Inverse Trigonometric Functions

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:

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

 

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

[latex]\begin{array}{c}\tan^{-1}(x)=y \,\, \text{if and only if} \, \tan (y)=x \, \text{and} \, -\frac{\pi}{2}

 

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:

[latex]\begin{array}{c}\csc^{-1}(x)=y \,\, \text{if and only if} \, \csc (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2}, \, y\ne 0;\hfill \\ \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}[/latex]

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).

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 16. The graph of each of the inverse trigonometric functions is a reflection about the line [latex]y=x[/latex] of the corresponding restricted trigonometric function.

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:

[latex]\sin (\sin^{-1}(\frac{\sqrt{2}}{2}))= \sin (\frac{\pi}{4})=\frac{\sqrt{2}}{2}[/latex]

For the second one, we have

[latex]\sin^{-1}( \sin (\pi))=\sin^{-1}(0)=0[/latex]

 

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,

[latex]\sin (\sin^{-1}y)=y \, \text{ if } \, -1 \le y \le 1[/latex]

 

and

[latex]\sin^{-1}( \sin x)=x \, \text{ if } \, -\frac{\pi}{2} \le x \le \frac{\pi}{2}[/latex]

 

Similarly, for the cosine function,

[latex]\cos (\cos^{-1}y)=y \, \text{ if } \, -1 \le y \le 1[/latex]

 

and

[latex]\cos^{-1}( \cos x)=x \, \text{ if } 0 \le x \le \pi[/latex]

 

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

Example: Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

  1. [latex]\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)[/latex]
  2. [latex]\tan \left(\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)[/latex]
  3. [latex]\cos^{-1}\left( \cos \left(\frac{5\pi}{4}\right)\right)[/latex]
  4. [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].

  1. 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?
    An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of the function “y = sin(x) + cos(x)”, a curved wave function. The graph of the function decreases until it hits the approximate point (-(3pi/4), -1.4), where it increases until the approximate point ((pi/4), 1.4), where it begins to decrease again. The x intercepts shown on this graph of the function are at (-(5pi/4), 0), (-(pi/4), 0), and ((3pi/4), 0). The y intercept is at (0, 1).

    Figure 17. The graph of [latex]y= \sin x + \cos x[/latex].

    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].

  2. 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?
  3. Repeat for [latex]A = 1, \, B = 2[/latex]. Is there any relationship to what you found in part (2)?
  4. 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
  5. Try to figure out the formula for the [latex]y[/latex]-values.
  6. 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.)
  7. 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.

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