# 5.3: Inverse Trigonometric Functions

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We have briefly mentioned the inverse trigonometric functions before, for example in Section 1.3 when we discussed how to use the \(\fbox{\(\sin^{-1}\)}\), \(\fbox{\(\cos^{-1}\)}\), and \(\fbox{\(\tan^{-1}\)}\) buttons on a calculator to find an angle that has a certain trigonometric function value. We will now define those inverse functions and determine their graphs.

**Figure 5.3.1**

Recall that a **function** is a rule that assigns a single object \(y \) from one set (the **range** to each object \(x \) from another set (the **domain**). We can write that rule as \(y = f(x) \), where \(f \) is the function (see Figure 5.3.1). There is a simple *vertical rule* for determining whether a rule \(y=f(x) \) is a function: \(f \) is a function if and only if every vertical line intersects the graph of \(y=f(x) \) in the \(xy\)-coordinate plane at most once (see Figure 5.3.2).

**Figure 5.3.2** Vertical rule for functions

Recall that a function \(f \) is **one-to-one** (often written as \(1-1\)) if it assigns distinct values of \(y \) to distinct values of \(x \). In other words, if \(x_1 \ne x_2 \) then \(f(x_1 ) \ne f(x_2 ) \). Equivalently, \(f \) is one-to-one if \(f(x_1 ) = f(x_2 ) \) implies \(x_1 = x_2 \). There is a simple *horizontal rule* for determining whether a function \(y=f(x) \) is one-to-one: \(f \) is one-to-one if and only if every horizontal line intersects the graph of \(y=f(x) \) in the \(xy\)-coordinate plane at most once (see Figure 5.3.3).

**Figure 5.3.3** Horizontal rule for one-to-one functions

If a function \(f \) is one-to-one on its domain, then \(f \) has an **inverse function**, denoted by \(f^{-1} \), such that \(y=f(x) \) if and only if \(f^{-1}(y) = x \). The domain of \(f^{-1} \) is the range of \(f \).

The basic idea is that \(f^{-1} \) "undoes'' what \(f \) does, and vice versa. In other words,

\[\nonumber \begin{alignat*}{3}

f^{-1}(f(x)) ~&=~ x \quad&&\text{for all \(x \) in the domain of \(f \), and}\\ \nonumber

f(f^{-1}(y)) ~&=~ y \quad&&\text{for all \(y \) in the range of \(f \).}

\end{alignat*}\]

We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. However, we can restrict those functions to *subsets* of their domains where they *are* one-to-one. For example, \(y=\sin\;x \) is one-to-one over the interval \(\left[ -\frac{\pi}{2},\frac{\pi}{2} \right] \), as we see in the graph below:

**Figure 5.3.4** \(y = \sin x\) with \(x\) restricted to \(\left [ − \frac{π}{ 2} , \frac{π}{ 2} \right ] \)

For \(-\frac{\pi}{2} \le x \le \frac{\pi}{2} \) we have \(-1 \le \sin\;x \le 1 \), so we can define the **inverse sine** function \(y=\sin^{-1} x \) (sometimes called the **arc sine** and denoted by \(y=\arcsin\;x\)) whose domain is the interval \([-1,1] \) and whose range is the interval \(\left[ -\frac{\pi}{2},\frac{\pi}{2} \right] \). In other words:

\[ \begin{alignat}{3}

\sin^{-1} (\sin\;y) ~&=~ y \quad&&\text{for \(-\tfrac{\pi}{2} \le y \le

\tfrac{\pi}{2}\)}\label{eqn:arcsin1}\\

\sin\;(\sin^{-1} x) ~&=~ x \quad&&\text{for \(-1 \le x \le 1\)}\label{eqn:arcsin2}

\end{alignat}\]

Example 5.13

Find \(\sin^{-1} \left(\sin\;\frac{\pi}{4}\right) \).

**Solution:**

Since \(-\frac{\pi}{2} \le \frac{\pi}{4} \le \frac{\pi}{2} \), we know that \(\sin^{-1} \left(\sin\;\frac{\pi}{4}\right) = \boxed{\frac{\pi}{4}}\; \), by Equation \ref{eqn:arcsin1}.

