Inverse Trigonometric Functions

Learning Outcomes

  • Understand and use the inverse sine, cosine, and tangent functions.
  • Find the exact value of expressions involving the inverse sine, cosine, and tangent functions.
  • Use a calculator to evaluate inverse trigonometric functions.
  • Use inverse trigonometric functions to solve right triangles.
  • Find exact values of composite functions with inverse trigonometric functions.

Understanding and Using the Inverse Sine, Cosine, and Tangent Functions

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 1.

A chart that says “Trig Functinos”, “Inverse Trig Functions”, “Domain: Measure of an angle”, “Domain: Ratio”, “Range: Ratio”, and “Range: Measure of an angle”.

Figure 1

For example, if [latex]f(x)=\sin x[/latex], then we would write [latex]f^{1}(x)={\sin}^{-1}{x}[/latex]. Be aware that [latex]{\sin}^{-1}x[/latex] does not mean [latex]\frac{1}{\sin{x}}[/latex]. The following examples illustrate the inverse trigonometric functions:

  • Since [latex]\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}[/latex], then [latex]\frac{\pi}{6}=\sin^{−1}(\frac{1}{2})[/latex].
  • Since [latex]\cos(\pi)=−1[/latex], then [latex]\pi=\cos^{−1}(−1)[/latex].
  • Since [latex]\tan\left(\frac{\pi}{4}\right)=1[/latex], then [latex]\frac{\pi}{4}=\tan^{−1}(1)[/latex].

In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if [latex]f(a)=b[/latex], then an inverse function would satisfy [latex]f^{−1}(b)=a[/latex].

Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. Figure 2 shows the graph of the sine function limited to [latex]\left[\frac{−\pi}{2}\text{, }\frac{\pi}{2}\right][/latex] and the graph of the cosine function limited to [0, π].

Two side-by-side graphs. The first graph, graph A, shows half of a period of the function sine of x. The second graph, graph B, shows half a period of the function cosine of x.

Figure 2. (a) Sine function on a restricted domain of [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex]; (b) Cosine function on a restricted domain of [0, π]

Figure 3 shows the graph of the tangent function limited to [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex].

A graph of one period of tangent of x, from -pi/2 to pi/2.

Figure 3. Tangent function on a restricted domain of [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex]

These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote.

On these restricted domains, we can define the inverse trigonometric functions.

  • The inverse sine function [latex]y=\sin^{−1}x[/latex] means [latex]x=\sin y[/latex]. The inverse sine function is sometimes called the arcsine function, and notated arcsin x.
    [latex]y=\sin^{−1}x[/latex] has domain [−1, 1] and range [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex]
  • The inverse cosine function [latex]y=\cos^{−1}x[/latex] means [latex]x=\cos y[/latex]. The inverse cosine function is sometimes called the arccosine function, and notated arccos x.
    [latex]y=\cos^{−1}x[/latex] has domain [−1, 1] and range [0, π]
  • The inverse tangent function [latex]y=\tan^{−1}x[/latex] means [latex]x=\tan y[/latex]. The inverse tangent function is sometimes called the arctangent function, and notated arctan x.
    [latex]y=\tan^{−1}x[/latex] has domain (−∞, ∞) and range [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex]

The graphs of the inverse functions are shown in Figure 4, Figure 5, and Figure 6. Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that [latex]\sin^{−1}x[/latex] has domain [−1, 1] and range [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], [latex]\cos^{−1}x[/latex] has domain [−1, 1] and range [0, π], and [latex]\tan^{−1}x[/latex] has domain of all real numbers and range [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex]. To find the domain and range of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line [latex]y=x[/latex].

A graph of the functions of sine of x and arc sine of x. There is a dotted line y=x between the two graphs, to show inverse nature of the two functions

Figure 4. The sine function and inverse sine (or arcsine) function

A graph of the functions of cosine of x and arc cosine of x. There is a dotted line at y=x to show the inverse nature of the two functions.

Figure 5. The cosine function and inverse cosine (or arccosine) function

 

A graph of the functions of tangent of x and arc tangent of x. There is a dotted line at y=x to show the inverse nature of the two functions.

Figure 6. The tangent function and inverse tangent (or arctangent) function

 

A General Note: Relations for Inverse Sine, Cosine, and Tangent Functions

For angles in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], if [latex]\sin y=x[/latex], then [latex]\sin^{−1}x=y[/latex].

