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Graph, Domain and Range of arcsin(x) Function
The definition, graph, and properties of the inverse trigonometric function \( \arcsin(x) \) are explored using graphs, examples with detailed solutions, and an interactive app.
Definition of arcsin(x) Functions
Let us examine the function \( \sin(x) \) that is shown below. On its implied domain, \( \sin(x) \) is not a one-to-one function as seen below; a horizontal line test will give several points of intersection. But if we limit the domain to \( [ -\dfrac{\pi}{2} , \dfrac{\pi}{2} ] \) (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 \( f(x) = \sin(x) , x \in [ -\dfrac{\pi}{2} , \dfrac{\pi}{2} ] \) is \( f^{-1} = \arcsin(x) \).
We define \( \arcsin(x) \) as follows:
\[ y = \arcsin(x) \iff x = \sin(y) \]
where \( -1 \le x \le 1 \) and \( -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2} \).
Let us make a table of values of \( y = \arcsin(x) \) and graph it along with \( y = \sin(x), [ -\dfrac{\pi}{2} , \dfrac{\pi}{2} ] \).
| \( x \) | \(-1\) | \(0\) | \(1\) |
| \( y = \arcsin(x) \) | \( -\dfrac{\pi}{2} \) | \( 0 \) | \( \dfrac{\pi}{2} \) |
| Reason | since \( \sin\left(-\dfrac{\pi}{2}\right) = -1 \) | since \( \sin(0) = 0 \) | since \( \sin\left(\dfrac{\pi}{2}\right) = 1 \) |
The graphs of \( y = \arcsin(x) \) and \( y = \sin(x) , x \in [ -\dfrac{\pi}{2} , \dfrac{\pi}{2} ] \) are shown below. Being inverses of each other, each of the two graphs is the reflection of the other across the line \( y = x \).
-and-arcsin(x).gif)
Examples & Step-by-Step Solutions
Example 1: Evaluating Arcsin Values
Problem: Evaluate \( \arcsin(x) \) given the value of \( x \).
Solution:
Special values related to special angles:
- \( \arcsin(0) = 0 \) because \( \sin(0) = 0 \)
- \( \arcsin(-1) = -\dfrac{\pi}{2} \; \text{ or } -90^{\circ} \) because \( \sin\left(-\dfrac{\pi}{2}\right) = -1 \)
- \( \arcsin(-3) = \) undefined because \( -3 \) is not in the domain of \( \arcsin(x) \) which is \( -1 \le x \le 1 \) (there is no angle that has sine equal to \(-3\)).
- \( \arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6} \; \text{ or } -30^{\circ} \) because \( \sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2} \)
- \( \arcsin(1) = \dfrac{\pi}{2} \; \text{ or } 90^{\circ} \) because \( \sin\left(\dfrac{\pi}{2}\right) = 1 \)
- \( \arcsin\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3} \; \text{ or } 60^{\circ} \) because \( \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} \)
Use of calculator:
- \( \arcsin(0.1) \approx 0.10 \; \text{ rad or } 5.74^{\circ} \)
- \( \arcsin(-0.4) \approx 0.41 \; \text{ rad or } 23.58^{\circ} \)
Properties of \( y = \arcsin(x) \)
- Domain: \( [-1 , 1] \)
- Range: \( [-\dfrac{\pi}{2} , \dfrac{\pi}{2}] \)
- \( \arcsin(-x) = - \arcsin(x) \), hence \( \arcsin(x) \) is an odd function.
- \( \arcsin(x) \) is a one-to-one function.
- \( \sin(\arcsin(x)) = x \), for \( x \) in the interval \( [-1, 1] \), due to the property of a function and its inverse: \( f(f^{-1}(x)) = x \).
- \( \arcsin(\sin(x)) = x \), for \( x \) in the interval \( [-\dfrac{\pi}{2},\dfrac{\pi}{2}] \), due to the property of a function and its inverse: \( f^{-1}(f(x)) = x \).
Example 2: Finding the Domain of Transformed Arcsin Functions
Problem: Find the domain of the functions:
a) \( y = \arcsin(2x) \)
b) \( y = \arcsin(-3x + 2) \)
c) \( y = -4 \arcsin(x/2) + \pi/4 \)
Solution:
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a) The domain is found by setting the argument \( 2x \) within the domain of the arcsine function (\( [-1, 1] \)):
\[ -1 \le 2x \le 1 \]
Divide all terms by \( 2 \):
\[ -\dfrac{1}{2} \le x \le \dfrac{1}{2} \]
The domain is \( \left[-\dfrac{1}{2}, \dfrac{1}{2}\right] \).