Example 5.14

Find \(\sin^{-1} \left(\sin\;\frac{5\pi}{4}\right) \).

**Solution:**

Since \(\frac{5\pi}{4} > \frac{\pi}{2} \), we can not use Equation \ref{eqn:arcsin1}. But we know that \(\sin\;\frac{5\pi}{4} = -\frac{1}{\sqrt{2}} \). Thus, \(\sin^{-1} \left(\sin\;\frac{5\pi}{4}\right) = \sin^{-1} \left( -\frac{1}{\sqrt{2}} \right) \) is, by definition, the angle \(y \) such that \(-\frac{\pi}{2} \le y \le \frac{\pi}{2} \) and \(\sin\;y = -\frac{1}{\sqrt{2}} \). That angle is \(y=-\frac{\pi}{4} \), since

\[\sin\;\left( -\tfrac{\pi}{4} \right) ~=~ -\sin\;\left( \tfrac{\pi}{4} \right) ~=~

-\tfrac{1}{\sqrt{2}} ~. \nonumber \]

Thus, \(\sin^{-1} \left(\sin\;\frac{5\pi}{4}\right) = \boxed{-\tfrac{\pi}{4}}\; \).

Example 5.14 illustrates an important point: \(\sin^{-1} x \) should *always *be a number between \(-\frac{\pi}{2} \) and \(\frac{\pi}{2} \). If you get a number outside that range, then you made a mistake somewhere. This why in Example 1.27 in Section 1.5 we got \(\sin^{-1}(-0.682) = -43^\circ \) when using the \(\fbox{\(\sin^{-1}\)}\) button on a calculator. Instead of an angle between \(0^\circ \) and \(360^\circ \) (i.e. \(0 \) to \(2\pi \) radians) we got an angle between \(-90^\circ \) and \(90^\circ \) (i.e. \(-\frac{\pi}{2} \) to \(\frac{\pi}{2} \) radians).

In general, the graph of an inverse function \(f^{-1} \) is the reflection of the graph of \(f \) around the line \(y=x \). The graph of \(y=\sin^{-1} x \) is shown in Figure 5.3.5. Notice the symmetry about the line \(y=x \) with the graph of \(y=\sin\;x \).

**Figure 5.3.5** Graph of \(y = \sin^{−1} x\)

The **inverse cosine** function \(y=\cos^{-1} x \) (sometimes called the **arc cosine** and denoted by \(y=\arccos\;x\)) can be determined in a similar fashion. The function \(y=\cos\;x \) is one-to-one over the interval \([0,\pi] \), as we see in the graph below:

**Figure 5.3.6** \(y = \cos x\) with \(x\) restricted to \([0,π]\)

Thus, \(y=\cos^{-1} x \) is a function whose domain is the interval \([-1,1] \) and whose range is the interval \([0,\pi] \). In other words:

\[\begin{alignat}{3}

\cos^{-1} (\cos\;y) ~&=~ y \quad&&\text{for \(0 \le y \le \pi\)}\label{eqn:arccos1}\\

\cos\;(\cos^{-1} x) ~&=~ x \quad&&\text{for \(-1 \le x \le 1\)}\label{eqn:arccos2}

\end{alignat}\]

The graph of \(y=\cos^{-1} x \) is shown below in Figure 5.3.7. Notice the symmetry about the line \(y=x \) with the graph of \(y=\cos\;x \).

**Figure 5.3.7** Graph of \(y = \cos^{−1} x\)

Example 5.15

Find \(\cos^{-1} \left(\cos\;\frac{\pi}{3}\right) \).

**Solution:**

Since \(0 \le \frac{\pi}{3} \le \pi \), we know that \(\cos^{-1} \left(\cos\;\frac{\pi}{3}\right) = \boxed{\frac{\pi}{3}}\; \), by Equation \ref{eqn:arccos1}.

Example 5.16

Find \(\cos^{-1} \left(\cos\;\frac{4\pi}{3}\right) \).

