# Graph, Domain and Range of arccos(x) function

The graph, domain and range and other properties of the inverse trigonometric function $\arccos(x)$ are explored using graphs, examples with detailed solutions and an interactive app.

## Definition of arccos(x) Functions

The function $\cos(x)$ is shown below. On its implied domain, cos(x) is not a one to one function as seen below; a horizontal line test for a one to one function would fail. But we limit the domain to $[ 0 , \pi ]$, blue graph below, we obtain a one to one function that has an inverse which cannot be obtained algebraically.

The inverse function of $f(x) = \cos(x) , x \in [ 0 , \pi]$ is $f^{-1} = \arccos(x)$
We define $\arccos(x)$ as follows $y = \arccos(x) \iff x = \cos(y)$ where $-1 \le x \le 1$ and $0 \le y \le \pi$

Let us make a table of values of $y = \arccos(x)$ and graph it.

 $x$ -1 0 1 $y = \arccos(x)$ $\pi$ $\dfrac{\pi}{2}$ $0$ since $\cos(\pi) = - 1$ since $\cos(\dfrac{\pi}{2}) = 0$ since $\cos(0) = 1$

The graphs of $y = \arccos(x)$ and $y = \cos(x) , x \in [ 0 , \pi]$ are shown below. Each graph is the reflection of the other on the line $y = x$; this is one of the graphical properties of inverse functions .

Example 1
Evaluate $arccos(x)$ given the value of $x$.
Special values related to special angles
$\arccos(1) = 0$ because $\cos(0) = 1$
$\arccos(-1) = \pi \; \text{ or } 180^{o}$ because $\cos(\pi) = - 1$
$\arccos(2) =$ undefined because $2$ is not in the domain of $\arccos(x)$ which is $-1 \le x \le 1$ ( there is no angle that has cosine equal to 2 ).
$\arccos(-\dfrac{1}{2}) = \dfrac{2\pi}{3} \; \text{ or } 120^{o}$ because $\cos(\dfrac{2\pi}{3}) = -\dfrac{1}{2}$
$\arccos( - \dfrac{\sqrt3}{2}) = \dfrac{5\pi}{6} \; \text{ or } 150^{o}$ because $\cos(\dfrac{5\pi}{6}) = - \dfrac{\sqrt3}{2}$
$\arccos(\dfrac{\sqrt 3}{2}) = \dfrac{\pi}{6} \; \text{ or } 30^{o}$ because $\sin(\dfrac{\pi}{6}) = \dfrac{\sqrt 3}{2}$
Use of calculator
$\arccos(0.25) = 1.32 \; \text{ or } 75.52^{o}$
$\arccos(-0.77) = 2.45 \; \text{ or } 140.35^{o}$

## Properties of $y = arccos(x)$

1. Domain: $[-1 , 1]$
2. Range: $[0 , \pi]$
3. $\arccos(x)$ is a one to one function
4. $\cos(\arccos(x)) = x$ , for x in the interval $[-1,1]$ , due to property of a function and its inverse :$f(f^{-1}(x) = x$ where $x$ is in the domain of $f^{-1}$
5. $\arccos(\cos(x)) = x$ , for x in the interval $[0 , \pi ]$ , due to property of a function and its inverse :$f^{-1}(f(x) = x$ where $x$ is in the domain of $f$

Example 2
Find the domain of the functions:
a) $y = \arccos(-3x)$       b) $y = \arccos(2 x - 1)$       c) $y = 3 \arccos(x/2 + 1) + \pi/4$

Solution to Example 2
a)
the domain is found by first writing that the argument $-3x$ of the given function $y = \arccos(-3x)$ is within the domain of the arccos function which one of the properties given above. Hence the need to solve the double inequality
$-1 \le -3x \le 1$
divide all terms of the double inequality by - 3 and change the symbols of inequality to obtain
$- 1/3 \le x \le 1/3$ , which is the domain of the function $y = \arccos(-3x)$.
b)
Solve the inequality
$-1 \le 2 x - 1 \le 1$
$0 \le 2 x \le 2$
$0 \le x \le 1$ , which is the domain of the function $y = \arccos(2 x - 1)$.
c)
$-1 \le x/2 + 1 \le 1$
Solve the above inequality
$- 4 \le x \le 0$ , which is the domain of the function $y = 3 \arccos(x/2 + 1) + \pi/4$.