For angles in the interval [0, π], if [latex]\cos y=x[/latex], then [latex]\cos^{−1}x=y[/latex].

For angles in the interval [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex], if [latex]\tan y=x[/latex], then [latex]\tan^{−1}x=y[/latex].

Example 1: Writing a Relation for an Inverse Function

Given [latex]\sin\left(\frac{5\pi}{12}\right)\approx 0.96593[/latex], write a relation involving the inverse sine.

Try It

Given [latex]\cos(0.5)\approx 0.8776[/latex], write a relation involving the inverse cosine.

Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions

Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically [latex]\frac{\pi}{ 6} (30^\circ)\text{, }\frac{\pi}{ 4} (45^\circ),\text{ and } \frac{\pi}{ 3} (60^\circ)[/latex], and their reflections into other quadrants.

How To: Given a “special” input value, evaluate an inverse trigonometric function.

  1. Find angle x for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.
  2. If x is not in the defined range of the inverse, find another angle y that is in the defined range and has the same sine, cosine, or tangent as x, depending on which corresponds to the given inverse function.

Example 2: Evaluating Inverse Trigonometric Functions for Special Input Values

Evaluate each of the following.

a. [latex]\sin−1\left(\frac{1}{2}\right)[/latex]

b. [latex]\sin−1\left(−\frac{2}{\sqrt{2}}\right)[/latex]

c. [latex]\cos−1\left(−\frac{3}{\sqrt{2}}\right)[/latex]

d. [latex]\tan^{− 1}(1)[/latex]

Try It

Evaluate each of the following.

  1. [latex]\sin^{−1}(−1)[/latex]
  2. [latex]\tan^{−1}(−1)[/latex]
  3. [latex]\cos^{−1}(−1)[/latex]
  4. [latex]\cos^{−1}(\frac{1}{2})[/latex]

Try It

Using a Calculator to Evaluate Inverse Trigonometric Functions

To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN.

In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.

In these examples and exercises, the answers will be interpreted as angles and we will use θ as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.

Example 3: Evaluating the Inverse Sine on a Calculator

Evaluate [latex]\sin^{−1}(0.97)[/latex] using a calculator.

Try It

Evaluate [latex]\cos^{−1}(−0.4)[/latex] using a calculator.

Try It

How To: Given two sides of a right triangle like the one shown in Figure 7, find an angle.

An illustration of a right triangle with an angle theta. Adjacent to theta is the side a, opposite theta is the side p, and the hypoteneuse is side h.

Figure 7

  1. If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use the equation [latex]\theta=\cos^{−1}\left(\frac{a}{h}\right)[/latex].
  2. If one given side is the hypotenuse of length h and the side of length p opposite to the desired angle is given, use the equation [latex]\theta=\sin^{−1}\left(\frac{p}{h}\right)[/latex].
  3. If the two legs (the sides adjacent to the right angle) are given, then use the equation [latex]\theta=\tan^{−1}\left(\frac{p}{a}\right)[/latex].

Example 4: Applying the Inverse Cosine to a Right Triangle

Solve the triangle in Figure 8 for the angle θ.

An illustration of a right triangle with the angle theta. Adjacent to the angle theta is a side with a length of 9 and a hypoteneuse of length 12.

Figure 8

Try It

Solve the triangle in Figure 9 for the angle θ.

An illustration of a right triangle with the angle theta. Opposite to the angle theta is a side with a length of 6 and a hypoteneuse of length 10.

Figure 9

Try It

Finding Exact Values of Composite Functions with Inverse Trigonometric Functions

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f(x) and g(x) be two different trigonometric functions belonging to the set {sin(x), cos(x), tan(x)} and let [latex]f^{−1}(y)[/latex] and [latex]g^{−1}(y)[/latex] be their inverses.

Evaluating Compositions of the Form [latex]f\left(f^{−1}(y)\right)[/latex] and [latex]f^{−1}(f(x))[/latex]

For any trigonometric function, [latex]f(f^{−1}(y))=y[/latex] for all y in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f was defined to be identical to the domain of [latex]f^{−1}[/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{−1}(f(x))[/latex].