**Solution:**

Since \(\frac{4\pi}{3} > \pi \), we can not use Equation \ref{eqn:arccos1}. But we know that \(\cos\;\frac{4\pi}{3} = -\frac{1}{2} \). Thus, \(\cos^{-1} \left(\cos\;\frac{4\pi}{3}\right) = \cos^{-1} \left( -\frac{1}{2} \right) \) is, by definition, the angle \(y \) such that \(0 \le y \le \pi \) and \(\cos\;y = -\frac{1}{2} \). That angle is \(y=\frac{2\pi}{3} \) (i.e. \(120^\circ\)). Thus, \(\cos^{-1} \left(\cos\;\frac{4\pi}{3}\right) = \boxed{\tfrac{2\pi}{3}}\; \).

Examples 5.14 and 5.16 may be confusing, since they seem to violate the general rule for inverse functions that \(f^{-1}(f(x)) = x \) for all \(x \) in the domain of \(f \). But that rule only applies when the function \(f \) is one-to-one over its *entire* domain. We had to restrict the sine and cosine functions to very small subsets of their entire domains in order for those functions to be one-to-one. That general rule, therefore, only holds for \(x \) in those small subsets in the case of the inverse sine and inverse cosine.

The **inverse tangent** function \(y=\tan^{-1} x \) (sometimes called the **arc tangent** and denoted by \(y=\arctan\;x\)) can be determined similarly. The function \(y=\tan\;x \) is one-to-one over the interval \(\left( -\frac{\pi}{2},\frac{\pi}{2} \right) \), as we see in Figure 5.3.8:

**Figure 5.3.8** \(y = \tan x \text{ with }x \text{ restricted to }\left ( − \frac{π}{ 2} , \frac{π}{ 2} \right ) \)

The graph of \(y=\tan^{-1} x \) is shown below in Figure 5.3.9. Notice that the vertical asymptotes for \(y=\tan\;x \) become horizontal asymptotes for \(y=\tan^{-1} x \). Note also the symmetry about the line \(y=x \) with the graph of \(y=\tan\;x \).

**Figure 5.3.9** Graph of \(y = \tan^{−1} x\)

Thus, \(y=\tan^{-1} x \) is a function whose domain is the set of all real numbers and whose range is the interval \(\left( -\frac{\pi}{2},\frac{\pi}{2} \right) \). In other words:

\[\begin{alignat}{3}

\tan^{-1} (\tan\;y) ~&=~ y \quad&&\text{for \(-\tfrac{\pi}{2} < y <

\tfrac{\pi}{2}\)}\label{eqn:arctan1}\\

\tan\;(\tan^{-1} x) ~&=~ x \quad&&\text{for all real \(x\)}\label{eqn:arctan2}

\end{alignat}\]

Example 5.17

Find \(\tan^{-1} \left(\tan\;\frac{\pi}{4}\right) \).

**Solution:**

Since \(-\tfrac{\pi}{2} \le \tfrac{\pi}{4} \le \tfrac{\pi}{2} \), we know that \(\tan^{-1} \left(\tan\;\frac{\pi}{4}\right) = \boxed{\frac{\pi}{4}}\; \), by Equation \ref{eqn:arctan1}.

Example 5.18

Find \(\tan^{-1} \left(\tan\;\pi\right) \).

**Solution:**

Since \(\pi > \tfrac{\pi}{2} \), we can not use Equation \ref{eqn:arctan1}. But we know that \(\tan\;\pi = 0 \). Thus, \(\tan^{-1} \left(\tan\;\pi\right) = \tan^{-1} 0 \) is, by definition, the angle \(y \) such that \(-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2} \) and \(\tan\;y = 0 \). That angle is \(y=0 \). Thus, \(\tan^{-1} \left(\tan\;\pi \right) = \boxed{0}\; \).

Example 5.19

Find the exact value of \(\cos\;\left(\sin^{-1}\;\left(-\frac{1}{4}\right)\right) \).

**Solution:**

Let \(\theta = \sin^{-1}\;\left(-\frac{1}{4}\right) \). We know that \(-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2} \), so since \(\sin\;\theta = -\frac{1}{4} < 0 \), \(\theta \) must be in QIV. Hence \(\cos\;\theta > 0 \). Thus,

\[ \cos^2 \;\theta ~=~ 1 ~-~ \sin^2 \;\theta ~=~ 1 ~-~ \left( -\frac{1}{4} \right)^2 ~=~\frac{15}{16}

\quad\Rightarrow\quad \cos\;\theta ~=~ \frac{\sqrt{15}}{4} ~. \nonumber \]

Note that we took the positive square root above since \(\cos\;\theta > 0 \). Thus, \(\cos\;\left(\sin^{-1}\;\left(-\frac{1}{4}\right)\right) = \boxed{\frac{\sqrt{15}}{4}}\; \).

