5.5 – Inverse Trigonometric Functions and their Graphs

5.5 Video 1 of 3:Inverse Sine Function

5.5 Video 2 of 3:Inverse Cosine Function

5.5 Video 3 of 3:Inverse Tangent Function

Section Notes

Inverse Trigonometric Functions and their Graphs

Recall that the inverse function $f^{-1}(x)$ is a function which “undoes” the function $f(x)$ . That is, the composite of a function and its inverse results in $x$ . Mathematically, that is $f(f^{-1}(x))=f^{-1}(f(x))=x$ . For a function to have an inverse, it is important that it is a one-to-one function. None of the six trigonometric functions are one-to-one, so they can not have an inverse function. However, if the domain of a trigonometric function is strategically limited, then we can create a one-to-one function on that domain. Such a function would have an inverse.

Sine

We saw previously that the sine function is periodic, and continues forever in both the positive and negative $x$ direction:

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But if we choose a domain in such a way that no $y$ value output appears twice, we are limited to half of a period of the sine function, selected in blue:

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Thus, we will consider the $sin$ function on the domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ :

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The graph of the inverse sine function is shown here:

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The Inverse Sine Function
On the domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , the sine function has an inverse, noted either $\sin^{-1}(x)$ or $\arcsin(x)$ . The domain of $\sin^{-1}(x)$ is $[-1,1]$ and its range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , is defined by $\sin^{-1}(x)=y \Leftrightarrow \sin(y)=x$ And holds the properties $\begin{align}\sin(\sin^{-1}(x))=x & & \text{ for } &&-1\le x\le 1\\ \sin^{-1}(\sin(x))=x & & \text{ for } & &-\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align}$

The Inverse Sine Function

On the domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , the sine function has an inverse, noted either $\sin^{-1}(x)$ or $\arcsin(x)$ . The domain of $\sin^{-1}(x)$ is $[-1,1]$ and its range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , is defined by

$\sin^{-1}(x)=y \Leftrightarrow \sin(y)=x$
And holds the properties

$\begin{align*}\sin(\sin^{-1}(x))=x & & \text{ for } &&-1\le x\le 1\\ \sin^{-1}(\sin(x))=x & & \text{ for } & &-\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align*}$

To evaluate the inverse sine function, it is often easier to convert it into a sine equation, and solve:

Find the value of $\sin^{-1}\left(-\frac{1}{2}\right)$
We can start by writing an equation. The expression is equal to an unknown value, so we can write $\sin^{-1}\left(-\frac{1}{2}\right)=x$ Which is equivalent to the sine equation $\sin(x)=-\frac{1}{2}$ Referring to the unit circle if necessary, we can see that $\sin(x)=-\frac{1}{2}$ for values such as $x=\left\{...,-\frac{5\pi}{6},-\frac{\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6},...\right\}$ , but only one of these values lies between $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , so we have a unique solution $x=\frac{-\pi}{6}$

Find the value of $\sin^{-1}\left(-\frac{1}{2}\right)$

We can start by writing an equation. The expression is equal to an unknown value, so we can write

$\sin^{-1}\left(-\frac{1}{2}\right)=x$
Which is equivalent to the sine equation
$\sin(x)=-\frac{1}{2}$

Referring to the unit circle if necessary, we can see that $\sin(x)=-\frac{1}{2}$ for values such as $x=\left\{...,-\frac{5\pi}{6},-\frac{\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6},...\right\}$ , but only one of these values lies between $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , so we have a unique solution $x=\frac{-\pi}{6}$

Evaluate the expression $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$
Proceeding similarly to the previous, we have: $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=x \ \ \Leftrightarrow \ \ \sin(x)=\frac{\sqrt{2}}{2}$ Thus, the only $x-$ value between $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which the sine value is $\frac{\sqrt{2}}{2}$ is $x=\frac{\pi}{4}$ .

Evaluate the expression $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$

Proceeding similarly to the previous, we have:

$\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=x \ \ \Leftrightarrow \ \ \sin(x)=\frac{\sqrt{2}}{2}$

Thus, the only $x-$ value between $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which the sine value is $\frac{\sqrt{2}}{2}$ is $x=\frac{\pi}{4}$ .

