Arcsin as an Integral/Lemma 2

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Let, $\sin_G$ be the Geometric Sine from Definition:Sine/Definition from Circle. $\arcsin_G$ is the inverse of this function.

$\displaystyle \map {\arcsin_G} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$


This result will be used in proving Derivative of Sine Function in the geometric case. So we can not use the same reasoning as Arcsin as an Integral/Lemma 1 because our logic would be circular.


Let $\theta$ be the length of the arc associated with the angle on the circle of radius $1$.

By definition of arcsine:

$y = \sin \theta \iff \theta = \arcsin y$

We have that arc length is always positive.

For negative $y$, the $\arcsin$ function is defined as being the negative of the arc length.

This makes the $\arcsin$ function and the $\sin$ function odd, and puts us in line with mathematical convention:

Inverse Sine is Odd Function.
Sine Function is Odd

Without this convention, the derivative of the $\sin$ function would not be continuous.


\(\text {(1)}: \quad\) \(\displaystyle x^2 + y^2\) \(=\) \(\displaystyle 1\) Equation of Circle
\(\displaystyle \dfrac {\d x} {\d y}\) \(=\) \(\displaystyle -\dfrac y x\) Implicit Differentiation
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac y {\sqrt {1 - y^2} }\) substituting for $x$


\(\displaystyle \arcsin_G y\) \(=\) \(\displaystyle \int_0^y \sqrt {1 + \paren {\dfrac {\d x} {\d y} }^2} \rd y\) Definition of Arc Length
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \sqrt {1 + \paren {-\dfrac y x}^2}\) substituting for $\dfrac {\d x} {\d y}$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \sqrt {1 + \dfrac {y^2} {x^2} } \rd y\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \sqrt {\dfrac {x^2} {x^2} + \dfrac {y^2} {x^2} } \rd y\) rewriting $1$ to create common denominator
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \sqrt {\dfrac {x^2 + y^2} {x^2} } \rd y\) combining terms with common denominator
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \sqrt {\dfrac 1 {x^2} } \rd y\) Equation of Circle $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \dfrac 1 x \rd y\) in Quadrant $\text I$ and Quadrant $\text {IV}$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y\) substituting for $x$ in Quadrant $\text I$ and Quadrant $\text {IV}$