Example 5.20

Show that \(\tan\;(\sin^{-1} x) = \dfrac{x}{\sqrt{1 - x^2}} \) for \(-1 < x < 1 \).

**Solution:**

When \(x=0 \), the Equation holds trivially, since

\[\nonumber \tan\;(\sin^{-1} 0) ~=~ \tan\;0 ~=~ 0 ~=~ \dfrac{0}{\sqrt{1 - 0^2}} ~.\]

Now suppose that \(0 < x < 1 \). Let \(\theta = \sin^{-1} x \). Then \(\theta \) is in QI and \(\sin\;\theta = x \). Draw a right triangle with an angle \(\theta \) such that the opposite leg has length \(x \) and the hypotenuse has length \(1 \), as in Figure 5.3.10 (note that this is possible since \(0 < x < 1\)). Then \(\sin\;\theta = \frac{x}{1} = x \). By the Pythagorean Theorem, the adjacent leg has length \(\sqrt{1 - x^2} \). Thus, \(\tan\;\theta = \frac{x}{\sqrt{1 - x^2}} \).

**Figure 5.3.10**

If \(-1 < x < 0 \) then \(\theta = \sin^{-1} x \) is in QIV. So we can draw the same triangle except that it would be "upside down'' and we would again have \(\tan\;\theta = \frac{x}{\sqrt{1 - x^2}} \), since the tangent and sine have the same sign (negative) in QIV. Thus, \(\tan\;(\sin^{-1} x) = \dfrac{x}{\sqrt{1 - x^2}} \) for \(-1 < x < 1 \).

The inverse functions for cotangent, cosecant, and secant can be determined by looking at their graphs. For example, the function \(y=\cot\;x \) is one-to-one in the interval \((0,\pi) \), where it has a range equal to the set of all real numbers. Thus, the **inverse cotangent** \(y=\cot^{-1} x \) is a function whose domain is the set of all real numbers and whose range is the interval \((0,\pi) \). In other words:

\[\begin{alignat}{3}

\cot^{-1} (\cot\;y) ~&=~ y \quad&&\text{for \(0 < y < \pi\)}\label{eqn:arccot1}\\

\cot\;(\cot^{-1} x) ~&=~ x \quad&&\text{for all real \(x\)}\label{eqn:arccot2}

\end{alignat}\]

The graph of \(y=\cot^{-1} x \) is shown below in Figure 5.3.11.

**Figure 5.3.11** Graph of \(y = \cot^{−1} x\)

Similarly, it can be shown that the **inverse cosecant** \(y=\csc^{-1} x \) is a function whose domain is \(|x| \ge 1 \) and whose range is \(-\frac{\pi}{2} \le y \le \frac{\pi}{2} \), \(y \ne 0 \). Likewise, the **inverse secant** \(y=\sec^{-1} x \) is a function whose domain is \(|x| \ge 1 \) and whose range is \(0 \le y \le \pi \), \(y \ne \frac{\pi}{2} \).

\[\begin{alignat}{3}

\csc^{-1} (\csc\;y) ~&=~ y \quad&&\text{for \(-\frac{\pi}{2} \le

y \le \frac{\pi}{2} \), \(y \ne 0\)}\label{eqn:arccsc1}\\

\csc\;(\csc^{-1} x) ~&=~ x \quad&&\text{for \(|x| \ge 1\)}\label{eqn:arccsc2}

\end{alignat}\]

\[\begin{alignat}{3}

\sec^{-1} (\sec\;y) ~&=~ y \quad&&\text{for \(0 \le y \le \pi \), \(y \ne

\frac{\pi}{2}\)}\label{eqn:arcsec1}\\

\sec\;(\sec^{-1} x) ~&=~ x \quad&&\text{for \(|x| \ge 1\)}\label{eqn:arcsec2}

\end{alignat}\]

It is also common to call \(\cot^{-1} x \), \(\csc^{-1} x \), and \(\sec^{-1} x \) the **arc cotangent**, **arc cosecant**, and **arc secant**, respectively, of \(x \). The graphs of \(y=\csc^{-1} x \) and \(y=\sec^{-1} x \) are shown in Figure 5.3.12:

**Figure 5.3.12**

Example 5.21

Prove the identity \(\tan^{-1} x \;+\; \cot^{-1} x ~=~ \frac{\pi}{2} \).

