Graph, Domain and Range of arccos(x) Function
The graph, domain, 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 if 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.
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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.
| \( x \) | \(-1\) | \(0\) | \(1\) |
| \( y = \arccos(x) \) | \( \pi \) | \( \dfrac{\pi}{2} \) | \( 0 \) |
| Reason | since \( \cos(\pi) = -1 \) | since \( \cos\left(\dfrac{\pi}{2}\right) = 0 \) | 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 across the line \( y = x \); this is one of the graphical properties of inverse functions.
-and-arccos(x).gif)
Examples & Step-by-Step Solutions
Example 1: Evaluating arccos(x) Values
Problem: Evaluate \( \arccos(x) \) given various values of \( x \).
Solution:
- \( \arccos(1) = 0 \) because \( \cos(0) = 1 \)
- \( \arccos(-1) = \pi \; \text{ or } 180^{\circ} \) 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 whose cosine equals 2).
- \( \arccos\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3} \; \text{ or } 120^{\circ} \) because \( \cos\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{2} \)
- \( \arccos\left(-\dfrac{\sqrt{3}}{2}\right) = \dfrac{5\pi}{6} \; \text{ or } 150^{\circ} \) because \( \cos\left(\dfrac{5\pi}{6}\right) = -\dfrac{\sqrt{3}}{2} \)
- \( \arccos\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6} \; \text{ or } 30^{\circ} \) because \( \cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2} \)
- Use of calculator:
\( \arccos(0.25) \approx 1.32 \; \text{ rad or } 75.52^{\circ} \)
\( \arccos(-0.77) \approx 2.45 \; \text{ rad or } 140.35^{\circ} \)
Properties of \( y = \arccos(x) \)
- Domain: \( [-1 , 1] \)
- Range: \( [0 , \pi] \)
- \( \arccos(x) \) is a one-to-one function.
- \( \cos(\arccos(x)) = x \), for \( x \) in the interval \( [-1, 1] \), due to the property of a function and its inverse: \( f(f^{-1}(x)) = x \).
- \( \arccos(\cos(x)) = x \), for \( x \) in the interval \( [0 , \pi] \), due to the property of a function and its inverse: \( f^{-1}(f(x)) = x \).
Example 2: Finding the Domain of Transformed arccos Functions
Problem: Find the domain of the functions:
a) \( y = \arccos(-3x) \)
b) \( y = \arccos(2x - 1) \)
c) \( y = 3 \arccos(x/2 + 1) + \pi/4 \)
Solution:
-
a) The domain is found by setting the argument \( -3x \) within the domain of the arccos function (\( [-1, 1] \)):
\[ -1 \le -3x \le 1 \]
Dividing all terms by \( -3 \) and reversing the inequality symbols yields:
\[ -\dfrac{1}{3} \le x \le \dfrac{1}{3} \]
The domain is \( \left[-\dfrac{1}{3}, \dfrac{1}{3}\right] \).
-
b) Solve the double inequality for \( 2x - 1 \):
\[ -1 \le 2x - 1 \le 1 \implies 0 \le 2x \le 2 \implies 0 \le x \le 1 \]
The domain is \( [0, 1] \).
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c) Set up the inequality for the argument \( x/2 + 1 \):
\[ -1 \le \dfrac{x}{2} + 1 \le 1 \implies -2 \le \dfrac{x}{2} \le 0 \implies -4 \le x \le 0 \]
The domain is \( [-4, 0] \).
Example 3: Finding the Range of Transformed arccos Functions
Problem: Find the range of the functions:
a) \( y = -2 \arccos(x) \)
b) \( y = -\arccos(x) + \pi/4 \)
c) \( y = -\arccos(x-1) - \pi \)
Solution:
-
a) Start with the range of \( \arccos(x) \):
\[ 0 \le \arccos(x) \le \pi \]
Multiply by \( -2 \), reversing the inequality symbols:
\[ -2\pi \le -2\arccos(x) \le 0 \]
The range is \( [-2\pi, 0] \).
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b) Start with \( 0 \le \arccos(x) \le \pi \). Multiply by \( -1 \):
\[ -\pi \le -\arccos(x) \le 0 \]
Add \( \dfrac{\pi}{4} \) to all terms:
\[ -\pi + \dfrac{\pi}{4} \le -\arccos(x) + \dfrac{\pi}{4} \le \dfrac{\pi}{4} \implies -\dfrac{3\pi}{4} \le y \le \dfrac{\pi}{4} \]
The range is \( \left[-\dfrac{3\pi}{4}, \dfrac{\pi}{4}\right] \).
