Arcsine as Integral/Lemma 2
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Lemma
Let $\sin_G$ be the geometric sine.
$\arcsin_G$ is the inverse of this function.
- $\ds \map {\arcsin_G} x = \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$
Proof
This result will be used in proving Derivative of Sine Function in the geometric case.
So we can not use the same reasoning as Lemma 1 because our logic would be circular.
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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:
Without this convention, the derivative of the $\sin$ function would not be continuous.
Now:
| \(\text {(1)}: \quad\) | \(\ds x^2 + y^2\) | \(=\) | \(\ds 1\) | Equation of Circle | ||||||||||
| \(\ds \dfrac {\d x} {\d y}\) | \(=\) | \(\ds -\dfrac y x\) | Implicit Differentiation | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac y {\sqrt {1 - y^2} }\) | substituting for $x$ |
Then:
| \(\ds \arcsin_G y\) | \(=\) | \(\ds \int_0^y \sqrt {1 + \paren {\dfrac {\d x} {\d y} }^2} \rd y\) | Definition of Arc Length | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \sqrt {1 + \paren {-\dfrac y x}^2}\) | substituting for $\dfrac {\d x} {\d y}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \sqrt {1 + \dfrac {y^2} {x^2} } \rd y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \sqrt {\dfrac {x^2} {x^2} + \dfrac {y^2} {x^2} } \rd y\) | rewriting $1$ to create common denominator | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \sqrt {\dfrac {x^2 + y^2} {x^2} } \rd y\) | combining terms with common denominator | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \sqrt {\dfrac 1 {x^2} } \rd y\) | Equation of Circle $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \dfrac 1 x \rd y\) | in Quadrant $\text I$ and Quadrant $\text {IV}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y\) | substituting for $x$ in Quadrant $\text I$ and Quadrant $\text {IV}$ |
$\Box$