**Solution: **

Let \(\theta = \cot^{-1} x \). Using relations from Section 1.5, we have

\[\nonumber \tan\;\left( \tfrac{\pi}{2} - \theta \right) ~=~ -\tan\;\left( \theta - \tfrac{\pi}{2} \right)

~=~ \cot\;\theta ~=~ \cot\;(\cot^{-1} x) ~=~ x ~,\]

by Equation \ref{eqn:arccot2}. So since \(\tan\;(\tan^{-1} x) = x \) for all \(x \), this means that \(\tan\;(\tan^{-1} x) = \tan\;\left( \tfrac{\pi}{2} - \theta \right) \). Thus, \(\tan\;(\tan^{-1} x) = \tan\;\left( \tfrac{\pi}{2} - \cot^{-1} x \right) \). Now, we know that \(0 < \cot^{-1} x < \pi \), so \(-\tfrac{\pi}{2} < \tfrac{\pi}{2} - \cot^{-1} x < \tfrac{\pi}{2} \), i.e. \(\tfrac{\pi}{2} - \cot^{-1} x \) is in the restricted subset on which the tangent function is one-to-one. Hence, \(\tan\;(\tan^{-1} x) = \tan\;\left( \tfrac{\pi}{2} - \cot^{-1} x \right)\) implies that \(\tan^{-1} x = \tfrac{\pi}{2} - \cot^{-1} x \), which proves the identity.

Example 5.22

Is \(\;\tan^{-1} a \;+\; \tan^{-1} b ~=~ \tan^{-1} \left( \dfrac{a+b}{1-ab} \right)\; \) an identity?

**Solution:**

In the tangent addition Equation \(\tan\;(A+B) = \dfrac{\tan\;A \;+\; \tan\;B}{1 \;-\; \tan\;A~\tan\;B} \), let \(A = \tan^{-1} a \) and \(B = \tan^{-1} b \). Then

\[\nonumber \begin{align*}

\tan\;(\tan^{-1} a \;+\; \tan^{-1} b ) ~&=~ \dfrac{\tan\;(\tan^{-1} a) \;+\; \tan\;(\tan^{-1}

b)}{1 \;-\; \tan\;(\tan^{-1} a)~\tan\;(\tan^{-1} b)}\\ \nonumber

&=~ \dfrac{a+b}{1-ab}\qquad\text{by Equation \ref{eqn:arctan2}, so it seems that we have}\\ \nonumber

\tan^{-1} a \;+\; \tan^{-1} b ~&=~ \tan^{-1} \left( \dfrac{a+b}{1-ab} \right)

\end{align*}\]

by definition of the inverse tangent. However, recall that \(-\tfrac{\pi}{2} < \tan^{-1} x < \tfrac{\pi}{2} \) for all real numbers \(x \). So in particular, we must have \(-\tfrac{\pi}{2} < \tan^{-1} \left( \frac{a+b}{1-ab} \right) < \tfrac{\pi}{2} \). But it is possible that \(\tan^{-1} a \;+\; \tan^{-1} b \) is *not* in the interval \(\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) \). For example,

\[ \tan^{-1} 1 \;+\; \tan^{-1} 2 ~=~ 1.892547 ~>~ \tfrac{\pi}{2} \approx 1.570796 ~.\nonumber \]

And we see that \(\tan^{-1} \left( \frac{1+2}{1-(1)(2)} \right) = \tan^{-1} (-3) = -1.249045 \ne \tan^{-1} 1 \;+\; \tan^{-1} 2 \). So the Equation is only true when \(-\tfrac{\pi}{2} < \tan^{-1} a \;+\; \tan^{-1} b < \tfrac{\pi}{2} \).

## Contributors

Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2.