3.4 Inverse Trigonometric Functions

Trigonometric functions are not one-to-one because their values repeat periodically and that the horizontal lines $y=c$ intersect the graphs in an infinite number of points, if at all, as we at once see from Figure 1 (recall the horizontal line test in Section One-to-One Functions). Therefore, they cannot have inverses unless we restrict their domains to intervals on which they are one-to-one.

(a) Graph of $y=\sin x$	(b) Graph of $y=\tan x$

Figure 1: Trigonometric functions are not one-to-one as they do not pass the horizontal line test.

Inverse of sine

If we look at the graph of $y=\sin x$ or if we consider the unit circle, we realize that the sine function on the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$ is increasing from $-1$ to $1$. So by restricting its domain to this interval, we make it a one-to-one function whose domain is $[-\frac{\pi }{2},\frac{\pi }{2}]$ and its range is $[-1,1]$. The inverse of the sine function, denoted by “${\sin }^{-1}x$” or  “$\arcsin x$” , is a one-to-one function whose domain is $[-1,1]$ and its range is $[-\frac{\pi }{2},\frac{\pi }{2}]$. The graph of $y=\arcsin x$ is obtained by reflecting the graph of $y=\sin x$ (restricted to the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$) in the line $y=x$ (see Figure 2).

•  The two symbols “${\sin }^{-1}x$” and “$\arcsin x$” are equivalent and can be used interchangeably. The first one is read “the inverse sine of $x$” and the second “the arc sine of $x$.”

• Again note that ${\sin }^{-1}x\ne \frac{1}{\sin x}$.

•  The graph of $y=\arcsin x$ is symmetric about the origin, which shows $y=\arcsin x$ is an odd function. To prove it algebraically, we need to show $\arcsin (-x)=-\arcsin x$. Let
  $$y=\arcsin (-x).$$ We know it means
  $$\begin{array}{rl}\sin y& =-x\\ \Rightarrow -\sin y& =x\\ \Rightarrow \sin (-y)& =x& \text{the sine is an odd function}\\ \Rightarrow -y& =\arcsin x\\ \Rightarrow y& =-\arcsin x\\ \Rightarrow \arcsin (-x)& =-\arcsin x\end{array}$$

Figure 2: The graph of $y=\arcsin x$ is obtained by reflecting the graph of $y=\sin x$ restricted to the interval $[-\pi /2,\pi /2]$ in the line $y=x$.

Inverse of cosine

The cosine and tangent can be inverted in a similar fashion. By considering the unit circle or looking at the graph of $y=\cos x$, we realize that $y=\cos x$ is not one-to-one on $[-\pi /2,\pi /2]$. So we had to choose a different interval for the cosine function. If we restrict the domain of the cosine function to the interval $[0,\pi ]$, we can make it one-to-one, so that it has an inverse function denoted by ${\cos }^{-1}x$ or $\arccos x$. The graph of $y=\arccos x$ is shown in Figure 3.

• The domain of $y=\arccos x$ is $[-1,1]$ and its range is $[0,\pi ]$.

• The graph of $y=\arccos x$ is neither symmetric about the $y$-axis nor is symmetric about the origin, which means that $y=\arccos x$ is neither odd nor even.

Figure 3: The graph of $y=\arccos x$ is obtained by reflecting the graph of $y=\cos x$ restricted to the interval $[0,\pi ]$ in the line $y=x$.

Inverse of Tangent 

For the tangent function, we choose the open interval $(-\frac{\pi }{2},\frac{\pi }{2})$ to perform the inversion. The resulting function is denoted by “${\tan }^{-1}x$” or “$\arctan x$.”

• The domain of $y=\arctan x$ is $(-\mathrm{\infty },\mathrm{\infty })$ and its range is $(-\frac{\pi }{2},\frac{\pi }{2})$.

The graph of $y=\arctan x$ is shown in Figure 4. This figure shows that the inverse tangent function is an odd function.

Figure 4: The graph of $y=\arctan x$ is obtained by reflecting the graph of $y=\tan x$ restricted to the open interval $(-\frac{\pi }{2},\frac{\pi }{2})$ in the line $y=x$.

• Recall that if $f$ and $g$ are inverse functions of each other then
  $$f(g(x))=x,g(f(x))=x$$ for every $x$ in the domain of the inside function, which are $g$ and $f$, respectively (Theorem 1 in Section on Inverse Functions). The following table summarizes some properties of the inverse trigonometric functions. Note that here we deal with the restricted domains  of the trigonometric functions; otherwise, their inverses do not exist.

  Function	Domain	Range	Cancelation euqations
  $y=\arcsin x$	$[-1,1]$	$[-\frac{\pi }{2},\frac{\pi }{2}]$	$\sin (\arcsin x)=x\text{if }-1\le x\le 1$ $\arcsin (\sin x)=x\text{if}-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
  $y=\arccos x$	$[-1,1]$	$[0,\pi ]$	$\cos (\arccos x)=x\text{if }-1\le x\le 1$ $\arccos (\cos x)=x\text{if }0\le x\le \pi$
  $y=\arctan x$	$(-\mathrm{\infty },\mathrm{\infty })$	$(-\frac{\pi }{2},\frac{\pi }{2})$	$\tan (\arctan x)=x\text{if }-\mathrm{\infty }<x<\mathrm{\infty }$ $\arctan (\tan x)=x\text{if }-\frac{\pi }{2}<x<\frac{\pi }{2}$

  Table 1: Properties of the inverse trigonometric function
  

Inverse of the Secondary Trigonometric Functions

Read more on the inverse of the secondary trigonometric functions (Optional)

Hide the inverse of the secondary trigonometry functions

The inverses of the cotangent, secant, and cosecant can be defined in a similar fashion but they are of lesser importance. Most calculators do not have special keys for arccot $x$, arcsec $x$, or arccsc $x$ (equivalent to ${\cot }^{-1}x$, ${\sec }^{-1}x$, or ${\csc }^{-1}x$), but we can say
$$\begin{array}{rl}\text{arccot }x& =\arctan \left(\frac{1}{x}\right)\\ \text{arcsec }x& =\arccos \left(\frac{1}{x}\right)\\ \text{arccsc }x& =\arcsin \left(\frac{1}{x}\right)\end{array}$$
The graphs of these inverses of cotangent, secant and cosecant are depicted in Figure 5.

• There is no universal agreement on how to restrict the domains of the secondary trigonometric functions. For example, in many books (especially the older ones) you may see that the inverse cotangent function is defined by restricting the domain of the cotangent function to the intervals $(0,\pi )$, but nowadays in most computer packages such as MATLAB, Mathematica, Sympy, and Maple, it is defined by restricting the domain of cotangent to the interval $(-\frac{\pi }{2},\frac{\pi }{2})$. In Figure 5, we have used the definitions that conform with the conventions used by these computer packages as they are consistent with Equations 1.

(a) Graph of $y=\text{arccot }x$	(b) Graph of $y=\text{arcsec }x$
(c) Graph of $y=\text{arccsc }x$

Figure 5

Examples

  When evaluating the inverse trigonometric functions, do not forget that their outputs are angles in radian measure.

Solution

Let $\alpha =\arccos (-\frac{1}{2})$. To find $\alpha$, we use the unit circle and draw a vertical line passing through $-\frac{1}{2}$ (recall the $x$-axis is also called the cosine axis). As we at once see from Figure 6, $\alpha$ lies in the second quadrant (there is another angle in the third quadrant for which the cosine is $-1/2$, but we are not interested in that angle as we have restricted the domain of the cosine function to the first and second quadrants). Considering the right triangle $\mathrm{△}OHP$ in Figure 6, it is obvious that
$$\cos (\mathrm{\angle }HOP)=\frac{1}{2}.$$ Because $\cos \frac{\pi }{3}=\frac{1}{2},$ we get
$$\mathrm{\angle }HOP=\frac{\pi }{3}.$$ From Figure 6,
$$\begin{array}{rl}\alpha & =\pi -\mathrm{\angle }HOP\\ & =\pi -\frac{\pi }{3}\\ & =\frac{2\pi }{3}.\end{array}$$
That is,
$$\arccos (-\frac{1}{2})=\frac{2\pi }{3}.$$

Figure 6

Solution

Because $-\frac{3\pi }{2}<-\frac{4\pi }{3}<-\pi$, the angle whose radian measure is $-\frac{4\pi }{3}$ is in the second quadrant. Because $-\frac{4\pi }{3}$ does not lie between $0$ and $\pi$, we cannot use the cancellation equation (see Equation 4 in Table 1). So the first step is to find an angle $\alpha$ such that $\alpha$ lies between 0 and $\pi$ and
$$\cos \alpha =\cos (-\frac{4\pi }{3}),$$ then we can use the cancellation equation. We at once see from the following figure that
$$\alpha =2\pi -\frac{4\pi }{3}=\frac{2\pi }{3}.$$ Therefore,
$$\arccos (\cos (-\frac{4\pi }{3}))=\arccos (\cos \frac{2\pi }{3}).$$ and now we can use the cancellation equation
$$\arccos (\cos \frac{2\pi }{3})=\frac{2\pi }{3}.$$

Figure 7

Solution

Because the domain the arc sine function $y=\arcsin x$ is $[-1,1]$;
therefore

$$\begin{array}{rl}Dom(f)& =\{x|-1\le x^2-3\le 1\}\\ & =\{x|2\le x^2\le 4\}& \text{(adding 3 to each side of the inequalities)}\end{array}$$
Here we have two inequalities $x^2\le 4$ and $2\le x^2$, and we need to find all $x$ for which the both inequalities hold:
$$x^2\le 4\Rightarrow |x|\le 4\Rightarrow -2\le x\le 2$$ In other words, $x\in [-2,2]$. Now the second inequality:

$$2\le x^2\Rightarrow \sqrt{2}\le |x|\Rightarrow x\ge \sqrt{2}\text{ or }x\le -\sqrt{2}$$ or $x\in (-\mathrm{\infty },-\sqrt{2}]\cup [\sqrt{2},\mathrm{\infty }).$ Therefore
$$\begin{array}{rl}Dom(f)& =\{x|-2\le x\le -\sqrt{2}\text{ or }\sqrt{2}\le x\le 2\}\\ & =[-2,-\sqrt{2}]\cup [\sqrt{2},2].\end{array}$$
To show that this is the correct domain of $f$, we can graph $f$ using a graphing calculator or a computer package (see the following figure).

Figure 7: Graph $y=\arcsin (x^2-3)$