## Initial Derivation (Right Triangle) [[Basic Trigonometric Functions]] take an angle and give the ratio between two sides of a right triangle. *Inverse* trig functions map from this ratio back to the angle. That is ![[Inverse Trig Functions.png]] [^1] | Basic Function | Formula | Inverse function | Formula for inverse | Defined range for inverse | | -------------- | ---------------------------------- | ---------------- | --|:----------------------------------------------:| | $\sin$ | $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ | $\sin^{-1}$, $\arcsin$ | $\arcsin\left( \frac{\text{opposite}}{\text{hypotenuse}} \right) = \theta$ | $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ | | $\cos$ | $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ | $\cos^{-1}$, $\arccos$ | $\arccos\left( \frac{\text{adjacent}}{\text{hypotenuse}} \right) = \theta$ | $0 \leq \theta \leq \pi$ | $\tan$ | $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ | $\tan^{-1}$, $\arctan$ | $\arctan\left( \frac{\text{opposite}}{\text{adjacent}} \right) = \theta$ | $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ *Note*: Even though $\sin^2 \theta$ denotes $\sin \theta \cdot \sin \theta$, the notation $\sin^{-1} \theta$ etc does not mean $\frac{1}{\sin \theta}$. It is an *inverse* of the function, not an *exponent*. The alternative nomenclature $\arcsin$, $\arccos$, $\arctan$ for $\sin^{-1} \theta$ and friends avoids this confusion and is a tad easier to type in latex because you don't need to use braces around the exponent, but seems generally to be out of favour for some reason. See Khan Academy review articles [here](https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/trig-solve-for-an-angle/a/inverse-trig-functions-intro) and [here](https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:inverse-trig/a/inverse-trigonometric-functions-review). ## Domain of Trigonometric Functions Once we work with the [[Unit Circle Definitions of Trigonometric Functions|unit circle]] definition and [[Trigonometric Identities from Rotation|rotations]] in particular, it becomes apparent that we need to restrict the domain of trigonometric functions for the inverse functions to be defined, as a function cannot map one input to multiple outputs. To understand the problem, consider for example: $ \begin{align*} \tan \pi = \tan 0 = 0 \end{align*} $ ...and in fact $\tan n \pi = 0 \; \forall n \in \mathbf{Z}.$So since there isn't a single $\theta$ for which $\tan \theta = 0$, what should we say is the value of $\arctan 0$? We can only have an inverse of a function where each input maps to a distinct output. To solve this problem, it is conventional to restrict the domain of trig functions (and therefore range of inverse trig functions) such that full rotations are not possible and the inverse still has the greatest possible range. In the case of $\arcsin$ and $\arctan$ this means to quadrants I and IV in the unit circle, ie $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and in the case of $\arccos$ it means to quadrants I and II (ie $0 \geq \theta \geq \frac{\pi}{2}$). In the case of $\tan$ specifically, the slope becomes vertical at $\frac{\pi}{2}$ and $\frac{-\pi}{2}$, so $\arctan \theta$ is only defined for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. One way to determine whether we are correctly restricting the domain of some function to make it invertible is to look at a graph of the function and see whether there is any place where it is possible to make a horizontal line that crosses the graph twice in that domain. If one can go through the whole domain without such a horizontal line then the function is invertible over that domain[^2]. # Definitions ## $\sin^{-1}$ / $\arcsin$ The inverse sine function ($\sin{-1}$) is defined as the function with the rule $\sin^{-1} x = y \qquad x \in [-1,1]$ such that $\sin y = x \qquad y\in[-\frac{\pi}{2},\frac{\pi}{2}]$ ![[Inverse Sine.png]] As might be expected, these are symmetrical about $y=x$. Note: - $\sin^{-1}$ is conventionally defined over the acute angles, so if solving for an obtuse or reflex angle, identities would need to be used to find the correct one. - both this restricted version of $\sin$ and the resulting inverse, $\sin^{-1}$ are increasing everywhere in their respective domains. This is not the case for the full $\sin$ function. ## $\cos^{-1}$/$\arccos$ The inverse cosine function $\cos^{-1}$ is defined as $\cos^{-1} x = y \qquad x \in [-1,1]$ such that $y = \cos x \qquad y \in [0,\pi]$ Just as $\cos x$ is decreasing everywhere in $[0,\pi]$, so $\cos^{-1} x$ decreases everywhere in its range $[-1,1]$ (going from $\pi$ to $0$). ![[Inverse Cosine.png]] ## $\tan^{-1}$/$\arctan$ The inverse tangent function $tan^{-1}$ is the function $\tan^{-1} x = y \qquad x \in \mathbf{R}$ such that $\tan y=x \qquad y\in\left( \frac{-\pi}{2}, \frac{\pi}{2}\right)$ Note that as $\tan$ is undefined at $\pm\frac{\pi}{2}$/$90\degree$, $\tan^{-1}$ can only return values in the closed interval between these values. ![[Inverse Tangent.png]] ### Latex Note: - when using $\sin^{-1}$ and friends, you need to put the whole "-1" inside braces. ie `\sin^{-1}`. If you want to avoid this, It is also a little simpler to type `\arcsin`. - When using the inverse trig functions, if you want to disambiguate operator associativity you will often need `\left(` and `\right)` if you want to put big parentheses around the ratio. For example, something like $\tan^{-1}\left( 1 - \frac{BC}{AB} \right) = \theta$ ... is written `\tan^{-1}\left( 1 - \frac{BC}{AB} \right) = \theta` in latex. This will look much nicer than if you just use `()` ie $\tan^{-1}( 1-\frac{BC}{AB} ) = \theta$ which gives you these small parentheses and I don't really know what to say except 🤮. If you don't need parentheses to make operator order clear, then something like $\tan^{-1} \frac{BC}{AB} = \theta$ is of course fine. [^1:] Diagram available on [geogebra](https://www.geogebra.org/calculator/wr5zgmhg) [^2:] This is checking whether the inverse would pass the "[vertical line test](https://en.wikipedia.org/wiki/Vertical_line_test)" because if you graph a curve, for it to be a function there cannot be any points where a vertical line would cross the graph twice. For the inverse of a function to pass the vertical line test within a given domain, the function itself must not have any points where a *horizontal* line would cross the graph twice in that range as the axes are reversed.