0.1 Inverse Trigonometric Functions

Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska

0.1 Inverse Trigonometric Functions

Learning Objectives

Review the concept of an inverse function.
Define inverse trigonometric functions.
Learn how to simplify expressions involving both trigonometric and inverse trigonometric functions.

The inverse to a given function reverses the action of this function. In other words, the inverse function undoes whatever the function does. In this section, we recall the formal definition of an inverse function, state the necessary conditions for an inverse function to exist, and use this to define inverse trigonometric functions. We then discuss how to simlify expressions involving both trigonometric and inverse trigonometric functions.

Definition

Given a function $f$ with domain $D$ and range $R$ , its inverse function (if it exists) is the function $f^{-1}$ with domain $R$ and range $D$ such that

$f^{-1}(y)=x$ whenever $f(x)=y$ , $y\in R$ .

It follows that, for a function $f$ and its inverse $f^{-1}$ ,

$f^{-1}(f(x))=x$ for all $x$ in $D$ , and $f(f^{-1}(y))=y$ for all $y$ in $R$ .
These are commonly referred to as cancellation equations .

Note that the -1 is not used as an exponent here, that is, $\displaystyle f^{-1}(x) \ne \frac1{f(x)}$ .
The figure below shows the relationship between the domain and range of $f$ and the domain and range of $f^{-1}$ .

An image of two bubbles. The first bubble is orange and has two labels: the top label is “Domain of f” and the bottom label is “Range of f inverse”. Within this bubble is the variable “x”. An orange arrow with the label “f” points from this bubble to the second bubble. The second bubble is blue and has two labels: the top label is “range of f” and the bottom label is “domain of f inverse”. Within this bubble is the variable “y”. A blue arrow with the label “f inverse” points from this bubble to the first bubble. — **Figure 1.** The relationship between domain and range of a function and those of its inverse.

Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for $f(x)=x^2$ . Solving the equation $y=x^2$ for $x,$ we obtain $x= \pm \sqrt{y}$ . This formula does not describe $x$ as a function of $y$ , because there are two values of $x$ corresponding to every $y \symbol{"3E} 0$ . So the problem with trying to find an inverse function for $f(x)=x^2$ is that two inputs can be sent to the same output. In contrast, there is no such issue with $f(x)=x^3$ because, for this function, each input is sent to a different output. A function that sends each input to a different output is called a one-to-one function.

Definition

We say that $f$ is a one-to-one function if $f(x_1) \ne f(x_2)$ whenever $x_1 \ne x_2$ .

One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the $xy$ -plane it cannot intersect the graph more than once. This establishes the horizontal line test to determine whether a function is one-to-one.

Horizontal Line Test

A function $f$ is one-to-one if and only if every horizontal line intersects the graph of $f$ no more than once.

An image of two graphs. Both graphs have an x axis that runs from -3 to 3 and a y axis that runs from -3 to 4. The first graph is of the function “f(x) = x squared”, which is a parabola. The function decreases until it hits the origin, where it begins to increase. The x intercept and y intercept are both at the origin. There are two orange horizontal lines also plotted on the graph, both of which run through the function at two points each. The second graph is of the function “f(x) = x cubed”, which is an increasing curved function. The x intercept and y intercept are both at the origin. There are three orange lines also plotted on the graph, each of which only intersects the function at one point. — **Figure 2.** (a) The function $f(x)=x^2$ is not one-to one as it doesn’t pass the horizontal line test. (b) The function $f(x)=x^3$ is one-to-one.

**Figure 2.** (a) The function $f(x)=x^2$ is not one-to one as it doesn’t pass the horizontal line test. (b) The function $f(x)=x^3$ is one-to-one.

Do not confuse the horizontal line test with the vertical line test that determines whether a given curve is the graph of a function.

Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

Solution

Since the horizontal line $y=n$ for any integer $n\ge 0$ intersects the graph more than once, this function is not one-to-one. (The function represented by this graph is called the floor function, it is denoted by $\lfloor\cdot\rfloor$ , and $\lfloor x\rfloor$ is equal to the biggest integer that does not exceed $x$ .)
Since every horizontal line intersects the graph at most once, this function is one-to-one.

Is the function $f$ graphed in the following image one-to-one?

An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of the function “f(x) = (x cubed) - x” which is a curved function. The function increases, decreases, then increases again. The x intercepts are at the points (-1, 0), (0,0), and (1, 0). The y intercept is at the origin.

Answer

No.

Hint

Use the horizontal line test.

Graphing Inverse Functions

We now discuss the relationship between the graph of a function $f$ and the graph of its inverse. Consider the graph of $f$ shown below and a point $(a,b)$ on the graph. Since $b=f(a)$ , we have that $f^{-1}(b)=a$ . Therefore, when we graph $f^{-1}$ , the point $(b,a)$ is on the graph. As a result, the graph of $f^{-1}$ is a reflection of the graph of $f$ about the line $y=x$ .

An image of two graphs. The first graph is of “y = f(x)”, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs “y = f(x)” with the point (a, b), but also graphs the function “y = f inverse (x)”, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”. — **Figure 3.** The graph of the inverse function $f^{-1}(x)$ is symmetric to the graph of $f(x)$ with respect to the line $y=x$ .

Sketching Graphs of Inverse Functions

For the graph of $f$ in the following image, sketch a graph of $f^{-1}$ by sketching the line $y=x$ and using symmetry. Identify the domain and range of $f^{-1}$ .

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4).

Solution

Reflect the graph about the line $y=x$ . The domain of $f^{-1}$ coincides with the range of $f$ , which is $[0,\infty)$ , and the range of $f^{-1}$ coincides with the domain of $f$ , which is $[-2,\infty)$ . In fact, $f^{-1}(x)=x^2-2$ , $x\ge 0$ , that can be verified by solving $f(x)=y$ for $x$ and interchanging the variables $x$ and $y$ in the formula obtained.

Sketch the graph of $f(x)=2x+3$ and the graph of its inverse using the symmetry property of inverse functions.

Answer

Restricted Domains

As we have seen, $f(x)=x^2$ does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of $f$ such that the function still has the same range over this subset and is one-to-one on it. Such subset is called a restricted domain. By restricting the domain of $f$ , we can define a new function $g$ such that the domain of $g$ is the restricted domain of $f$ and $g(x)=f(x)$ for all $x$ in the domain of $g$ . Then we can define an inverse function for $g$ on that domain. For example, since $f(x)=x^2$ is one-to-one on the interval $[0,\infty)$ and the function $f(x)$ still takes all non-negative values (which is the range of $f(x)=x^2$ ) when $x$ changes in this interval, we can define a new function $g$ such that the domain of $g$ is $[0,\infty)$ and $g(x)=x^2$ for all $x$ in its domain. Since $g$ is a one-to-one function, it has an inverse function, given by the formula $g^{-1}(x)=\sqrt{x}$ . On the other hand, the function $f(x)=x^2$ is also one-to-one and takes all non-negative values on the domain $(-\infty,0]$ . Therefore, we could also define a new function $h$ such that the domain of $h$ is $(-\infty,0]$ and $h(x)=x^2$ for all $x$ in the domain of $h$ . Then $h$ is a one-to-one function and must also have an inverse. Its inverse is given by the formula $h^{-1}(x)=-\sqrt{x}$ .

An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is “g(x) = x squared”, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is “g inverse (x) = square root of x”, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is “h(x) = x squared”, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is “h inverse (x) = -(square root of x)”, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”. — **Figure 4.** (a) $\ds g^{-1}(x)=\sqrt x$ , where $g$ is the restriction of $f$ to $[0,\infty)$ .
(b) $\ds h^{-1}(x)=-\sqrt x$ , where $h$ is the restriction of $f$ to $(-\infty,0]$ .

**Figure 4.** (a) $\ds g^{-1}(x)=\sqrt x$ , where $g$ is the restriction of $f$ to $[0,\infty)$ .
(b) $\ds h^{-1}(x)=-\sqrt x$ , where $h$ is the restriction of $f$ to $(-\infty,0]$ .

Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Let us start with the sine function. The sine function is one-to-one and takes all values from its range $[-1,1]$ on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $\displaystyle\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ . By doing so, we define the inverse sine function on the domain $[-1,1]$ such that for any $x$ in the interval $[-1,1]$ , the inverse sine function tells us which angle $\theta$ in the interval $\displaystyle\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ satisfies $\sin (\theta) =x$ . Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Definition

The inverse sine function, denoted $\sin^{-1}$ or $\mathrm{arcsin}$ , and the inverse cosine function, denoted $\cos^{-1}$ or $\mathrm{arccos},$ are defined on the domain $D=\{x|-1 \le x \le 1\}$ as follows:

$\begin{array}{c}\sin^{-1}(x)=y \,\, \text{if and only if} \, \sin (y)=x \, \text{and} \, -\displaystyle\frac{\pi}{2} \le y \le \frac{\pi}{2};\hfill \\[3mm] \cos^{-1}(x)=y \,\, \text{if and only if} \, \cos (y)=x \, \text{and} \, 0 \le y \le \pi. \hfill \end{array}$

The inverse tangent function, denoted $\tan^{-1}$ or $\mathrm{arctan}$ , and inverse cotangent function, denoted $\cot^{-1}$ or $\mathrm{arccot}$ , are defined on the domain $D=(-\infty,\infty)=\mathbb{R}$ as follows:

$\begin{array}{c}\tan^{-1}(x)=y \,\, \text{if and only if} \, \tan (y)=x \, \text{and} \, \displaystyle-\frac{\pi}{2}<y<\frac{\pi}{2};\hfill \\[3mm] \cot^{-1}(x)=y \,\, \text{if and only if} \, \cot (y)=x \, \text{and} \, 0<y<\pi. \hfill \end{array}$

The inverse cosecant function, denoted $\csc^{-1}$ or $\mathrm{arccsc}$ , and inverse secant function, denoted $\sec^{-1}$ or $\mathrm{arcsec}$ , are defined on the domain $D=\{x: \, |x| \ge 1\}$ as follows:

$\begin{array}{c}\csc^{-1}(x)=y \,\, \text{if and only if} \, \csc (y)=x \, \text{and} \, \displaystyle-\frac{\pi}{2} \le y \le \frac{\pi}{2}, \, y\ne 0;\hfill \\[3mm] \sec^{-1}(x)=y \,\, \text{if and only if} \, \sec (y)=x \, \text{and} \, 0 \le y \le \pi, \, y \ne \frac{\pi}2.\hfill \end{array}$

Finding Domains of Functions Involving Inverse Trigonometric Functions

Determine the domain of the function $f(x)=\sin^{-1}(x^2+3x+1)$ .

Solution

Since the domain of inverse sine is $[-1,1]$ , we have that $x$ must satisfy $-1\le x^2+3x+1\le1$ . We solve each of the inequalities separately.

$-1\le x^2+3x+1\ \to\ 0\le x^2+3x+2=(x+1)(x+2)$

Zeroes of the right hand side are -2 and -1, and they divide the $x$ -axis into subintervals $(-\infty,-2)$ , $(-2,-1)$ , and $(-1,\infty)$ . Using sample points or any other appropriate method, we determine that the inequality is satisfied for $x\in(-\infty,-2]\cup[-1,\infty)$ , where the endpoints of the intervals are included since the inequality is not strict. Likewise,

$x^2+3x+1\le1\ \to\ x^2+3x=x(x+3)\le0\ \to\ x\in[-3,0]$ .

Since $x$ must satisfy both inequalities, our domain is going to be the intersection of $(-\infty,-2]\cup[-1,\infty)$ and $[-3,0]$ , which is $[-3,-2]\cup[-1,0]$ .

Find the domain of the function $\ds\sec^{-1}\left(\frac{2}{x+1}\right)$ .

Answer

$x\in[-3,-1)\cup(-1,1]$

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains specified earlier and reflect the graphs about the line $y=x$ to obtain the curves shown below.

An image of six graphs. The first graph is of the function “f(x) = sin inverse(x)”, which is an increasing curve function. The function starts at the point (-1, -(pi/2)) and increases until it ends at the point (1, (pi/2)). The x intercept and y intercept are at the origin. The second graph is of the function “f(x) = cos inverse (x)”, which is a decreasing curved function. The function starts at the point (-1, pi) and decreases until it ends at the point (1, 0). The x intercept is at the point (1, 0). The y intercept is at the point (0, (pi/2)). The third graph is of the function f(x) = tan inverse (x)”, which is an increasing curve function. The function starts close to the horizontal line “y = -(pi/2)” and increases until it comes close the “y = (pi/2)”. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The x intercept and y intercept are both at the origin. The fourth graph is of the function “f(x) = cot inverse (x)”, which is a decreasing curved function. The function starts slightly below the horizontal line “y = pi” and decreases until it gets close the x axis. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The fifth graph is of the function “f(x) = csc inverse (x)”, a decreasing curved function. The function starts slightly below the x axis, then decreases until it hits a closed circle point at (-1, -(pi/2)). The function then picks up again at the point (1, (pi/2)), where is begins to decrease and approach the x axis, without ever touching the x axis. There is a horizontal asymptote at the x axis. The sixth graph is of the function “f(x) = sec inverse (x)”, an increasing curved function. The function starts slightly above the horizontal line “y = (pi/2)”, then increases until it hits a closed circle point at (-1, pi). The function then picks up again at the point (1, 0), where is begins to increase and approach the horizontal line “y = (pi/2)”, without ever touching the line. There is a horizontal asymptote at the “y = (pi/2)”. — **Figure 5.** The graphs of basic inverse trigonometric functions.