Evaluate $\arcsin\left(\frac{-\sqrt{3}}{2}\right)$
$\arcsin\left(\frac{-\sqrt{3}}{2}\right)=x \ \ \Leftrightarrow \ \ \sin(x)=\frac{-\sqrt{3}}{2}$ Again, paying attention to the range of $\arcsin(x)$ we have $x=-\frac{\pi}{3}$

Evaluate $\arcsin\left(\frac{-\sqrt{3}}{2}\right)$

$\arcsin\left(\frac{-\sqrt{3}}{2}\right)=x \ \ \Leftrightarrow \ \ \sin(x)=\frac{-\sqrt{3}}{2}$

Again, paying attention to the range of $\arcsin(x)$ we have $x=-\frac{\pi}{3}$

Evaluate the expression $\sin^{-1}\left(\frac{5}{2}\right)$
This expression has no value, as $\frac{5}{2}$ is outside the domain of $\sin^{-1}(x)$ . This can further be illustrated by rewriting the expression as a $\sin$ equation: $\sin^{-1}\left(\frac{5}{2}\right)=x \ \ \Leftrightarrow \ \ \sin(x)=\frac{5}{2}$ But there are no such $x$ values, since $\sin(x)$ has a range of $[-1,1]$ .

Evaluate the expression $\sin^{-1}\left(\frac{5}{2}\right)$

This expression has no value, as $\frac{5}{2}$ is outside the domain of $\sin^{-1}(x)$ . This can further be illustrated by rewriting the expression as a $\sin$ equation:

$\sin^{-1}\left(\frac{5}{2}\right)=x \ \ \Leftrightarrow \ \ \sin(x)=\frac{5}{2}$

But there are no such $x$ values, since $\sin(x)$ has a range of $[-1,1]$ .

Similar to the $\sin$ function, there are values of $\sin^{-1}(x)$ that require the use of a calculator to evaluate. Many calculators have a $\sin^{-1}$ , while others have an $\arcsin$ button, and others still might have an inverse button that, when pressed, triggers the inverse of the next button pushed. It is important to know how to use your calculator for this function.

Use a calculator to evaluate $\sin^{-1}(0.25)$ . Round your answer to four decimal places.
$\sin^{-1}(0.25)\approx 0.2527$

Use a calculator to evaluate $\sin^{-1}(0.25)$ . Round your answer to four decimal places.

$\sin^{-1}(0.25)\approx 0.2527$

Use a calculator to evaluate $\sin^{-1}(0.9)$
$\sin^{-1}(0.9)\approx 1.1198$

Use a calculator to evaluate $\sin^{-1}(0.9)$

$\sin^{-1}(0.9)\approx 1.1198$

Some expressions have both a $\sin$ and a $\sin^{-1}$ in them. They key is to be aware of the domains involved!

Evaluate the exrpession $\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right)$
Because $\frac{\pi}{3}$ is within the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , then the $\sin$ and $\sin^{-1}$ cancel each other, and we have $\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right)=\frac{\pi}{3}$

Evaluate the expression $\sin^{-1}\left(\sin\left(\frac{11\pi}{6}\right)\right)$
First note that $\frac{11\pi}{6}$ is not in the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , so the $\sin$ and $\sin^{-1}$ functions do not simply cancel each other out. But note that $\sin\left(\frac{11\pi}{6}\right)=-\frac{1}{2}$ , so we can substitute this value into the expression: $\sin^{-1}\left(\sin\left(\frac{11\pi}{6}\right)\right)=x$ $=\sin^{-1}\left(-\frac{1}{2}\right)=x$ rewriting the expression as a $\sin$ expression gives: $\sin(x)=-\frac{1}{2}$ The value of $x$ within the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ that solves this equation is $\frac{\pi}{6}$ , thus, $\sin^{-1}\left(\sin\left(\frac{11\pi}{6}\right)\right)=\frac{\pi}{6}$

Evaluate the expression $\sin^{-1}\left(\sin\left(\frac{11\pi}{6}\right)\right)$

First note that $\frac{11\pi}{6}$ is not in the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , so the $\sin$ and $\sin^{-1}$ functions do not simply cancel each other out. But note that $\sin\left(\frac{11\pi}{6}\right)=-\frac{1}{2}$ , so we can substitute this value into the expression:

$\sin^{-1}\left(\sin\left(\frac{11\pi}{6}\right)\right)=x$
$=\sin^{-1}\left(-\frac{1}{2}\right)=x$
rewriting the expression as a $\sin$ expression gives:
$\sin(x)=-\frac{1}{2}$
The value of $x$ within the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ that solves this equation is $\frac{\pi}{6}$ , thus,

$\sin^{-1}\left(\sin\left(\frac{11\pi}{6}\right)\right)=\frac{\pi}{6}$

Such expressions can be simplified by using the reference angles that lie within the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ :

Evaluate the expression $\sin^{-1}\left(\sin\left(\frac{17\pi}{4}\right)\right)$
$\sin^{-1}\left(\sin\left(\frac{17\pi}{4}\right)\right)$ $=\sin^{-1}\left(\sin\left(\frac{\pi}{4}\right)\right)$ $=\frac{\pi}{4}$

Evaluate the expression $\sin^{-1}\left(\sin\left(\frac{17\pi}{4}\right)\right)$

$\sin^{-1}\left(\sin\left(\frac{17\pi}{4}\right)\right)$
$=\sin^{-1}\left(\sin\left(\frac{\pi}{4}\right)\right)$
$=\frac{\pi}{4}$

Cosine

Similar to the sine function, the cosine function is periodic:

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Again, we can select a half-period of cosine which will make the function one-to-one, however, this interval is not the same as the sine. We can see the half-period blow, highlighted in blue.

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Thus, we consider the $\cos$ function on the domain $\left[0,\pi\right]$ :

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The graph of the inverse cosine function is graphed here:

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The Inverse Cosine Function
On the domain $\left[0,\pi\right]$ , the cosine function has an inverse, noted either $\cos(x)$ or $\arccos(x)$ . The domain of $\cos^{-1}(x)$ is $\left[-1,1\right]$ , and has range $\left[0,\pi\right]$ , is defined by $\cos^{-1}(x)=y \ \ \Leftrightarrow \ \ \cos(y)=x$ And holds the properties $\begin{align} \cos(\cos^{-1}(x))=x & & \text{ for } & & -1\le x \le 1 \\ \sin^{-1}(\sin(x))=x & & \text{ for } & & 0 \le x \le \pi\end{align}$

The Inverse Cosine Function

On the domain $\left[0,\pi\right]$ , the cosine function has an inverse, noted either $\cos(x)$ or $\arccos(x)$ . The domain of $\cos^{-1}(x)$ is $\left[-1,1\right]$ , and has range $\left[0,\pi\right]$ , is defined by
$\cos^{-1}(x)=y \ \ \Leftrightarrow \ \ \cos(y)=x$
And holds the properties

$\begin{align*} \cos(\cos^{-1}(x))=x & & \text{ for } & & -1\le x \le 1 \\ \sin^{-1}(\sin(x))=x & & \text{ for } & & 0 \le x \le \pi\end{align*}$

Evaluating the cosine function can be similar to the inverse sine functions:

Find the value of $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = x \ \ \Leftrightarrow \ \ \cos(x)=\frac{\sqrt{3}}{2}$ And the $x-$ value between $\left[0,\pi\right]$ that satisfies this equation is $x=\frac{\pi}{6}$

Find the value of $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = x \ \ \Leftrightarrow \ \ \cos(x)=\frac{\sqrt{3}}{2}$

And the $x-$ value between $\left[0,\pi\right]$ that satisfies this equation is $x=\frac{\pi}{6}$

Evaluate $\cos^{-1}\left(-\frac{1}{2}\right)$
$\cos^{-1}\left(-\frac{1}{2}\right)=x \ \ \Leftrightarrow \ \ \cos(x)=-\frac{1}{2}$ So $x=\frac{2\pi}{3}$

Evaluate $\cos^{-1}\left(-\frac{1}{2}\right)$

$\cos^{-1}\left(-\frac{1}{2}\right)=x \ \ \Leftrightarrow \ \ \cos(x)=-\frac{1}{2}$

So $x=\frac{2\pi}{3}$

Evaluate $\cos^{-1}(1.5)$
1.5 is outside the interval $\left[0,\pi\right]$ ,so there is no solution for this expression.

And expressions using both $\cos$ and $\cos^{-1}$ are treated similarly to ones with s

Evaluate $\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)$
Because $-\frac{\pi}{3}$ is outside the interval $\left[0,\pi\right]$ , we can replace the cosine expression: $\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)$ $\cos^{-1}\left(\frac{1}{2}\right)=x \ \ \Leftrightarrow \ \ \cos(x)=\frac{1}{2}$ Thus, we see that the value in the interval $\left[0,\pi\right]$ that satisfies this equation is $x=\frac{\pi}{3}$ .

Evaluate $\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)$

Because $-\frac{\pi}{3}$ is outside the interval $\left[0,\pi\right]$ , we can replace the cosine expression:

$\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)$
$\cos^{-1}\left(\frac{1}{2}\right)=x \ \ \Leftrightarrow \ \ \cos(x)=\frac{1}{2}$

Thus, we see that the value in the interval $\left[0,\pi\right]$ that satisfies this equation is $x=\frac{\pi}{3}$ .

Evaluate the expression $\cos^{-1}\left(\cos\left(\frac{23\pi}{6}\right)\right)$
Because $\frac{23\pi}{6}$ is outside the interval $\left[0,\pi\right]$ , we can simplify the inside cosine expression: $\cos^{-1}\left(\cos\left(\frac{23\pi}{6}\right)\right)$ $=\cos^{-1}\left(\cos\left(\frac{11\pi}{6}\right)\right)$ But $\frac{11\pi}{6}$ is still outside the interval $\left[0,\pi\right]$ , so simplifying to an angle within the interval using the fact that $\cos\left(\frac{11\pi}{6}\right)=\cos\left(\frac{\pi}{6}\right)$ gives: $=\cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right)$ Thus, $x=\frac{\pi}{6}$

Evaluate the expression $\cos^{-1}\left(\cos\left(\frac{23\pi}{6}\right)\right)$

Because $\frac{23\pi}{6}$ is outside the interval $\left[0,\pi\right]$ , we can simplify the inside cosine expression:

$\cos^{-1}\left(\cos\left(\frac{23\pi}{6}\right)\right)$
$=\cos^{-1}\left(\cos\left(\frac{11\pi}{6}\right)\right)$
But $\frac{11\pi}{6}$ is still outside the interval $\left[0,\pi\right]$ , so simplifying to an angle within the interval using the fact that $\cos\left(\frac{11\pi}{6}\right)=\cos\left(\frac{\pi}{6}\right)$ gives:
$=\cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right)$

Thus, $x=\frac{\pi}{6}$

Tangent

Tangent has a very natural domain restriction between the two asymptotes that creates a one-to-one function. Thus, we will restrict tangent to the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ in order to consider its inverse function.

A period of tangent is graphed here:

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Note that the range of the tangent function is $\mathbb{R}$ . Thus, the domain of the inverse tangent function is also $\mathbb{R}$ . The graph of the inverse tangent is graphed here:

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The Inverse Tangent Function
On the domain $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , the tangent function has an inverse, noted either $\tan^{-1}(x)$ or $\arctan(x)$ . The domain of $\tan^{-1}(x)$ is all real numbers, and its range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , is defined by $\tan^{-1}(x)=y \ \ \Leftrightarrow \ \ \tan(y)=x$ And holds the properties $\begin{align}\tan(\tan^{-1}(x))=x & & \text{ for } & & x\in\mathbb{R} \\ \tan^{-1}(\tan(x))=x & & \text{ for } & & -\frac{\pi}{2}<x<\frac{\pi}{2}\end{align}$

The Inverse Tangent Function

On the domain $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , the tangent function has an inverse, noted either $\tan^{-1}(x)$ or $\arctan(x)$ . The domain of $\tan^{-1}(x)$ is all real numbers, and its range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , is defined by

$\tan^{-1}(x)=y \ \ \Leftrightarrow \ \ \tan(y)=x$

And holds the properties

$\begin{align*}\tan(\tan^{-1}(x))=x & & \text{ for } & & x\in\mathbb{R} \\ \tan^{-1}(\tan(x))=x & & \text{ for } & & -\frac{\pi}{2}<x<\frac{\pi}{2}\end{align*}$

Evaluate $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$
$\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)=x \ \ \Leftrightarrow \ \ \tan(x)=\frac{\sqrt{3}}{3}$ The tangent equation on the right, of course, has solution $x=\frac{\pi}{6}$

Evaluate $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$

$\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)=x \ \ \Leftrightarrow \ \ \tan(x)=\frac{\sqrt{3}}{3}$

The tangent equation on the right, of course, has solution $x=\frac{\pi}{6}$

Evaluate $\tan^{-1}(-\sqrt{3})$
$\tan^{-1}(-\sqrt{3})=x \ \ \Leftrightarrow \ \ \tan(x)=-\sqrt{3}$ and the value in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ that solves this is $x=-\frac{\pi}{3}$ .

Evaluate $\tan^{-1}(-\sqrt{3})$