Inverse Trigonometric Functions

What is an Inverse Trigonometric Function?

From our Inverse Functions article, we remember that the inverse of a function can be found algebraically by switching the x- and y-values and then solving for y. We also remember that we can find the graph of the inverse of a function by reflecting the graph of the original function over the line \(y=x\).

We already know about inverse operations. For instance, addition and subtraction are inverses, and multiplication and division are inverses.

In trigonometry, this idea is the same. Inverse trigonometric functions do the opposite of the normal trigonometric functions. More specifically,

• Inverse sine, \(sin^{-1}\) or \(arcsin\), does the opposite of the sine function.

• Inverse cosine, \(cos^{-1}\) or \(arccos\) , does the opposite of the cosine function.

• Inverse tangent, \(tan^{-1}\) or \(arctan\), does the opposite of the tangent function.

• Inverse cotangent, \(cot^{-1}\) or \(arccot\), does the opposite of the cotangent function.

• Inverse secant, \(sec^{-1}\) or \(arcsec\), does the opposite of the secant function.

• Inverse cosecant, \(csc^{-1}\) or \(arccsc\), does the opposite of the cosecant function.

Using the right triangle below, let's define the inverse trig functions!

Fig. 1. A right triangle with the sides labeled.

A Note on Notation

As you might have noticed, the notation used to define the inverse trig functions makes it look like they have exponents. While it may seem like it, the \(-1\) superscript is NOT an exponent! In other words, \(\sin^{-1}(x)\) is not the same as \(\dfrac{1}{\sin(x)}\)! The \(-1\) superscript simply means “inverse.”

Inverse Trigonometric Functions: Formulas

The main inverse trigonometric formulas are listed in the table below.

Let's explore these with an example!

Inverse Trigonometric Function Graphs

What do the inverse trigonometric functions look like? Let's check out their graphs.

Domain and Range of Inverse Trigonometric Functions

But, before we can graph the inverse trigonometric functions, we need to talk about their domains. Because the trigonometric functions are periodic, and therefore not one-to-one, they don't have inverse functions. So then, how can we have inverse trigonometric functions?

To find inverses of the trigonometric functions, we must either restrict or specify their domains so that they are one-to-one! Doing so allows us to define a unique inverse of either sine, cosine, tangent, cosecant, secant, or cotangent.

In general, we use the following convention when evaluating inverse trigonometric functions:

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains specified in the table above and reflect those graphs about the line \(y=x\), just as we did for finding Inverse Functions.

Below are the 6 main inverse trigonometric functions and their graphs, domain, range (also known as the principal interval), and any asymptotes.

Inverse Trigonometric Functions: Unit Circle

When we deal with inverse trigonometric functions, the unit circle is still a very helpful tool. While we typically think about using the unit circle to solve trigonometric functions, the same unit circle can be used to solve, or evaluate, the inverse trigonometric functions.

Before we get to the unit circle itself, let's take a look at another, simpler tool. The diagrams below can be used to help us remember from which quadrants the inverse trigonometric functions on the unit circle will come.

Fig. 3. A diagram that shows in which quadrants cosine, secant, and cotangent (and therefore their inverses) return values.

Just as the cosine, secant, and cotangent functions return values in Quadrants I and II (between 0 and 2π), their inverses, arc cosine, arc secant, and arc cotangent, do as well.

Fig. 4. A diagram that shows in which quadrants sine, cosecant, and tangent (and therefore their reciprocals) return values.

Just as the sine, cosecant, and tangent functions return values in Quadrants I and IV (between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\)), their inverses, arc sine, arc cosecant, and arc tangent, do as well. Note that the values from Quadrant IV will be negative.

We might recall that we can use the Unit Circle to solve trigonometric functions of special angles: angles that have trigonometric values that we evaluate exactly.

Fig. 5. The unit circle.

When using the unit circle to evaluate inverse trigonometric functions, there are several things we need to keep in mind:

• If the answer is in Quadrant IV, it must be a negative answer (in other words, we go clockwise from the point (1, 0) instead of counterclockwise).
  • For example, if we want to evaluate \(\sin^{-1}\left( -\dfrac{1}{2} \right)\) , our first instinct is to say the answer is \(330^o\) or \(\dfrac{11\pi}{6}\). However, since the answer must be between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\) (the standard domain for inverse sine), we need to change our answer to the co-terminal angle \(-30^o\), or \(-\dfrac{\pi}{6}\).
• To use the unit circle to get the inverses for the reciprocal functions (secant, cosecant, and cotangent), we can take the reciprocal of what is in the parentheses and use the trigonometric functions.
  • For example, if we want to evaluate \(\sec^{-1}(-\sqrt{2})\), we would look for \(\cos^{-1} \left( - \dfrac{1}{\sqrt{2}} \right)\) on the unit circle, which is the same as \(\cos^{-1} \left( - \dfrac{\sqrt{2}}{2} \right)\), which gives us \(\dfrac{3\pi}{4}\) or \(135^o\).
• Remember to check your work!
  • Given any trigonometric function with a positive argument (assuming the conventional restricted domain), we should get an angle that is in Quadrant I \( 0 \leq \theta \leq \left( \dfrac{\pi}{2} \right) \).
  • For the arcsin, arccsc, and arctan functions:
    • If we are given a negative argument, our answer will be in Quadrant IV \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\).
  • For the arccos, arcsec, and arccot functions:
    • If we are given a negative argument, our answer will be in Quadrant II \(\dfrac{\pi}{2} \leq \theta \leq \pi\).
  • For any argument that is outside the domains of the trigonometric functions for arcsin, arccsc, arccos, and arcsec, we will get no solution.

The Calculus of Inverse Trigonometric Functions

In calculus, we will be asked to find derivatives and integrals of inverse trigonometric functions. In this article, we present a brief overview of these topics.

For a more in-depth analysis, please refer to our articles on Derivatives of Inverse Trigonometric Functions and Integrals Resulting in Inverse Trigonometric Functions.

Derivatives of Inverse Trigonometric Functions

A surprising fact about the Derivatives of Inverse Trigonometric Functions is that they are algebraic functions, not trigonometric functions. The derivatives of inverse trigonometric functions are defined as:

Integrals Resulting in Inverse Trigonometric Functions

Previously, we have developed the formulas for the derivatives of inverse trigonometric functions. These formulas are what we use to develop the Integrals Resulting in Inverse Trigonometric Functions. These integrals are defined as:

There are 6 inverse trigonometric functions, so why are there only three integrals? The reason for this is that the remaining three integrals are just negative versions of these three. In other words, the only difference between them is whether the integrand is positive or negative.

• Rather than memorizing three more formulas, if the integrand is negative, we can factor out -1 and evaluate using one of the three formulas above.

Inverse Trigonometric Integrals

Other than the integrals that result in the inverse trigonometric functions, there are integrals that involve the inverse trigonometric functions. These integrals are:

• The inverse trigonometric integrals that involve arc sine.

  • \(\int sin^{-1} u du = sin^{-1}(u)+\sqrt{1-u^2}+C\)

  • \(\int u \sin^{-1}u du=\dfrac{2u^2-1}{4} \sin^{-1}(u)+\dfrac{u\sqrt{1-u^2}}{4}+C\)

  • \(\\int u^n sin^{-1}u du \dfrac{1}{n+1} \left[ u^{n+1} \sin^{-1}(u) - \int \dfrac{u^{n+1}du}{\sqrt{1-u^2}}, n \neq -1 \right]\)

• The inverse trigonometric integrals that involve arc cosine.

  • \(\int cos^{-1}udu =cos^{-1}(u)-\sqrt{1-u^2}+C\)

  • \(\int cos^{-1} u du = \dfrac{1}{n+1}\left[ u^{n+1} \cos^{-1} (u)+ \int \dfrac{u^{n+1}du}{\sqrt{1-u^2}} \right], n \neq -1\)

• The inverse trigonometric integrals that involve arc tangent.

  • \(\int tan^{-1}udu=tan^{-1}(u)-\dfrac{1}{2}ln(1+u^2)+C\)

  • \(\int u \tan^{-1} u du = \dfrac{u^2-1}{2}\tan^{-1}(u)+C\)

  • \(\int u^n tan^{-1} udu = \dfrac{1}{n+1}\left[ \dfrac{u^{n+1} du}{1+u^2}\right], n \neq -1\)

Solving Inverse Trigonometric Functions: Examples

When we solve, or evaluate, inverse trigonometric functions, the answer we get is an angle.

What about the composition of a trigonometric function and its inverse?

Let's think about the answer for the second expression in the example above.

• Isn't the inverse of a function supposed to undo the original function? Why isn't \( \sin^{-1} ( \sin (\pi) )= \pi \)?

  • Remembering the definition of inverse functions: a function \(f\) and its inverse \(f^{-1}\) satisfy the conditions \( f (f^{-1}(y))=y\)for all y in the domain of \( f^{-1}\) , and \(f^{-1}(f(x))=x\) for all \(x\) in the domain of \(f\).

So, what happened in this example?

• The issue here is that the inverse sine function is the inverse of the restricted sine function on the domain \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] \) . Therefore, for \(x\) in the interval \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] \), it is true that \(\sin^{-1}(\sin(x))=x\). However, for values of x outside this interval, this equation does not hold true, even though \(\sin^{-1}(\sin(x))\)is defined for all real numbers of \(x\).

Then, what about \(\sin(\sin^{-1}(y))\)? Does this expression have a similar issue?

• This expression does not have the same issue because the domain of \(\sin^{-1}\) is the interval \([-1, 1]\).

  • So, \(\sin(\sin^{-1}(y))=y\) if \(-1 \leq y \leq 1\). This expression is not defined for any other values of \(y\).

Let's summarize these findings:

When it is not explicitly specified, we restrict the inverse trigonometric functions to the standard bounds specified in the section “inverse trigonometric functions in a table”. We saw this restriction in place in the first example.

However, there may be cases where we want to find an angle corresponding to a trigonometric value evaluated within a different specified bound. In such cases, it is useful to remember the trigonometric quadrants:

Fig. 6. The trigonometric quadrants and where which trig (and therefore inverse trig) functions are positive.