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b) Solve the double inequality for \( -3x + 2 \):
\[ -1 \le -3x + 2 \le 1 \]
Subtract \( 2 \) from all terms:
\[ -3 \le -3x \le -1 \]
Divide by \( -3 \) and reverse the inequality signs:
\[ \dfrac{1}{3} \le x \le 1 \]
The domain is \( \left[\dfrac{1}{3}, 1\right] \).
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c) Set up the inequality for the argument \( x/2 \):
\[ -1 \le \dfrac{x}{2} \le 1 \]
Multiply all terms by \( 2 \):
\[ -2 \le x \le 2 \]
The domain is \( [-2, 2] \).
Example 3: Finding the Range of Transformed Arcsin Functions
Problem: Find the range of the functions:
a) \( y = 2 \arcsin(x) \)
b) \( y = - \arcsin(x) + \pi/2 \)
c) \( y = \arcsin(x-1) \)
Solution:
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a) Start with the range of \( \arcsin(x) \):
\[ -\dfrac{\pi}{2} \le \arcsin(x) \le \dfrac{\pi}{2} \]
Multiply all terms by \( 2 \):
\[ -\pi \le 2 \arcsin(x) \le \pi \]
The range is \( [-\pi, \pi] \).
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b) Start with \( -\dfrac{\pi}{2} \le \arcsin(x) \le \dfrac{\pi}{2} \). Multiply by \( -1 \) and reverse the inequality signs:
\[ -\dfrac{\pi}{2} \le -\arcsin(x) \le \dfrac{\pi}{2} \]
Add \( \dfrac{\pi}{2} \) to all terms:
\[ 0 \le - \arcsin(x) + \dfrac{\pi}{2} \le \pi \]
The range is \( [0, \pi] \).
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c) The graph of \( \arcsin(x-1) \) is shifted 1 unit to the right. Shifting a graph horizontally does not affect the range. Hence, the range remains \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \).
Example 4: Evaluating Composite Expressions
Problem: Evaluate if possible:
a) \( \sin(\arcsin(-0.99)) \)
b) \( \arcsin\left(\sin\left(-\dfrac{\pi}{9}\right)\right) \)
c) \( \sin(\arcsin(1.4)) \)
d) \( \arcsin\left(\sin\left(\dfrac{7\pi}{6}\right)\right) \)
Solution:
- a) \( \sin(\arcsin(-0.99)) = -0.99 \) using property 5.
- b) \( \arcsin\left(\sin\left(-\dfrac{\pi}{9}\right)\right) = -\dfrac{\pi}{9} \) using property 6.
- c) Property 5 cannot be used because \( 1.4 \) is not in the domain of \( \arcsin(x) \) (\( [-1, 1] \)). Therefore, \( \sin(\arcsin(1.4)) \) is undefined.
- d) Property 6 cannot be directly applied because \( \dfrac{7\pi}{6} \) is outside the interval \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \). First, evaluate:
\[ \sin\left(\dfrac{7\pi}{6}\right) = -\dfrac{1}{2} \]
Substitute this back into the expression:
\[ \arcsin\left(\sin\left(\dfrac{7\pi}{6}\right)\right) = \arcsin\left(-\dfrac{1}{2}\right) \]
Using the definition of arcsin:
\[ \arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6} \text{ because } \sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2} \]
Interactive Tutorial to Explore the Transformed arcsin(x)
The exploration is carried out by analyzing the effects of the parameters \( a, b, c \) and \( d \) included in the more general arcsin function given by:
\[ f(x) = a \arcsin(b x + c) + d \]
Change parameters \( a, b, c \) and \( d \) and click on the button 'draw' in the left panel below.
- Set the parameters to \( a = 1, b = 1, c = 0 \) and \( d = 0 \) to obtain:
\[ f(x) = \arcsin(x) \]
Check that the domain of \( \arcsin(x) \) is given by the interval \( [-1 , 1] \) and the range is given by the interval \( [-\dfrac{\pi}{2} , +\dfrac{\pi}{2} ] \) (\( \dfrac{\pi}{2} \approx 1.57 \)).
