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.

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].

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].

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].

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

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

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

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.

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.

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

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.

Figure 8

Show Solution

Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

[latex]\begin{align}&\cos\theta=\frac{9}{12}\\ &\theta=\cos^{−1}\left(\frac{9}{12}\right) && \text{Apply definition of the inverse.} \\ &\theta\approx0.7227\text{ or about }41.4096^{\circ} && \text{Evaluate.} \end{align}[/latex]

Try It

Solve the triangle in Figure 9 for the angle θ.

Figure 9

Show Solution

[latex]\sin^{−1}(0.6)=36.87^{\circ}=0.6435[/latex] radians

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].

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.

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.

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 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].

Show Solution

While we could use a similar technique as in Example 6, we will demonstrate a different technique here. From the inside, we know there is an angle such that [latex]\tan\theta=\frac{7}{4}[/latex]. We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure 12.

Figure 12. A right triangle with two sides known

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.

[latex]\begin{gathered}4^{2}+7^{2}=\text{hypotenuse}^{2} \\ \text{hypotenuse}=\sqrt{65} \end{gathered}[/latex]

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is [latex]2\pi[/latex]. In other words, every [latex]2\pi[/latex] units, the y-values repeat. If we need to find all possible solutions, then we must add [latex]2\pi k[/latex], where [latex]k[/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\pi :[/latex]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example 14: Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation [latex]\sec \theta =-4[/latex], giving your answer in radians.

Show Solution

We can begin with some algebra.

[latex]\begin{gathered}\sec \theta =-4\\ \frac{1}{\cos \theta }=-4\\ \cos \theta =-\frac{1}{4}\end{gathered}[/latex]

Check that the MODE is in radians. Now use the inverse cosine function.

[latex]\begin{gathered}{\cos }^{-1}\left(-\frac{1}{4}\right)\approx 1.8235 \\ \theta \approx 1.8235+2\pi k \end{gathered}[/latex]

Since [latex]\frac{\pi }{2}\approx 1.57[/latex] and [latex]\pi \approx 3.14[/latex], 1.8235 is between these two numbers, thus [latex]\theta \approx \text{1}\text{.8235}[/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.

Figure 13.

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\theta \text{ }\text{ }\text{'}\approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}[/latex] The other solution in quadrant III is [latex]\theta \text{ }\text{ }\text{'}\approx \pi +\text{1}\text{.3181}\approx \text{4}\text{.4597}\text{.}[/latex]

The solutions are [latex]\theta \approx 1.8235\pm 2\pi k[/latex] and [latex]\theta \approx 4.4597\pm 2\pi k[/latex].

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\pi[/latex], not [latex]2\pi[/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\frac{\pi }{2}[/latex], unless, of course, a problem places its own restrictions on the domain.

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[/latex] or [latex]u[/latex]. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example 21: Solving a Trigonometric Equation Using Algebra

Solve exactly:

[latex]2{\sin }^{2}\theta +\sin \theta =0;0\le \theta <2\pi[/latex]

Show Solution

This problem should appear familiar as it is similar to a quadratic. Let [latex]\sin \theta =x[/latex]. The equation becomes [latex]2{x}^{2}+x=0[/latex]. We begin by factoring:

[latex]\begin{gathered}2{x}^{2}+x=0\\ x\left(2x+1\right)=0\end{gathered}[/latex]

Set each factor equal to zero.

[latex]\begin{gathered}x=0 \\ 2x+1=0 \\ x=-\frac{1}{2} \end{gathered}[/latex]

Then, substitute back into the equation the original expression [latex]\sin \theta[/latex] for [latex]x[/latex]. Thus,

[latex]\begin{gathered}\sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

The solutions within the domain [latex]0\le \theta <2\pi[/latex] are [latex]\theta =0,\pi ,\frac{7\pi }{6},\frac{11\pi }{6}[/latex]. If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

[latex]\begin{gathered}2{\sin }^{2}\theta +\sin \theta =0 \\ \sin \theta \left(2\sin \theta +1\right)=0 \\ \sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ 2\sin \theta +1=0 \\ 2\sin \theta =-1 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

Analysis of the Solution

We can see the solutions on the graph in Figure 3. On the interval [latex]0\le \theta <2\pi[/latex], the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Figure 14

We can verify the solutions on the unit circle in Sum and Difference Identities as well.

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.
• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
• Equations involving a single trigonometric function can be solved or verified using the unit circle.
• We can also solve trigonometric equations using a graphing calculator.
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.