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate $\cos^{-1}\left(\displaystyle\frac{1}{2}\right)$ , we need to find an angle $\theta$ such that $\cos (\theta) =\displaystyle\frac{1}{2}$ . Clearly, many angles have this property. However, given the definition of $\cos^{-1}$ , we need the angle $\theta$ that not only solves this equation, but also lies in the interval $[0,\pi].$ We conclude that $\displaystyle\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$ .

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions $\displaystyle\sin \left(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)$ and $\sin^{-1}(\sin(\pi))$ . For the first one, we simplify as follows: $\displaystyle\sin \left(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)= \sin \left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$ .

For the second one, we have $\displaystyle\sin^{-1}( \sin (\pi))=\sin^{-1}(0)=0$ .

The inverse function is supposed to “undo” the original function, so why isn’t $\sin^{-1}(\sin (\pi))=\pi$ ? Recalling our definition of inverse functions, a function $f$ and its inverse $f^{-1}$ satisfy the conditions $f(f^{-1}(y))=y$ for all $y$ in the domain of $f^{-1}$ and $f^{-1}(f(x))=x$ for all $x$ in the domain of $f$ , so what happened here? The issue is that the inverse sine function, $\sin^{-1},$ is the inverse of the restricted sine function defined on the domain $\left[\displaystyle-\frac{\pi}{2},\frac{\pi}{2}\right]$ . Therefore, for $x$ in the interval $\displaystyle\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , it is true that $\sin^{-1}(\sin (x))=x$ . However, for values of $x$ outside this interval, the equation does not hold, even though $\sin^{-1}(\sin (x))$ is defined for all real numbers $x$ .

What about $\sin (\sin^{-1}(y))$ ? Does that have a similar issue? The answer is no . Since the domain of $\sin^{-1}$ is the interval $[-1,1]$ , we conclude that $\sin (\sin^{-1}(y))=y$ if $-1 \le y \le 1$ and the expression is not defined for other values of $y$ . To summarize,

$\sin (\sin^{-1}(y))=y \ \text{if} \ -1 \le y \le 1$ and $\sin^{-1}( \sin (x))=x \ \text{if} \displaystyle-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ .

Likewise, for the cosine function, $\cos (\cos^{-1}(y))=y \ \text{if} \ -1 \le y \le 1$ and $\cos^{-1}( \cos (x))=x \ \text{if} \ 0 \le x \le \pi$ .

Similar properties hold for the other trigonometric functions and their inverses.

Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

$\displaystyle\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
$\displaystyle \cot \left(\cot^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right)$
$\displaystyle\sin^{-1}\left( \sin \left(\frac{32\pi}{7}\right)\right)$
$\displaystyle\sec\left( \tan^{-1} \left(\frac34\right)\right)$
$\ds\csc\left(\cos^{-1}\left(\dfrac{\pi}x\right)\right),\quad x\in(\pi,\infty)$

Solution

Evaluating $\displaystyle\cos^{-1}\left(-\frac{\sqrt{3}}2\right)$ is equivalent to finding the angle $\theta$ such that $\cos(\theta) =-\displaystyle\frac{\sqrt{3}}2$ and $\displaystyle0 \le \theta \le \pi$ . Using the unit circle, we see that the angle $\theta =\displaystyle\frac{5\pi}6$ satisfies these two conditions. Therefore, $\displaystyle\cos^{-1}\left(-\frac{\sqrt{3}}2\right)=\frac{5\pi}6$ .
Since $\displaystyle-\frac{1}{\sqrt{2}}$ is not equal to a cotangent of a standard angle, we cannot evaluate $\displaystyle\cot^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ without a calculator. However, since the functions $\cot$ and $\cot^{-1}$ are mutually inverse and $\displaystyle-\frac{1}{\sqrt2}$ is in the domain of $\cot^{-1},$ which is $\mathbb{R}$ , we conclude that $\displaystyle\cot\left(\cot^{-1}\left(-\frac1{\sqrt2}\right)\right)=-\frac1{\sqrt2}$ .
Let $\displaystyle\sin^{-1}\left( \sin \left(\frac{32\pi}{7}\right)\right)=\theta$ . Then the angle $\theta$ we are looking for should satisfy $\displaystyle\sin(\theta)=\sin \left(\frac{32\pi}7\right)$ and $\displaystyle-\frac{\pi}2 \le \theta \le \frac{\pi}2$ . We have that