- Change coefficient \( a \) and note how the graph of \( a \arcsin(x) \) changes (Hint: vertical compression, stretching, reflection). How does it affect the range of the \( a \arcsin(x) \) function? Does coefficient \( a \) affect the domain of \( a \arcsin(x) \)?
- Change coefficient \( b \) and note how the graph of \( \arcsin(b x) \) changes (horizontal compression, stretching). Does a change in \( b \) affect the domain or/and the range of the function?
- Change coefficient \( c \) and note how the graph of \( a \arcsin(bx + c) \) changes (horizontal shift). Does a change of the coefficient \( c \) affect the domain or/and the range of the function?
- Change coefficient \( d \) and note how the graph of \( \arcsin(bx + c) + d \) changes (vertical shift). Does a change in coefficient \( d \) affect the range of the function? Does it affect its domain?
- If the range of \( \arcsin(x) \) is given by the interval \( [-\dfrac{\pi}{2} , \dfrac{\pi}{2}] \), what is the range of \( a \arcsin(x) \)? What is the range of \( a \arcsin(x) + d \)?
- What is the domain and range of \( a \arcsin(b x + c) + d \)?
Exercises & Step-by-Step Solutions
Exercise 1: Domain and Range of \( f(x) = \arcsin(x - 1) - \pi/2 \)
Problem: Find the domain and range of \( f(x) = \arcsin(x - 1) - \pi/2 \).
Solution:
- Domain: Set \( -1 \le x - 1 \le 1 \implies 0 \le x \le 2 \). Domain is \( [0, 2] \).
- Range: Since \( -\dfrac{\pi}{2} \le \arcsin(x-1) \le \dfrac{\pi}{2} \), subtract \( \dfrac{\pi}{2} \) from all parts:
\[ -\dfrac{\pi}{2} - \dfrac{\pi}{2} \le \arcsin(x-1) - \dfrac{\pi}{2} \le \dfrac{\pi}{2} - \dfrac{\pi}{2} \implies -\pi \le f(x) \le 0 \]
Range is \( [-\pi, 0] \).
Exercise 2: Domain and Range of \( g(x) = - \arcsin(2x - 2) + \pi \)
Problem: Find the domain and range of \( g(x) = - \arcsin(2x - 2) + \pi \).
Solution:
- Domain: Set \( -1 \le 2x - 2 \le 1 \implies 1 \le 2x \le 3 \implies \dfrac{1}{2} \le x \le \dfrac{3}{2} \). Domain is \( \left[\dfrac{1}{2}, \dfrac{3}{2}\right] \).
- Range: Since \( -\dfrac{\pi}{2} \le \arcsin(2x-2) \le \dfrac{\pi}{2} \), multiply by \( -1 \):
\[ -\dfrac{\pi}{2} \le -\arcsin(2x-2) \le \dfrac{\pi}{2} \]
Add \( \pi \):
\[ -\dfrac{\pi}{2} + \pi \le g(x) \le \dfrac{\pi}{2} + \pi \implies \dfrac{\pi}{2} \le g(x) \le \dfrac{3\pi}{2} \]
Range is \( \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \).
Exercise 3: Domain and Range of \( h(x) = - \dfrac{1}{2} \arcsin(x) - \pi / 4 \)
Problem: Find the domain and range of \( h(x) = - \dfrac{1}{2} \arcsin(x) - \pi / 4 \).
Solution:
- Domain: Standard domain of \( \arcsin(x) \) is \( [-1, 1] \).
- Range: Since \( -\dfrac{\pi}{2} \le \arcsin(x) \le \dfrac{\pi}{2} \), multiply by \( -\dfrac{1}{2} \):
\[ -\dfrac{\pi}{4} \le -\dfrac{1}{2}\arcsin(x) \le \dfrac{\pi}{4} \]
Subtract \( \dfrac{\pi}{4} \):
\[ -\dfrac{\pi}{4} - \dfrac{\pi}{4} \le h(x) \le \dfrac{\pi}{4} - \dfrac{\pi}{4} \implies -\dfrac{\pi}{2} \le h(x) \le 0 \]
Range is \( \left[-\dfrac{\pi}{2}, 0\right] \).
More References and Links to Inverse Trigonometric Functions
Explore additional resources on inverse trigonometric functions: