Recall that a function like $f\left(x\right)=x^2$ does not have an inverse that is a function, but by restricting the domain of f to nonnegative numbers, we have a new squaring function, $f(x)=x^2,x\ge 0$, that has an inverse, $f^{-1}\left(x\right)=\sqrt{x}$. This is equivalent to restricting the range of the inverse relation to exclude ordered pairs that contain negative numbers.

In a similar manner, we can define new trigonometric functions whose inverses are functions. We can do this by restricting either the domains of the basic trigonometric functions or the ranges of their inverse relations. This can be done in many ways, but the restrictions illustrated below with solid red curves are fairly standard in mathematics.

For the inverse sine function, we choose a range close to the origin that allows all inputs on the interval $\left[-1,1\right]$ to have function values. Thus we choose the interval $\left[-\pi /2,\pi /2\right]$ for the range (Fig. 1). For the inverse cosine function, we choose a range close to the origin that allows all inputs on the interval $\left[-1,1\right]$ to have function values. We choose the interval $\left[0,\pi \right]$ (Fig. 2). For the inverse tangent function, we choose a range close to the origin that allows all real numbers to have function values. The interval $\left(-\pi /2,\pi /2\right)$ satisfies this requirement (Fig. 3).

The notation arcsin x arises because the function value, y, is the length of an arc on the unit circle for which the sine is x. Either of the two kinds of notation above can be read “the inverse sine of x” or “the arcsine of x” or “the number (or angle) whose sine is x.”

The graphs of the inverse trigonometric functions are as follows.

The following diagrams show the restricted ranges for the inverse trigonometric functions on a unit circle. Compare these graphs with the graphs on the preceding page. The ranges of these functions should be memorized. The missing endpoints in the graph of the arctangent function indicate inputs that are not in the domain of the original function.

We can also use a calculator to find inverse trigonometric function values. The cosecant, the secant, and the cotangent functions are the reciprocals of the sine, the cosine, and the tangent functions, respectively. Thus we have

On most graphing calculators, we can find inverse function values in either radians or degrees simply by selecting the appropriate mode. The keystrokes involved in finding inverse function values vary with the calculator. Be sure to read the instructions for the particular calculator that you are using.

Various compositions of trigonometric functions and their inverses often occur in practice. For example, we might want to try to simplify an expression such as

In the expression on the left, we are finding “the sine of a number whose sine is x.” Recall from Section 5.1 that if a function f has an inverse that is also a function, then

and

Thus, if $f\left(x\right)=\text{sin}x$ and $f^{-1}\left(x\right)={\text{sin}}^{-1}x$, then

which is any number on the interval $\left[-1,1\right]$. Similar results hold for the other trigonometric functions.

Now let’s consider an expression like ${\text{sin}}^{-1}\left(\text{sin}x\right)$. We might also suspect that this is equal to x for any x in the domain of sin x, but this is not true unless x is in the range of the ${\text{sin}}^{-1}$ function. Note that in order to define ${\text{sin}}^{-1}$, we had to restrict the domain of the sine function. In doing so, we restricted the range of the inverse sine function. Thus,

Similar results hold for the other trigonometric functions.

Now we find some other function compositions.

Next, let’s consider

Without using a calculator, we cannot find ${\text{sin}}^{-1}{\displaystyle \frac{3}{5}}$. However, we can still evaluate the entire expression by sketching a reference triangle. We are looking for angle $\theta$ such that ${\text{sin}}^{-1}{\displaystyle \frac{3}{5}}=\theta$, or $\text{sin}\theta ={\displaystyle \frac{3}{5}}$. Since ${\text{sin}}^{-1}$ is defined in $\left[-\pi /2,\pi /2\right]$ and ${\displaystyle \frac{3}{5}}>0$, we know that $\theta$ is in quadrant I. We sketch a reference right triangle, as shown at left. The angle $\theta$ in this triangle is an angle whose sine is ${\displaystyle \frac{3}{5}}$. We wish to find the cosine of this angle. Since the triangle is a right triangle, we can find the length of the base, b. It is 4. Thus we know that $\text{cos}\theta =b/5$, or ${\displaystyle \frac{4}{5}}$. Therefore,

Example 6

Find $\text{sin}\left({\text{cot}}^{-1}{\displaystyle \frac{x}{2}}\right)$.

Solution

Since ${\text{cot}}^{-1}$ is defined in $\left(0,\pi \right)$, we consider quadrants I and II. We draw right triangles, as shown at left, whose legs have lengths x and 2, so that $\text{cot}\theta =x/2$.

In each, we find the length of the hypotenuse and then read off the sine ratio. We get

$$\text{sin}\left({\text{cot}}^{-1}{\displaystyle \frac{x}{2}}\right)={\displaystyle \frac{2}{\sqrt{x^2+4}}}.$$

Now Try Exercise 55.

In the following example, we use a sum identity to evaluate an expression.

Example 7

Evaluate:

$$\text{sin}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}+{\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right).$$

Solution

Since ${\text{sin}}^{-1}{\displaystyle \frac{1}{2}}$ and ${\text{cos}}^{-1}{\displaystyle \frac{5}{13}}$ are both angles, the expression is the sine of a sum of two angles, so we use the identity

$$\text{sin}\left(u+v\right)=\text{sin}u\text{cos}v+\text{cos}u\text{sin}v.$$

Thus,

$$\begin{array}{c}\text{sin}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}+{\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right)\hfill \\ =\text{sin}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}\right)\cdot \text{cos}\left({\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right)+\text{cos}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}\right)\cdot \text{sin}\left({\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right)\hfill & \hfill \\ ={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{5}{13}}+\text{cos}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}\right)\cdot \text{sin}\left({\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right).\hfill & \mathbf{\text{Using composition identities}}\hfill \end{array}$$

Now since ${\text{sin}}^{-1}{\displaystyle \frac{1}{2}}=\pi /6,\text{cos}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}\right)$ simplifies to $\text{cos}\pi \text{/}6$, or $\sqrt{3}/2$. We can illustrate this with a reference triangle in quadrant I. (See Fig. 1.)

To find $\text{sin}\left({\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right)$, we use a reference triangle in quadrant I and determine that the sine of the angle whose cosine is ${\displaystyle \frac{5}{13}}$ is ${\displaystyle \frac{12}{13}}$. (See Fig. 2.)

Our expression now simplifies to

$${\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{5}{13}}+{\displaystyle \frac{\sqrt{3}}{2}}\cdot {\displaystyle \frac{12}{13}},\text{or}{\displaystyle \frac{5+12\sqrt{3}}{26}}.$$

Thus,

$$\text{sin}\left({\text{sin}}^{-1}{\displaystyle \frac{1}{2}}+{\text{cos}}^{-1}{\displaystyle \frac{5}{13}}\right)={\displaystyle \frac{5+12\sqrt{3}}{26}}.$$

Now Try Exercise 63.