Example 3
Find the range of the functions:
a) $y = - 2 \arccos(x)$       b) $y = - \arccos(x) + \pi/4$       c) $y = - \arccos(x-1) - \pi$

Solution to Example 3
a)
We start with the range of $\arccos(x)$ as a double inequality
$0 \le \arccos(x) \le \pi$
multiply all terms of the above inequality by - 2, change the symbols of inequality and simplify
$- 2 \pi \le - 2\arccos(x) \le 0$
the range of the given function $2 \arccos(x)$ is given by the interval $[ - 2 \pi , 0 ]$.

b)
we start with the range of $\arccos(x)$
$0 \le \arccos(x) \le \pi$
Multiply all terms of the above inequality by -1 and change symbol of the double inequality
$-\pi \le -\arccos(x) \le 0$
add $\dfrac{\pi}{4}$ to all terms of the above inequality and simplify
$- 3 \dfrac{\pi}{4} \le - \arccos(x) + \dfrac{\pi}{4} \le \dfrac{\pi}{4}$
the range of the given function $y = - \arccos(x) + \pi/4$ is given by the interval $[ - 3 \dfrac{\pi}{4} , \dfrac{\pi}{4} ]$.

c)
The graph of the given function $\arccos(x-1)$ is the graph of $\arccos(x)$ shifted 1 unit to the right. Shifting a graph to the left or to the right does not affect the range. Hence the range of $\arccos(x-1)$ is given by the interval $[ 0 , \pi ]$ and may be written as a double inequality
$0 \le \arccos(x-1) \le \pi$
Mutliply all terms of the inequality by -1 and change the symbol of inequality
$-\pi \le -\arccos(x-1)\le 0$
Add $-\pi$ to all terms of the inequality and simplify to obtain the range of the function $y = - \arccos(x-1) - \pi$
$- 2 \pi \le - \arccos(x-1) -\pi \le -\pi$

Example 4
Evaluate if possible
a) $\cos(\arccos(-0.5))$       b) $\arccos(\cos(\dfrac{\pi}{6}) )$       c) $\cos(\arccos(-2.1))$       d) $\arccos(\cos(\dfrac{- \pi}{3}) )$

Solution to Example 4
a)
$\cos(\arccos(-0.5)) = -0.5$ using property 4 above
b)
$\arccos(\cos(\dfrac{\pi}{6}) ) = \dfrac{\pi}{6}$ using property 5 above
c)
NOTE that we cannot use property 4 because -2.1 is not in the domain of $\arccos(x)$
$\cos(\arccos(-2.1)$ is undefined
d)
NOTE that we cannot use property 5 because $- \dfrac{\pi}{3}$ is not in the domain of that property. We will first transform $\cos(-\dfrac{\pi}{3})$ as follows
$\cos(- \dfrac{\pi}{3}) = \cos(\dfrac{\pi}{3})$ , cosine function is even
We now substitute $\cos( - \dfrac{ \pi}{3})$ by \cos(\dfrac{\pi}{3}) in the given expression
$\arccos(\cos( - \dfrac{\pi}{3}) ) = \arccos(\cos(\dfrac{\pi}{3}) )$
We now use the property 5 because $\dfrac{\pi}{3}$ is in the domain of the property. Hence
$\arccos(\cos( - \dfrac{\pi}{3}) ) = \arccos(\cos(\dfrac{\pi}{3})) = \dfrac{\pi}{3}$

## Interactive Tutorial to Explore the Transformed arccos(x)

The exploration is carried out by analyzing the effects of the parameters $a, b, c$ and $d$ included in the more general arccos function given by $f(x) = a \arccos(b x + c) + d$
Change parameters $a, b, c$ and $d$ and click on the button 'draw' in the left panel below. Zoom in and out for better viewing.

 a = 1 -10+10 b = 1 -10+10 c = 0 -10+10 d = 0 -10+10

1. Set the parameters to $a = 1, b = 1, c = 0$ and $d = 0$ to obtain $f(x) = \arccos(x)$ Check that the domain of $\arccos(x)$ is given by the interval $[-1 , 1]$ and the range is given by the interval $[0 , \pi ]$ , $(\pi \approx 3.14)$.
2. Change coefficient $a$ and explore its affect on the range of $a \arccos(x)$ function. (Hint: vertical compression, stretching, reflection).
Does the domain of $a \arccos(x)$ change with $a$.?
3. Does a change in $b$ affect the domain or/and the range of $a \arccos(b x)$? (Hint: horizontal compression, stretching).
4. Does a change in $c$ affect the domain or/and the range of $a\arccos(b x + c)$? How about horizontal shift?
5. Does a change in $d$ affect the range of $a\arccos(bx + c) + d$?
6. What is the domain and range of $a \arccos(b x + c) + d$ in terms of $a, b, c$ and $d$?

## Exercises

1. Find the domain and range of $f(x) = \arccos(2 x + 5) - \pi/2$.
2. Find the domain and range of $g(x) = - 0.5 \arccos(x + 4) - \pi$.
3. Find the domain and range of $h(x) = - \dfrac{1}{2} \arccos(x) - \pi / 4$.
Answers to Above Questions
1. Domain: $[-3 , -2]$ , Range: $[- \pi/2 , \pi/2]$.
2. Domain: $[-5 , -3]$ , Range: $[ -3\pi / 2 , -\pi]$.
3. Domain: $[-1 , 1]$ , Range: $[- 3\pi/4 , \pi/4]$.

## More References and Links to Inverse Trigonometric Functions

Find Domain and Range of Arccosine Functions
Inverse Trigonometric Functions
Graph, Domain and Range of Arcsin function
Graph, Domain and Range of Arctan function
Find Domain and Range of Arcsine Functions
Solve Inverse Trigonometric Functions Questions