5.3: Inverse Trigonometric Functions

Last updated
Save as PDF

Page ID: 3337

$guichard.png$
Contributed by Michael Corral
Professor (Mathematics) at Schoolcraft College

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

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.