A General Note: Compositions of a trigonometric function and its inverse

[latex]\begin{align} &\sin(\sin^{−1}x)=x\text{ for }−1\leq x\leq1\\ &\cos(\cos^{−1}x)=x\text{ for }−1\leq x\leq1 \\ &\tan(\tan^{−1}x)=x\text{ for }−\infty\text{ < }x\text{ < }\infty \end{align}[/latex]

[latex]\begin{align} \hfill &\sin^{−1}(\sin x)=x\text{ only for }−\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \hfill \\ &\cos^{−1}(\cos x)=x\text{ only for }0\leq x\leq\pi \hfill \\ &\tan^{−1}(\tan x)=x\text{ only for }−\frac{\pi}{2}\text{ < }x\text{ < }\frac{\pi}{2} \end{align}[/latex]

Q & A

Is it correct that [latex]\sin^{−1}(\sin x)=x[/latex]?

No. This equation is correct if x belongs to the restricted domain [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][\latex]. The situation is similar for cosine and tangent and their inverses. For example, [latex]\sin^{−1}\left(\sin\left(\frac{3\pi}{4}\right)\right)=\frac{\pi}{4}[/latex].

How To:

Given an expression of the form [latex]f^{−1}(f(\theta))[/latex] where [latex]f(\theta)=\sin\theta\text{, }\cos\theta\text{, or }\tan\theta[/latex], evaluate.

  1. If θ is in the restricted domain of f, then [latex]f^{−1}(f(\theta))=\theta[/latex].
  2. If not, then find an angle ϕ within the restricted domain of f such that [latex]f(\phi)=f(\theta)[/latex]. Then [latex]f^{−1}(f(\theta))=\phi[/latex].

Example 5: Using Inverse Trigonometric Functions

Evaluate the following:

  1. [latex]\sin^{−1}(\sin(\frac{\pi}{3}))[/latex]
  2. [latex]\sin^{−1}(\sin(\frac{2\pi}{3}))[/latex]
  3. [latex]\cos^{−1}(\cos(\frac{2\pi}{3}))[/latex]
  4. [latex]\cos^{−1}(\cos(−\frac{\pi}{3}))[/latex]

Try It

Evaluate [latex]\tan^{−1}\left(\tan\left(\frac{\pi}{8}\right)\right)[/latex] and [latex]\tan^{−1}\left(\tan\left(\frac{11\pi}{9}\right)\right)[/latex].

Evaluating Compositions of the Form [latex]f^{−1}(g(x))[/latex]

Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form [latex]f^{−1}(g(x))[/latex]. For special values of x, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is θ, making the other [latex]\frac{\pi}{2}−\theta[/latex]. Consider the sine and cosine of each angle of the right triangle in Figure 10.

An illustration of a right triangle with angles theta and pi/2 - theta. Opposite the angle theta and adjacent the angle pi/2-theta is the side a. Adjacent the angle theta and opposite the angle pi/2 - theta is the side b. The hypoteneuse is labeled c.

Figure 10. Right triangle illustrating the cofunction relationships

Because [latex]\cos\theta=\frac{b}{c}=\sin\left(\frac{\pi}{2}−\theta\right)[/latex], we have [latex]\sin^{−1}(\cos\theta)=\frac{\pi}{2}−\theta\text{ if }0\leq\theta\leq\pi[/latex]. If θ is not in this domain, then we need to find another angle that has the same cosine as θ and does belong to the restricted domain; we then subtract this angle from [latex]\frac{\pi}{2}[/latex]. Similarly, [latex]\sin\theta=\frac{a}{c}=\cos\left(\frac{\pi}{2}−\theta\right)[/latex], so [latex]\cos^{−1}(\sin\theta)=\frac{\pi}{2}−\theta\text{ if }−\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}[/latex]. These are just the function-cofunction relationships presented in another way.

How To: Given functions of the form [latex]\sin^{−1}(\cos x)\text{ and }\cos^{−1}(\sin x)[/latex], evaluate them.

