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The Derivative of Arc Versed Cosine


Versed cosine aka vercosin

Do not confuse with coversedsine.

Contents

The inverse of {$ \operatorname{vercosin} $}

is

{$$ \begin{gather} \operatorname{arcvercosin}\theta = \arccos(1-\theta) \cr 0\le\theta\le\pi \end{gather}$$}

A previous result is

{$$ \operatorname{vercosin}^\prime\theta = - \sin\theta $$}


{$ y = \operatorname{vercosin}(x) $} and {$ y=\operatorname{arcvercosin}(x) $}

The demonstration proceeds by implicit differential:

{$$ \begin{align} \operatorname{vercosin}(\operatorname{arcvercosin}\theta) &= \theta \tag{definition of inverse} \cr \text{Let } \quad y &= \operatorname{arcvercosin}\theta \cr \operatorname{vercosin}(y) &= \theta \cr {d \over {d\theta}}\left( \operatorname{vercosin}(y) \right) &= {d \over {d\theta}} \theta \tag{Chain rule} \cr {d \over {d\theta}}\operatorname{vercosin}(y){d \over {d\theta}}y &= 1 \cr -\sin(y){d \over {d\theta}}y &= 1 \tag{previous result} \cr {d \over {d\theta}} &= - {1 \over {\sin(y)}} \cr {d \over {d\theta}}y &= - {1 \over {\sqrt{1 - \cos^2(y)}}} \tag{Pythagorean} \end{align} $$}

In the first and second quadrant {$ \sin $} is positive, so we choose the positive root. Now to substitute the value of y back:

{$$ \begin{align} {d \over {d\theta}}\operatorname{arcvercosin}\theta &= - {1 \over {\sqrt{1 - \cos^2(\operatorname{vercosin}\theta)}}} \cr {d \over {d\theta}}\operatorname{arcvercosin}\theta &= - {1 \over {\sqrt{1 - \cos^2(\arccos(1- \theta))}}} \cr {d \over {d\theta}}\operatorname{arcvercosin}\theta &= - {1 \over {\sqrt{1 - (1- \theta)^2}}} \cr \therefore \quad {d \over {d\theta}}\operatorname{arcvercosin}\theta &= - {1 \over {\sqrt{ 2\theta - \theta^2}}} \end{align} $$}


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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