$\displaystyle\frac{32\pi}7=\frac{32}{7}\pi=\left(4+\frac{4}{7}\right)\pi =4\pi+\frac{4\pi}{7}$ .

Since sine function is $2\pi$ -periodic, $\displaystyle\sin \left(\frac{32\pi}7\right)=\sin\left(\frac{4\pi}7\right)$ . However, $\ds\frac{4\pi}{7}\symbol{"3E}\frac{\pi}{2}$ , which means that $\ds\frac{4\pi}7$ does not belong to the interval $\ds\left[-\frac{\pi}2,\frac{\pi}2\right]$ , and hence it is not our answer yet. To “bring” the angle into the desired interval, we can use either the unit circle or the properties of the sine function, in this case, $\sin(\pi-x)=\sin(x)$ . We have

$\ds\sin\left(\frac{4\pi}{7}\right)=\sin\left(\pi-\frac{4\pi}7\right)=\sin\left(\frac{3\pi}{7}\right)$ , and hence, $\ds\sin\left(\frac{32\pi}7\right)=\sin\left(\frac{3\pi}{7}\right)$ . Because $\ds\frac{3\pi}{7}\in\left[-\frac{\pi}2,\frac{\pi}2\right]$ , we obtain that $\displaystyle \sin^{-1} \left(\sin\left(\frac{32\pi}7\right)\right)=\frac{3\pi}7$ .
Let $\ds\tan^{-1}\left(\frac34\right)=\theta$ . Then $\ds\tan(\theta)=\frac34$ , $\ds\theta\in\left(-\frac{\pi}2,\frac{\pi}2\right)$ , and we need to find $\sec(\theta)$ . Note that since $\ds\frac34\symbol{"3E}0$ , we can even say that $\ds\theta\in\left(0,\frac{\pi}2\right)$ . We can solve this question either using right triangle trigonometry, or applying an appropriate trigonometric formula. With the geometric approach, we draw a right triangle with acute angle $\theta$ , opposite side 3 and adjacent side 4, as $\ds\tan(\theta)=\frac34=\frac{opp}{adj}$ . Then, by the Pythagorean theorem, the hypotenuse has length $\sqrt{3^2+4^2}=5$ . It follows that $\ds\sec(\theta)=\frac{hyp}{adj}=\frac54$ . Alternatively, we can use the formula $\sec^2(\theta)=1+\tan^2(\theta)$ (that follows from the Pythagorean trigonometric identity $\sin^2(\theta)+\cos^2(\theta)=1$ by expressing $\sec$ and $\tan$ in terms of $\sin$ and $\cos$ and bringing the right hand side to the common denominator). Since secant is positive on the interval $\ds\left(0,\frac{\pi}2\right)$ , we obtain that $\ds\sec(\theta)=\sqrt{1+\tan^2(\theta)}=\sqrt{1+\left(\frac34\right)^2}=\sqrt{1+\frac9{16}}=\sqrt{\frac{25}{16}}=\frac54.$
Note that since $x\symbol{"3E}\pi$ , we have that $\ds\frac{\pi}x\in(0,1)$ , and so $\ds\cos^{-1}\left(\frac{\pi}{x}\right)$ is well-defined. If $\ds\cos^{-1}\left(\frac{\pi}{x}\right)=\theta$ , then $\theta\in[0,\pi]$ , and we are looking for $\csc(\theta)$ . Because $\ds\frac{\pi}x\in(0,1)$ , we have that $\ds\theta\in\left(0,\frac{\pi}2\right)$ , and hence $\csc(\theta)$ is well-defined. Using a triangle diagram ( $\ds\cos(\theta)=\frac{\pi}{x}=\frac{adj}{hyp}$ , so we take adjacent side to be $\pi$ , hypotenuse to be $x$ , and then, applying the Pythagoren theorem, we obtain that the opposite side is going to be $\sqrt{x^2-\pi^2}$ ), we find that $\ds\csc(\theta)=\frac{hyp}{opp}=\frac{x}{\sqrt{x^2-\pi^2}}$ . Alternatively, $\ds\csc(\theta)=\frac1{\sin(\theta)}$ and $\sin(\theta)=\sqrt{1-\cos^2(\theta)}={\sqrt{1-\left(\frac{\pi}{x}}\right)^2},$ which follows from the Pythagorean trigonometric identity $\sin^2(\theta)+\cos^2(\theta)=1$ since $\sin(\theta)\symbol{"3E}0$ on $\ds\left(0,\frac{\pi}2\right)$ , implying that $\ds\csc(\theta)=\frac1{\sqrt{1-\left(\frac{\pi}{x}}\right)^2}$ . It is an easy, yet useful, exercise to verify that the two answers we obtained are actually the same.

