Inverse Trigonometric Functions

For any right triangle, given one other angle and the length of one side, we can figure out what the other angles and sides are. But what if we are given only two sides of a right triangle? We need a procedure that leads us from a ratio of sides to an angle. This is where the notion of an inverse to a trigonometric function comes into play. In this section, we will explore the 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 below.

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

• Since $\sin\left(\tfrac{\pi}{6}\right)=\tfrac12$, then $\tfrac{\pi}{6}=\sin^{-1}\left(\tfrac12\right)$.
• Since $\cos(\pi)=-1$, then $\pi=\cos^{-1}(-1)$.
• Since $\tan\left(\tfrac{\pi}{4}\right)=1$, then $\tfrac{\pi}{4}=\tan^{-1}(1)$.

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 $f(a)=b$, then an inverse function would satisfy $f^{-1}(b)=a$.

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 $0$. The figure below shows the graph of the sine function limited to $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ and the graph of the cosine function limited to $[0,\pi]$.

(a) Sine function on a restricted domain of $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.

(b) Cosine function on a restricted domain of $[0,\pi]$.

The figure below shows the graph of the tangent function limited to $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

Tangent function on a restricted domain of $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

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 $y=\sin^{-1}x$ means $x=\sin y$. The inverse sine function is sometimes called the arcsine function, and notated $\arcsin x$.

  $$y=\sin^{-1}x\ \text{has domain}\ [-1,1]\ \text{and range}\ \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$$

• The inverse cosine function $y=\cos^{-1}x$ means $x=\cos y$. The inverse cosine function is sometimes called the arccosine function, and notated $\arccos x$.

  $$y=\cos^{-1}x\ \text{has domain}\ [-1,1]\ \text{and range}\ [0,\pi]$$

• The inverse tangent function $y=\tan^{-1}x$ means $x=\tan y$. The inverse tangent function is sometimes called the arctangent function, and notated $\arctan x$.

  $$y=\tan^{-1}x\ \text{has domain}\ (-\infty,\infty)\ \text{and range}\ \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$$

The graphs of the inverse functions are shown below. Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that $\sin^{-1}x$ has domain $[-1,1]$ and range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, $\cos^{-1}x$ has domain $[-1,1]$ and range $[0,\pi]$, and $\tan^{-1}x$ has domain of all real numbers and range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. 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 $y=x$.

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

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

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

Example. Given $\sin\left(\tfrac{5\pi}{12}\right)\approx0.96593$, write a relation involving the inverse sine.

Solution. Use the relation for the inverse sine. If $\sin y=x$, then $\sin^{-1}x=y$.

In this problem, $x=0.96593$, and $y=\tfrac{5\pi}{12}$.

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 $\tfrac{\pi}{6}$ ($30^\circ$), $\tfrac{\pi}{4}$ ($45^\circ$), and $\tfrac{\pi}{3}$ ($60^\circ$), and their reflections into other quadrants.

Example. Evaluate each of the following.

ⓐ $\sin^{-1}\left(\tfrac12\right)$ ⓑ $\sin^{-1}\left(-\tfrac{\sqrt2}{2}\right)$ ⓒ $\cos^{-1}\left(-\tfrac{\sqrt3}{2}\right)$ ⓓ $\tan^{-1}(1)$

Solution. ⓐ Evaluating $\sin^{-1}\left(\tfrac12\right)$ is the same as determining the angle that would have a sine value of $\tfrac12$. In other words, what angle $x$ would satisfy $\sin(x)=\tfrac12$? There are multiple values that would satisfy this relationship, such as $\tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$, but we know we need the angle in the interval $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, so the answer will be $\sin^{-1}\left(\tfrac12\right)=\tfrac{\pi}{6}$. Remember that the inverse is a function, so for each input, we will get exactly one output.

ⓑ To evaluate $\sin^{-1}\left(-\tfrac{\sqrt2}{2}\right)$, we know that $\tfrac{5\pi}{4}$ and $\tfrac{7\pi}{4}$ both have a sine value of $-\tfrac{\sqrt2}{2}$, but neither is in the interval $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. For that, we need the negative angle coterminal with $\tfrac{7\pi}{4}$: $\sin^{-1}\left(-\tfrac{\sqrt2}{2}\right)=-\tfrac{\pi}{4}$.

