# 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 $h(x) = - \dfrac{1}{2} \arccos(x) - \pi / 4$.
1. Domain: $[-3 , -2]$ , Range: $[- \pi/2 , \pi/2]$.
2. Domain: $[-5 , -3]$ , Range: $[ -3\pi / 2 , -\pi]$.
3. Domain: $[-1 , 1]$ , Range: $[- 3\pi/4 , \pi/4]$.