Key Concepts and Equations

For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
For a function $f$ and its inverse $f^{-1}, \, f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$ and $f^{-1}(f(x))=x$ for all $x$ in the domain of $f$ .
Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
The graph of a function $f$ and its inverse $f^{-1}$ are symmetric about the line $y=x$ .

Exercises

For the following exercises, find the domains of the given functions.

1. $f(x)=1-\cos^{-1}(4x+5)$

Answer

$x\in[-1.5,-1]$

2. $f(x)=\sec^{-1}(2x^2-3)$

3. $\ds f(x)=\sin^{-1}\left(\frac x{x-2}\right)$

Answer

$x\in(-\infty,1]$

4. $f(x)=\cos^{-1}(x+1)+\csc^{-1}(3x+2)$

For the following exercises, evaluate the functions, when possible. If the function is undefined, explain why.

5. $\ds\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$

Answer

$\ds\frac{\pi}{6}$

6. $\ds\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)$

7. $\cot^{-1}(1)$

Answer

$\ds\frac{\pi}{4}$

8. $\sin^{-1}(\sqrt3)$

9. $\ds\cos^{-1}\left(\cos\left(-\frac{12\pi}{7}\right)\right)$

Answer

$\ds\frac{2\pi}{7}$

10. $\ds \sin^{-1} \left(\sin\left(\frac{36\pi}{11}\right)\right)$

11. $\sec^{-1} (\sin(2))$

Answer

Undefined

12. $\ds\tan\left( \sin ^{-1}\left(\frac{1}{3}\right)\right)$

13. $\ds\cos\left( \tan^{-1}\left(-\frac{1}{2}\right)\right)$

Answer

$\ds\frac{2\sqrt5}5$

14. $\ds\sin\left(\sec^{-1}\left(\frac{7}{4}\right)\right)$

For the following exercises, simplify the expressions. Your answer should not contain trigonometric functions.

15. $\ds\cot\left( \cos^{-1} \left(\frac{\sqrt x}{3}\right)\right),\ x\in[0,9)$

Answer

$\ds\sqrt{\frac{x}{9-x}}$

16. $\ds\sin\left(2 \csc^{-1} \left(-\frac{1}{x}\right)\right),\ x\in(0,1]$

Glossary

inverse function: for a function $f$ , the inverse function $f^{-1}$ is defined if $f$ is one-to-one by $f^{-1}(y)=x$ whenever $f(x)=y$

one-to-one function: a function $f$ is one-to-one if $f(x_1) \ne f(x_2)$ whenever $x_1 \ne x_2$

horizontal line test: a function $f$ is one-to-one if and only if every horizontal line intersects the graph of $f$ at most once

restricted domain: a subset of the domain of a function $f$

inverse trigonometric functions: the inverses of the trigonometric functions are defined on restricted domains where they are one-to-one functions

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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Learning Objectives

Definition

Definition

Horizontal Line Test

Determining Whether a Function Is One-to-One

Solution

Answer

Hint

Graphing Inverse Functions

Sketching Graphs of Inverse Functions

Solution

Answer

Restricted Domains

Inverse Trigonometric Functions

Definition

Finding Domains of Functions Involving Inverse Trigonometric Functions

Solution

Answer

Evaluating Expressions Involving Inverse Trigonometric Functions

Solution

Key Concepts and Equations

Exercises

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Glossary

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