Arctangent in terms of Arcsine
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Theorem
- $\arctan x = \map \arcsin {\dfrac x {\sqrt {1 + x^2} } }$
where $x$ is a real number.
Proof 1
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \arctan x\) |
$\blacksquare$
Proof 2
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:
- $\map \arctan x = \map \arcsin {\dfrac x {\sqrt {1 + x^2} } }$
$\blacksquare$