# Inverse Trigonometric Functions

## Definition

Inverse trigonometric functions, like any other inverse function, are mathematical operators that undo the function's operation. For the right triangle

we have seen the basic trigonometric functions

\[\begin{array}{lcr} \sin \theta = \frac{b}{c} & \qquad \cos \theta = \frac{a}{c} & \qquad \tan \theta = \frac{b}{a} \end{array} \]

where the *angle* is the input (or argument) to the function and the *ratio of sides* is the output of the function result. The inverse function reverses this operation, taking the *ratio of sides* as input and returning the *angle* as output:

\[\begin{align} \sin^{-1} \left( \frac{b}{c} \right) &= \theta \\ \cos^{-1} \left( \frac{a}{c} \right) &= \theta \\ \tan^{-1} \left( \frac{b}{a} \right) &= \theta. \end{align} \]

This means the inverse trigonometric functions are useful whenever we know the sides of a triangle and want to find its angles.

**Note:** The notation \( \sin^{-1} \) might be confusing, as we normally use a negative exponent to indicate the reciprocal. However, in this case, \( \sin^{-1} \alpha \neq \frac{1}{\sin \alpha} \). When we want the reciprocal of \( \sin \) we use \( \csc \) (see Reciprocal Trigonometric Functions for more details). In order to avoid this ambiguity, sometimes people might choose to write the inverse functions with an *arc* prefix. For example:

\[ \arcsin \beta = \sin^{-1} \beta.\]

## Composing Functions with inverses

Notice that by definition of the inverse function, we have the following relationships:

\[\begin{align} \sin( \sin^{-1}(x)) & = x\\ \cos( \cos^{-1}(x)) & = x\\ \tan( \tan^{-1}(x)) & = x. \end{align}\]

If we restrict \(x\) to be in the appropriate domain, the above function composition order can also be reversed:

\[\begin{align} \sin^{-1}( \sin(x)) & = x \qquad \text{ for } - \frac{\pi}{2} \leq x < \frac{\pi}{2}\\ \cos^{-1}(( \cos(x)) & = x \qquad \text{ for } 0 \leq x < \pi\\ \tan^{-1}(( \tan(x)) & = x \qquad \text{ for } - \frac{\pi}{2} \leq x < \frac{\pi}{2}. \end{align}\]

## Domain and Range

In the sine function, many different angles \(\theta\) map to the same value of \(\sin(\theta)\). For example, \[0 = \sin 0 = \sin (\pi) = \sin (2\pi) = \cdots = \sin (k \pi)\] for any integer \(k\). To overcome the problem of having multiple values map to the same angle for the inverse sine function, we will restrict our domain before finding the inverse. The domain and ranges of the basic inverse trigonometric functions are

\[\begin{array}{|c|c|c|} \hline \text{Function} & \text{Domain} & \text{Range}\\ \hline \sin^{-1}(x) & [-1, 1] & \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]\\ \hline \cos^{-1}(x) & [-1, 1] & \left[ 0, \pi \right]\\ \hline \tan^{-1}(x) & (-\infty, \infty) & \left(- \frac{\pi}{2}, \frac{\pi}{2} \right)\\ \hline \end{array}\]

The graphs of the inverse functions are the original function in the domain specified above, which has been flipped about the line \(y=x\). The effect of flipping the graph about the line \(y=x\) is to swap the roles of \(x\) and \(y\), so this observation is true for the graph of any inverse function. See inverse trigonometric graph for more details.

## Specific Values

There are specific values of the inverse function that are useful to remember. By remembering the specific values of trigonometric functions, these values for the inverse function are also easily remembered.

\[ \begin{array} {| c | c | c | c | c | c |} \hline \text{x} & \frac {\sqrt{0}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{4}} {2} \\ \hline \sin^{-1} (x) & 0 & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} \\ \hline \cos^{-1}(x) & \frac{\pi}{2} & \frac{\pi}{3} & \frac{\pi}{4} & \frac{\pi}{6} & 0\\ \hline \end{array}\]

## Examples

## Simplify the expression

\[ \sin \left(\cos^{-1} \left( \frac{x}{x+1} \right) \right).\]

Solution:Let \(\theta = \cos^{-1} \left( \frac{x}{x+1} \right) \). Then by definition, \(\cos(\theta) = \frac{x}{x+1}\). From the definition of cosine based on right triangles, construct the right triangleSince \(\cos \theta = \frac{a}{c},\) by letting \(a=x\) and \(c=x+1\), we have \(\cos(\theta) = \frac{x}{x+1}\). From the Pythagorean Theorem, we then have

\[b^2 = c^2 - a^2 = (x+1)^2-x^2 = x^2 + 2x + 1 - x^2 = 2x + 1.\]

This implies

\[ \sin \left( \cos^{-1} \frac{x}{x+1} \right) = \sin (\theta) = \frac{b}{c} = \frac{\sqrt{2x + 1}}{x+1}\].

## In the following equation, where \( a \) and \( b \) are coprime positive integers, what is the sum of \( a \) and \( b \):

\[ \cos^{-1} \frac{17}{\sqrt{1130}}= \tan^{-1} \frac{a}{b}?\]

Since we are trying to find \( a \) and \( b \), we should take the tangent of both sides of the equation:

\[\begin{align} \tan \left( \cos^{-1} \frac{17}{\sqrt{1130}} \right) &= \tan \left( \tan^{-1} \frac{a}{b} \right) \\ \tan \left( \underbrace{\cos^{-1} \frac{17}{\sqrt{1130}}}_{\large{\theta}} \right) &= \frac{a}{b}. \end{align}\]

Further, we we can use the ratio given to sketch the triangle with \( \theta \) in it, using the definition of \( \cos^{-1} \):

Now, using the Pythagorean theorem, we can see that \( 17^2 + a^2 = 1130 \), which implies \( a=\sqrt{1130-289}=29 \). Finally, we evaluate \( \tan \theta = \frac{29}{17} \), which implies \( a+b=46 \). \( _\square \)

[problem-https://brilliant.org/problems/arc-ed-up/?group=c5jcX5khUFtL&ref_id=571659]]

**Cite as:**Inverse Trigonometric Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/inverse-trigonometric-functions/