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{$$ {d \over {d\theta}}\operatorname{covercosin}\theta = \cos\theta $$}

The inverse of {$ \operatorname{covercosin} $}:

{$$ \begin{align} \operatorname{arccovercosin}\theta &= \arcsin(\theta - 1) \cr -{\pi \over 2}\lt\theta &\lt {\pi \over 2} \end{align} $$}

{$ y=\operatorname{covercosin}(x) $} and {$ y=\operatorname{arccovercosin}(x) $}

The demonstration proceeds by implicit differentiation:

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

As {$\cos$} is non-negative in the first and fourth quadrants, only the positive root is needed. To substitute back the value of *y*:

{$$ \begin{align} {d \over {d\theta}}\operatorname{arccoversosin}\theta &= {1 \over {\sqrt{1 - \sin^2(\operatorname{arccovercosin}\theta)}}} \cr {d \over {d\theta}}\operatorname{arccoversosin}\theta &= {1 \over {\sqrt{1 - \sin^2(\arcsin(\theta - 1))}}} \cr {d \over {d\theta}}\operatorname{arccoversosin}\theta &= {1 \over {\sqrt{1 - (\theta - 1)^2}}} \cr \cr \therefore \quad {d \over {d\theta}}\operatorname{arccoversosin}\theta &= {1 \over {\sqrt{2\theta - \theta^2}}}\end{align} $$}

{$ y=\operatorname{covercosin}(x) $}, {$ y=\operatorname{arccovercosin}(x) $}. {$ y^\prime=\operatorname{covercosin}(x) $} and {$ y^\prime =\operatorname{arccovercosin}(x) $}

**''Sources:**'

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- FooPlot: Online graphing calculator and function plotter

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**Category:** Math Calculus Trigonometry

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