}$

$\ds $

$=$

$\ds \dfrac {\sin \theta} {1 + \sqrt {1 - \sin^2 \theta} }$

$\ds $

$=$

$\ds \dfrac x {1 + \sqrt {1 - x^2} }$

$\ds \leadsto \ \ $

$\ds \dfrac \theta 2$

$=$

$\ds \map \arctan {\dfrac x {1 + \sqrt {1 - x^2} } }$

taking arctangent of both sides

$\ds \leadsto \ \ $

$\ds \arcsin x$

$=$

$\ds 2 \map \arctan {\dfrac x {1 + \sqrt {1 - x^2} } }$

substituting $\theta \gets \arcsin x$

$\blacksquare$