Arctangent in terms of Arcsine/Proof 2

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Theorem

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


Proof

From Pfaff's Transformation:

$\ds \map F {a, b; c; x} = \paren {1 - x}^{-a} \map F {a, c - b; c; \dfrac x {x - 1} }$

where $\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$.


We have:

\(\ds \map \arctan x\) \(=\) \(\ds x \map F {\dfrac 1 2, 1; \dfrac 3 2; -x^2}\) Arctangent Function in terms of Gaussian Hypergeometric Function
\(\ds \) \(=\) \(\ds x \paren {1 - \paren {-x^2} }^{-\frac 1 2} \map F {\dfrac 1 2, \dfrac 3 2 - 1; \dfrac 3 2; \dfrac {\paren {-x^2} } {\paren {-x^2} - 1} }\) Pfaff's Transformation
\(\ds \) \(=\) \(\ds \dfrac x {\sqrt {1 + x^2} } \map F {\dfrac 1 2, \dfrac 1 2; \dfrac 3 2; \dfrac {x^2} {1 + x^2} }\) simplifying
\(\ds \) \(=\) \(\ds \map \arcsin {\dfrac x {\sqrt {1 + x^2} } }\) Arcsine Function in terms of Gaussian Hypergeometric Function


Therefore:

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

$\blacksquare$


Sources

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