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The Derivative of Arc Coversed Sine

Versed cosine, aka CVS, aka coversine

Do not confuse with versed cosign.

Contents

A previous result found {$${d \over {d\theta}}\operatorname{cvs}\theta = - \cos\theta$$}

The inverse:

{$$\begin{gather} \operatorname{arccvs}\theta = \arcsin(1 - \theta) \cr -{\pi \over 2}\le\theta\le {\pi \over 2}\end{gather}$$}

{$y = \operatorname{cvs}(x)$} and {$y=\operatorname{arccvs}(x)$}

This demonstration proceeds by implicit differentiation.

{\begin{align} \operatorname{cvs} \left( \operatorname{arccvs}\theta \right) &= \theta \tag{definition of inverse} \cr \text{Let }\quad y &= \operatorname{arccvs}\theta \cr \operatorname{cvs}(y) &= \theta \cr \left( \operatorname{cvs}(y) \right) ^\prime = \theta^\prime \cr \operatorname{cvs}^\prime(y) y^\prime = 1 \tag{Chain rule}\cr -cos(y)y^\prime &= 1 \cr y^\prime &= - {1 \over {cos(y)}} \cr y^\prime &= - {1 \over {\sqrt{1 - \sin^2(y)}}}\tag{Pythagorean} \end{align}}

As we are confined to the first and fourth quadrant, {$\cos$} is positive, so we take only the positive root. Substituting back the value of y:

{\begin{align} \operatorname{arccvs}^\prime\theta &= - {1 \over {\sqrt{1 - \sin^2(\operatorname{arccvs}\theta)}}} \cr \operatorname{arccvs}^\prime\theta &= - {1 \over {\sqrt{1 - \sin^2(\arcsin(1-\theta))}}} \cr \operatorname{arccvs}^\prime\theta &= - {1 \over {\sqrt{1 - (1-\theta)^2}}} \cr \therefore \quad \operatorname{arccvs}^\prime\theta &= - {1 \over {\sqrt{2\theta - \theta^2}}} \end{align}}

Graph:

1. FooPlot: Online graphing calculator and function plotter

Sources:

Recommended:

Category: Math Calculus Trigonometry

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

March 16, 2018

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