ⓒ To evaluate $\cos^{-1}\left(-\tfrac{\sqrt3}{2}\right)$, we are looking for an angle in the interval $[0,\pi]$ with a cosine value of $-\tfrac{\sqrt3}{2}$. The angle that satisfies this is $\cos^{-1}\left(-\tfrac{\sqrt3}{2}\right)=\tfrac{5\pi}{6}$.

ⓓ Evaluating $\tan^{-1}(1)$, we are looking for an angle in the interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ with a tangent value of $1$. The correct angle is $\tan^{-1}(1)=\tfrac{\pi}{4}$.

Evaluate each of the following, as in the example above.

Evaluate the remaining two expressions.

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 $\theta$ 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. Evaluate $\sin^{-1}(0.97)$ using a calculator.

Solution. Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using.

In radian mode, $\sin^{-1}(0.97)\approx1.3252$. In degree mode, $\sin^{-1}(0.97)\approx75.93^\circ$. Note that in calculus and beyond we will use radians in almost all cases.

Example. Solve the triangle below for the angle $\theta$.

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

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 $f^{-1}(y)$ and $g^{-1}(y)$ be their inverses.

Evaluating Compositions of the Form $f(f^{-1}(y))$ and $f^{-1}(f(x))$

For any trigonometric function, $f(f^{-1}(y))=y$ 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 $f^{-1}$. However, we have to be a little more careful with expressions of the form $f^{-1}(f(x))$.

Example. Evaluate the following:

ⓐ $\sin^{-1}\left(\sin\left(\tfrac{\pi}{3}\right)\right)$ ⓑ $\sin^{-1}\left(\sin\left(\tfrac{2\pi}{3}\right)\right)$ ⓒ $\cos^{-1}\left(\cos\left(\tfrac{2\pi}{3}\right)\right)$ ⓓ $\cos^{-1}\left(\cos\left(-\tfrac{\pi}{3}\right)\right)$

Solution. ⓐ $\tfrac{\pi}{3}$ is in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, so $\sin^{-1}\left(\sin\left(\tfrac{\pi}{3}\right)\right)=\tfrac{\pi}{3}$.

ⓑ $\tfrac{2\pi}{3}$ is not in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, but $\sin\left(\tfrac{2\pi}{3}\right)=\sin\left(\tfrac{\pi}{3}\right)$, so $\sin^{-1}\left(\sin\left(\tfrac{2\pi}{3}\right)\right)=\tfrac{\pi}{3}$.

ⓒ $\tfrac{2\pi}{3}$ is in $[0,\pi]$, so $\cos^{-1}\left(\cos\left(\tfrac{2\pi}{3}\right)\right)=\tfrac{2\pi}{3}$.

ⓓ $-\tfrac{\pi}{3}$ is not in $[0,\pi]$, but $\cos\left(-\tfrac{\pi}{3}\right)=\cos\left(\tfrac{\pi}{3}\right)$ because cosine is an even function. $\tfrac{\pi}{3}$ is in $[0,\pi]$, so $\cos^{-1}\left(\cos\left(-\tfrac{\pi}{3}\right)\right)=\tfrac{\pi}{3}$.

Evaluating Compositions of the Form $f^{-1}(g(x))$

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 $f^{-1}(g(x))$. 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 $\theta$, making the other $\tfrac{\pi}{2}-\theta$. Consider the sine and cosine of each angle of the right triangle below.

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

Example. Evaluate $\sin^{-1}\left(\cos\left(\tfrac{13\pi}{6}\right)\right)$

ⓐ by direct evaluation. ⓑ by the method described previously.

Solution. ⓐ Here, we can directly evaluate the inside of the composition.

Now, we can evaluate the inverse function as we did earlier.

ⓑ We have $x=\tfrac{13\pi}{6}$, $y=\tfrac{\pi}{6}$, and

Evaluating Compositions of the Form $f(g^{-1}(x))$

To evaluate compositions of the form $f(g^{-1}(x))$, where $f$ and $g$ are any two of the functions sine, cosine, or tangent and $x$ is any input in the domain of $g^{-1}$, we have exact formulas, such as $\sin(\cos^{-1}x)=\sqrt{1-x^2}$. 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, $\sin^2x+\cos^2x=1$, to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.

Example. Find an exact value for $\sin\left(\cos^{-1}\left(\tfrac45\right)\right)$.

Solution. Beginning with the inside, we can say there is some angle such that $\theta=\cos^{-1}\left(\tfrac45\right)$, which means $\cos\theta=\tfrac45$, and we are looking for $\sin\theta$. We can use the Pythagorean identity to do this.