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c) The graph of \( \arccos(x-1) \) is shifted 1 unit right, which does not affect the range \( [0, \pi] \). Thus:
\[ 0 \le \arccos(x-1) \le \pi \]
Multiply by \( -1 \):
\[ -\pi \le -\arccos(x-1) \le 0 \]
Add \( -\pi \) to all terms:
\[ -2\pi \le -\arccos(x-1) - \pi \le -\pi \]
The range is \( [-2\pi, -\pi] \).
Example 4: Evaluating Composite Expressions
Problem: Evaluate if possible:
a) \( \cos(\arccos(-0.5)) \)
b) \( \arccos\left(\cos\left(\dfrac{\pi}{6}\right)\right) \)
c) \( \cos(\arccos(-2.1)) \)
d) \( \arccos\left(\cos\left(-\dfrac{\pi}{3}\right)\right) \)
Solution:
- a) \( \cos(\arccos(-0.5)) = -0.5 \) using property 4.
- b) \( \arccos\left(\cos\left(\dfrac{\pi}{6}\right)\right) = \dfrac{\pi}{6} \) using property 5.
- c) Property 4 cannot be used because \( -2.1 \) is not in the domain of \( \arccos(x) \). Therefore, \( \cos(\arccos(-2.1)) \) is undefined.
- d) Property 5 cannot be directly applied because \( -\dfrac{\pi}{3} \) is outside the interval \( [0, \pi] \). First, use the even property of cosine: \( \cos\left(-\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{3}\right) \). Then:
\[ \arccos\left(\cos\left(-\dfrac{\pi}{3}\right)\right) = \arccos\left(\cos\left(\dfrac{\pi}{3}\right)\right) = \dfrac{\pi}{3} \]
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.
- 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 \( [0 , \pi ] \) (\( \pi \approx 3.14 \)).
- Change coefficient \( a \) and explore its effect on the range of \( a \arccos(x) \). (Hint: vertical compression, stretching, reflection). Does the domain change with \( a \)?
- Does a change in \( b \) affect the domain or/and the range of \( a \arccos(b x) \)? (Hint: horizontal compression, stretching).
- Does a change in \( c \) affect the domain or/and the range of \( a\arccos(b x + c) \)? How about horizontal shift?
- Does a change in \( d \) affect the range of \( a\arccos(bx + c) + d \)?
- What is the domain and range of \( a \arccos(b x + c) + d \) in terms of \( a, b, c \) and \( d \)?
Exercises & Step-by-Step Solutions
Exercise 1: Domain and Range of \( f(x) = \arccos(2x + 5) - \pi/2 \)
Problem: Find the domain and range of \( f(x) = \arccos(2x + 5) - \pi/2 \).
Solution:
- Domain: Set \( -1 \le 2x + 5 \le 1 \implies -6 \le 2x \le -4 \implies -3 \le x \le -2 \). Domain is \( [-3, -2] \).
- Range: Since \( 0 \le \arccos(2x+5) \le \pi \), subtract \( \frac{\pi}{2} \) from all parts:
\[ -\dfrac{\pi}{2} \le \arccos(2x+5) - \dfrac{\pi}{2} \le \dfrac{\pi}{2} \]
Range is \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \).
Exercise 2: Domain and Range of \( g(x) = -0.5 \arccos(x + 4) - \pi \)
Problem: Find the domain and range of \( g(x) = -0.5 \arccos(x + 4) - \pi \).
Solution:
- Domain: Set \( -1 \le x + 4 \le 1 \implies -5 \le x \le -3 \). Domain is \( [-5, -3] \).
- Range: Since \( 0 \le \arccos(x+4) \le \pi \), multiply by \( -0.5 \):
\[ 0 \ge -0.5 \arccos(x+4) \ge -0.5\pi \]
Subtract \( \pi \):
\[ -\pi \ge -0.5 \arccos(x+4) - \pi \ge -1.5\pi \implies -\dfrac{3\pi}{2} \le y \le -\pi \]
Range is \( \left[-\dfrac{3\pi}{2}, -\pi\right] \).
Exercise 3: Domain and Range of \( h(x) = -0.5 \arccos(x) - \pi/4 \)
Problem: Find the domain and range of \( h(x) = -0.5 \arccos(x) - \pi/4 \).
Solution:
- Domain: Standard domain of \( \arccos(x) \) is \( [-1, 1] \).
- Range: Since \( 0 \le \arccos(x) \le \pi \), multiply by \( -0.5 \):
\[ -\dfrac{\pi}{2} \le -0.5 \arccos(x) \le 0 \]
Subtract \( \dfrac{\pi}{4} \):
\[ -\dfrac{\pi}{2} - \dfrac{\pi}{4} \le h(x) \le -\dfrac{\pi}{4} \implies -\dfrac{3\pi}{4} \le h(x) \le -\dfrac{\pi}{4} \]
Range is \( \left[-\dfrac{3\pi}{4}, -\dfrac{\pi}{4}\right] \).
More References and Links to Inverse Trigonometric Functions
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