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Finding Exact Values of Inverse Trig Functions

$\begin{equation*} \sin^{-1} \left( \frac{1}{2} \right) \end{equation*}$

The sine of what angle is $\frac{1}{2}$ ?

$\begin{equation*} \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \end{equation*}$

therefore:

$\begin{equation*} \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \end{equation*}$

$\sin^{-1}$ restricted to $\left[ -\frac{\pi}{2} \text{, } \frac{\pi}{2} \right]$

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$\cos^{-1}$ restricted to $\left[ 0 \text{, } \pi \right]$

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$\tan^{-1}$ restricted to $\left[ -\frac{\pi}{2} \text{, } \frac{\pi}{2} \right]$

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Example

$\begin{equation*} \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \end{equation*}$

$\begin{equation*} \cos \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2} \end{equation*}$

$\begin{equation*} \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \frac{5\pi}{6} \end{equation*}$

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$\begin{equation*} \tan^{-1} \left( -\sqrt{3} \right) \end{equation*}$

$\begin{equation*} \tan \left( \frac{5\pi}{3} \right) = -\sqrt{3} \end{equation*}$

$\begin{equation*} \tan^{-1} \left( -\sqrt{3} \right) = \frac{5\pi}{3} \end{equation*}$

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