Since $\theta=\cos^{-1}\left(\tfrac45\right)$ is in quadrant I, $\sin\theta$ must be positive, so the solution is $\tfrac35$. See the figure below.

We know that the inverse cosine always gives an angle on the interval $[0,\pi]$, so we know that the sine of that angle must be positive; therefore $\sin\left(\cos^{-1}\left(\tfrac45\right)\right)=\sin\theta=\tfrac35$.

Example. Find an exact value for $\sin\left(\tan^{-1}\left(\tfrac74\right)\right)$.

Solution. While we could use a similar technique as in the previous example, we will demonstrate a different technique here. From the inside, we know there is an angle such that $\tan\theta=\tfrac74$. We can envision this as the opposite and adjacent sides on a right triangle, as shown below.

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

Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse.

This gives us our desired composition.

Example. Find a simplified expression for $\cos\left(\sin^{-1}\left(\tfrac{x}{3}\right)\right)$ for $-3\le x\le3$.

Solution. We know there is an angle $\theta$ such that $\sin\theta=\tfrac{x}{3}$.

Because we know that the inverse sine must give an angle on the interval $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, we can deduce that the cosine of that angle must be positive.

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 $f(x)$, if $x=f^{-1}(y)$, then $f(x)=y$. However, $f(x)=y$ only implies $x=f^{-1}(y)$ if $x$ is in the restricted domain of $f$.
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, $\tfrac{\pi}{4}=\tan^{-1}(1)$ and $\tfrac{\pi}{6}=\sin^{-1}\left(\tfrac12\right)$.
• 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, $\sin\left(\cos^{-1}(x)\right)=\sqrt{1-x^2}$.
• If the inside function is a trigonometric function, then the only possible combinations are $\sin^{-1}(\cos x)=\tfrac{\pi}{2}-x$ if $0\le x\le\pi$ and $\cos^{-1}(\sin x)=\tfrac{\pi}{2}-x$ if $-\tfrac{\pi}{2}\le x\le\tfrac{\pi}{2}$.
• 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.

Key terms

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

Practice

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

Find exact values of composite functions with inverse trigonometric functions

This section is adapted from Precalculus 2e, Section 6.3: Inverse Trigonometric Functions by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the decorative “Trig Functions / Inverse Trig Functions” domain-range chart image (Figure 1), recreated as a two-by-two Markdown table with identical content; recreated all fifteen instructional figures as accessible spec-first SVGs built from exact coordinates — the restricted-domain sine and cosine panels and the restricted-domain tangent graph; the three function-and-inverse overlay graphs (sine/arcsine, cosine/arccosine, tangent/arctangent, each with the dashed line $y=x$), simplified to label only the endpoints of each inverse curve’s range ($\pm\tfrac{\pi}{2}$ or $\pi$) rather than every eighth-multiple of $\pi$ the source prints, since the shape and endpoints are the instructional content; the generic right-triangle diagram for the calculator How To; the two worked-example/Try-It right triangles solved for $\theta$; the cofunction-relationship triangle; and the three composition right triangles. Converted the two-column “Trig Functions/Inverse Trig Functions” relations into the book’s callout convention, and the Q&A into the book’s Q&A callout convention. Every retained Try It became a real fillin component; the four-part Try It after Example 2 was split into two consecutive pairs of fillins (separated by a sentence of connecting prose) to respect the section’s 2–3 consecutive-question limit, each carrying answerForm="evaluated-trig" so retyping the printed subject cannot pass. Every calculator-rounding Try It and exercise states the number of decimal places to make it gradable as a decimal. Omitted the “Access these online resources” media link. Adapted nine selected end-of-section exercises — one true/false identity, a domain-and-range graphing item (adapted to a static figure plus two fill-ins, with a graph-recognition multiple choice added for the inverse tangent alongside it), three exact-value evaluations, three calculator/right-triangle evaluations, and two exact-value compositions — into eleven interactive components in a closing Practice block, one group per objective. This module’s own “Key Concepts” and its module-scoped <glossary> (six terms: arccosine, arcsine, arctangent, and the three inverse function definitions) are transcribed as this section’s ## Key concepts and ## Key terms; the printed textbook interleaves them with the rest of chapter 6’s summary terms on its own consolidated Chapter Review pages (PDF pp. 678–679) rather than printing them immediately after this section, a print-layout choice that does not change which content belongs to module m49390.