Arcsin as an Integral/Lemma 2

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Lemma

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

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


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 Arcsin as an Integral/Lemma 1 because our logic would be circular.

Limit-arc.png


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.

Now:

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


Then:

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

$\Box$