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A previous result found {$$ {d \over {d\theta}}\operatorname{ver}\theta = \sin\theta $$}

The only reference I have found to the inverse function on the web has an obviously wrong formula for this inverse. Needless to say, all of this should be verified as this is only my work.

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

Obvious as this seems, a person offering a desktop calculator app for sale got it wrong (those pesky signs, you know) so it behooves me to verify that this is indeed the inverse of versed sine.

{$$ \begin{align} \operatorname{ver}\theta &= 1 - cos\theta \cr \operatorname{ver}\theta -1 &= - \cos\theta \cr 1 - \operatorname{ver}\theta &= \cos\theta \cr arccos \left( 1 - \operatorname{ver}\theta \right) &= \theta \end{align} $$}

{$ y = \operatorname{arcver}(x) $} and {$ y=\operatorname{ver}(x) $}

Demonstration via implicit differentiation:

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

In the first and second quadrant, to which we are restricted, {$ \sin(y) $} is always non-negative, so we need choose only the positive root. Now to substitute back the value of {$y$}:

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

Verify this result, as I have not found it in the literature. However this looks good in light of the graph: the slope is undefined at 0 and 2, is 1 at 1, and is positive everywhere it is defined.

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**Category:** Math Calculus Trigonometry

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