Arctangent in terms of Arcsine/Proof 1

Theorem

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

Proof

Let:

$\theta = \arctan x$

Then by the definition of arctangent:

$x = \tan \theta$

Then:

\(\ds \map \arcsin { \dfrac x {\sqrt {1 + x^2} } }\)

\(=\)

\(\ds \map \arcsin { \dfrac {\tan \theta} {\sqrt {1 + \tan^2 \theta} } }\)

\(\ds \)

\(=\)

\(\ds \map \arcsin { \dfrac {\tan \theta} {\sqrt {\sec^2 \theta} } }\)

Difference of Squares of Secant and Tangent

\(\ds \)

\(=\)

\(\ds \map \arcsin { \dfrac {\sin \theta} { \cos \theta \dfrac 1 {\cos \theta } } }\)

Definition of Real Tangent Function

\(\ds \)

\(=\)

\(\ds \map \arcsin {\sin \theta }\)

\(\ds \)

\(=\)

\(\ds \theta\)

Definition of Real Arcsine

\(\ds \)

\(=\)

\(\ds \arctan x\)

$\blacksquare$