Learning Outcomes
- Evaluate 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. Consider the sine function. The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [latex][-\frac{\pi}{2},\frac{\pi}{2}][/latex]. By doing so, we define the inverse sine function on the domain [latex][-1,1][/latex] such that for any [latex]x[/latex] in the interval [latex][-1,1][/latex], the inverse sine function tells us which angle [latex]\theta [/latex] in the interval [latex][-\frac{\pi}{2},\frac{\pi}{2}][/latex] satisfies [latex] \sin \theta =x[/latex]. 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 [latex] \sin^{-1}[/latex] or arcsin, and the inverse cosine function, denoted [latex]\cos^{-1}[/latex] or arccos, are defined on the domain [latex]D=\{x|-1 \le x \le 1\}[/latex] as follows:
The inverse tangent function, denoted [latex]\tan^{-1}[/latex] or arctan, and inverse cotangent function, denoted [latex]\cot^{-1}[/latex] or arccot, are defined on the domain [latex]D=\{x|-\infty <x<\infty \}[/latex] as follows:
The inverse cosecant function, denoted [latex]\csc^{-1}[/latex] or arccsc, and inverse secant function, denoted [latex]\sec^{-1}[/latex] or arcsec, are defined on the domain [latex]D=\{x| \, |x| \ge 1\}[/latex] as follows:
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line [latex]y=x[/latex] (Figure 16).
When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate [latex]\cos^{-1}(\frac{1}{2})[/latex], we need to find an angle [latex]\theta [/latex] such that [latex] \cos \theta =\frac{1}{2}[/latex]. Clearly, many angles have this property. However, given the definition of [latex]\cos^{-1}[/latex], we need the angle [latex]\theta [/latex] that not only solves this equation, but also lies in the interval [latex][0,\pi][/latex]. We conclude that [latex]\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}[/latex]. Review the following table of common sine and cosine values using reference angles, if necessary.
Recall: Commonly encountered angles in first quadrant of the unit circle
Angle | 0 | [latex]\frac{\pi }{6}[/latex], or 30° | [latex]\frac{\pi }{4}[/latex], or 45° | [latex]\frac{\pi }{3}[/latex], or 60° | [latex]\frac{\pi }{2}[/latex], or 90° |
Cosine | 1 | [latex]\frac{\sqrt{3}}{2}[/latex] | [latex]\frac{\sqrt{2}}{2}[/latex] | [latex]\frac{1}{2}[/latex] | 0 |
Sine | 0 | [latex]\frac{1}{2}[/latex] | [latex]\frac{\sqrt{2}}{2}[/latex] | [latex]\frac{\sqrt{3}}{2}[/latex] | 1 |
We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions [latex] \sin (\sin^{-1}(\frac{\sqrt{2}}{2}))[/latex] and [latex]\sin^{-1}(\sin(\pi))[/latex]. For the first one, we simplify as follows:
For the second one, we have
The inverse function is supposed to “undo” the original function, so why isn’t [latex]\sin^{-1}(\sin (\pi))=\pi [/latex]? Recalling our definition of inverse functions, a function [latex]f[/latex] and its inverse [latex]f^{-1}[/latex] satisfy the conditions [latex]f(f^{-1}(y))=y[/latex] for all [latex]y[/latex] in the domain of [latex]f^{-1}[/latex] and [latex]f^{-1}(f(x))=x[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex], so what happened here? The issue is that the inverse sine function, [latex]\sin^{-1}[/latex], is the inverse of the restricted sine function defined on the domain [latex][-\frac{\pi}{2},\frac{\pi}{2}][/latex]. Therefore, for [latex]x[/latex] in the interval [latex][-\frac{\pi}{2},\frac{\pi}{2}][/latex], it is true that [latex]\sin^{-1}(\sin x)=x[/latex]. However, for values of [latex]x[/latex] outside this interval, the equation does not hold, even though [latex]\sin^{-1}(\sin x)[/latex] is defined for all real numbers [latex]x[/latex].
What about [latex] \sin (\sin^{-1}y)[/latex]? Does that have a similar issue? The answer is no. Since the domain of [latex]\sin^{-1}[/latex] is the interval [latex][-1,1][/latex], we conclude that [latex] \sin (\sin^{-1}y)=y[/latex] if [latex]-1 \le y \le 1[/latex] and the expression is not defined for other values of [latex]y[/latex]. To summarize,
and
Similarly, for the cosine function,
and
Similar properties hold for the other trigonometric functions and their inverses.
Example: Evaluating Expressions Involving Inverse Trigonometric Functions
Evaluate each of the following expressions.
- [latex]\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)[/latex]
- [latex] \tan \left(\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)[/latex]
- [latex]\cos^{-1}\left( \cos \left(\frac{5\pi}{4}\right)\right)[/latex]
- [latex]\sin^{-1}\left( \cos \left(\frac{2\pi}{3}\right)\right)[/latex]
Watch the following video to see the worked solution to Example: Evaluating Expressions Involving Inverse Trigonometric Functions
Activity: The Maximum Value of a Function
In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.
This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable [latex]x[/latex].
- Consider the graph in Figure 17 of the function [latex]y= \sin x + \cos x[/latex]. Describe its overall shape. Is it periodic? How do you know?
Using a graphing calculator or other graphing device, estimate the [latex]x[/latex]– and [latex]y[/latex]-values of the maximum point for the graph (the first such point where [latex]x>0[/latex]). It may be helpful to express the [latex]x[/latex]-value as a multiple of [latex]\pi[/latex].
- Now consider other graphs of the form [latex]y=A \sin x + B \cos x[/latex] for various values of [latex]A[/latex] and [latex]B[/latex]. Sketch the graph when [latex]A = 2[/latex] and [latex]B = 1[/latex], and find the [latex]x[/latex]– and [latex]y[/latex]-values for the maximum point. (Remember to express the [latex]x[/latex]-value as a multiple of [latex]\pi[/latex], if possible.) Has it moved?
- Repeat for [latex]A = 1, \, B = 2[/latex]. Is there any relationship to what you found in part (2)?
- Complete the following table, adding a few choices of your own for [latex]A[/latex] and [latex]B[/latex]:
[latex]A[/latex] [latex]B[/latex] [latex]x[/latex] [latex]y[/latex] [latex]A[/latex] [latex]B[/latex] [latex]x[/latex] [latex]y[/latex] 0 1 [latex]\sqrt{3}[/latex] 1 1 0 1 [latex]\sqrt{3}[/latex] 1 1 12 5 1 2 5 12 2 1 2 2 3 4 4 3 - Try to figure out the formula for the [latex]y[/latex]-values.
- The formula for the [latex]x[/latex]-values is a little harder. The most helpful points from the table are [latex](1,1), \, (1,\sqrt{3}), \, (\sqrt{3},1)[/latex]. (Hint: Consider inverse trigonometric functions.)
- If you found formulas for parts (5) and (6), show that they work together. That is, substitute the [latex]x[/latex]-value formula you found into [latex]y=A \sin x + B \cos x[/latex] and simplify it to arrive at the [latex]y[/latex]-value formula you found.