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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:

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Sources:

Recommended:

Category: Math Calculus Trigonometry


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

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August 05, 2017

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