  1. If x is in [0,π], then [latex]\sin^{−1}(\cos x)=\frac{\pi}{2}−x[/latex].
  2. If x is not in [0,π], then find another angle y in [0,π] such that [latex]\cos y=\cos x[/latex].
    [latex]\sin^{−1}(\cos x)=\frac{\pi}{2}−y[/latex]
  3. If x is in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], then [latex]\cos^{−1}(\sin x)=\frac{\pi}{2}−x[/latex].
  4. If x is not in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], then find another angle y in [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex] such that [latex]\sin y=\sin x[/latex].
    [latex]\cos^{−1}(\sin x)=\frac{\pi}{2}−y[/latex]

Example 6: Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate [latex]\sin^{−1}(\cos(\frac{13\pi}{6}))[/latex]

  1. by direct evaluation.
  2. by the method described previously.

Try It

Evaluate [latex]\cos^{−1}(\sin(−\frac{11\pi}{4}))[/latex].

Try It

Evaluating Compositions of the Form [latex]f(g^{−1}(x))[/latex]

To evaluate compositions of the form [latex]f(g^{−1}(x))[/latex], where f and g are any two of the functions sine, cosine, or tangent and x is any input in the domain of [latex]g−1[/latex], we have exact formulas, such as [latex]\sin\left({\cos}^{−1}x\right)=\sqrt{1−{x}^{2}}[/latex]. When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity, [latex]\sin^{2}x+cos^{2}x=1[/latex], to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.

Example 7: Evaluating the Composition of a Sine with an Inverse Cosine

Find an exact value for [latex]\sin\left(\cos^{−1}\left(\frac{4}{5}\right)\right)[/latex].

Try It

Evaluate [latex]\cos(\tan^{−1}(\frac{5}{12}))[/latex].

Example 8: Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for [latex]\sin\left(\tan^{−1}\left(\frac{7}{4}\right)\right)[/latex].

Try It

Evaluate [latex]\cos(\sin^{−1}(\frac{7}{9}))[/latex].

Try It

Example 9: Finding the Cosine of the Inverse Sine of an Algebraic Expression

Find a simplified expression for [latex]\cos\left(\sin^{−1}\left(\frac{x}{3}\right)\right)[/latex] for [latex]−3\leq x\leq3[/latex].

Try It

Find a simplified expression for [latex]\sin\left(\tan^{−1}\left(4x\right)\right)\\[/latex] for [latex]−\frac{1}{4}\leq x \leq\frac{1}{4}[/latex].

Try It

Key Concepts

  • An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
  • Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
  • For any trigonometric function [latex]f(x)[/latex], if [latex]x=f^{−1}(y)[/latex], then [latex]f(x)=y[/latex]. However, [latex]f(x)=y[/latex] only implies [latex]x=f^{−1}(y)[/latex] if x is in the restricted domain of f.
  • Special angles are the outputs of inverse trigonometric functions for special input values; for example, [latex]\frac{\pi}{4}=\tan^{−1}( 1 )\text{ and }\frac{\pi}{6}=\sin^{−1}(\frac{1}{2})[/latex].
  • A calculator will return an angle within the restricted domain of the original trigonometric function.
  • Inverse functions allow us to find an angle when given two sides of a right triangle.
  • In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, [latex]\sin\left(\cos^{−1}\left(x\right)\right)=\sqrt{1−x^{2}}[/latex].
  • If the inside function is a trigonometric function, then the only possible combinations are [latex]\sin^{−1}\left(\cos x\right)=\frac{\pi}{2}−x[/latex] if [latex]0\leq x\leq\pi[/latex] and [latex]\cos^{−1}\left(\sin x\right)=\frac{\pi}{2}−x[/latex] if [latex]−\frac{\pi}{2}\leq x \leq\frac{\pi}{2}[/latex].
  • When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function.
  • When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides.

Glossary

arccosine
another name for the inverse cosine; [latex]\arccos x=\cos^{−1}x[/latex]
arcsine
another name for the inverse sine; [latex]\arcsin x=\sin^{−1}x[/latex]
arctangent
another name for the inverse tangent; [latex]\arctan x=\tan^{−1}x[/latex]
inverse cosine function
the function [latex]\cos^{−1}x[/latex], which is the inverse of the cosine function and the angle that has a cosine equal to a given number
inverse sine function
the function [latex]\sin^{−1}x[/latex], which is the inverse of the sine function and the angle that has a sine equal to a given number
inverse tangent function
the function [latex]\tan^{−1}x[/latex], which is the inverse of the tangent function and the angle that has a tangent equal to a given number