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Graph, Domain and Range of arcsin(x) function

The definition, graph and the 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.

graph of sin(x) with limited domain

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	$\dfrac{\pi}{2}$
	since $\sin(-\dfrac{\pi}{2}) = - 1$	since $\sin(0) = 0$	since $\sin(\dfrac{\pi}{2}) = 1$

The graphs of

$y = \arcsin(x)$ and

$y = \sin(x) , x \in [ -\dfrac{\pi}{2} , \dfrac{\pi}{2} ]$ are shown below. Being inverse of each other, each of the two graphs is the reflection of the other on the line

$y = x$ .

Example 1
Evaluate

$arcsin(x)$ given the value of

$x$ .
Special values related to special angles

$\arcsin(0) = 0$ because

$\sin(0) = 0$

$\arcsin(-1) = -\dfrac{\pi}{2} \; \text{ or } -90^{o}$ because

$\sin(-\dfrac{\pi}{2}) = -1$

$\arcsin(-3) =$ 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 ).

$\arcsin(-\dfrac{1}{2}) = -\dfrac{\pi}{6} \; \text{ or } -30^{o}$ because

$\sin(-\dfrac{\pi}{6}) = -\dfrac{1}{2}$

$\arcsin(1) = \dfrac{\pi}{2} \; \text{ or } 90^{o}$ because

$\sin(\dfrac{\pi}{2}) = 1$

$\arcsin(\dfrac{\sqrt 3}{2}) = \dfrac{\pi}{3} \; \text{ or } 60^{o}$ because

$\sin(\dfrac{\pi}{3}) = \dfrac{\sqrt 3}{2}$
Use of calculator

$\arcsin(0.1) = 0.10 \; \text{ or } 5.74^{o}$

$\arcsin(-0.4) = 0.41 \; \text{ or } 23.58^{o}$

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 property of a function and its inverse : $f(f^{-1}(x) = x$ where $x$ is in the domain of $f^{-1}$
$\arcsin(\sin(x)) = x$ , for x in the interval $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ , 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)

$y = \arcsin(2x)$ b)

$y = \arcsin(-3 x +2)$ c)

$y = -4 \arcsin(x/2) + \pi/4$

Solution to Example 2
a)
the domain is found by first writing that the argument

$2x$ of the given function is within the domain of the arcsine function given above in the properties. Hence we need to solve the double inequality

$-1 \le 2x \le 1$
divide all terms of the double inequality by 2 to obtain

$-1/2 \le x \le 1/2$ , which is the domain of the given function.
b)

$-1 \le -3 x +2 \le 1$
Solve the above inequality

$-3 \le -3 x \le -1$

$1/3 \le x \le 1$ , which is the domain of the given function.
c)

$-1 \le x/2 \le 1$
Solve the above inequality

$-2 \le x \le 2$ , which is the domain of the given function.

Example 3
Find the range of the functions:
a)

$y = 2 \arcsin(x)$ b)

$y = - \arcsin(x) + \pi/2$ c)

$y = \arcsin(x-1)$

Solution to Example 3
a)
the range is found by first writing the range of

$\arcsin(x)$ as a double inequality

$-\dfrac{\pi}{2} \le \arcsin(x) \le \dfrac{\pi}{2}$
multiply all terms of the above inequality by 2 and simplify

$-\pi \le 2 \arcsin(x) \le \pi$
the range of the given function

$2 \arcsin(x)$ is given by the interval

$[ -\pi , \pi ]$ .

b)
we start with the range of

$\arcsin(x)$

$-\dfrac{\pi}{2} \le \arcsin(x) \le \dfrac{\pi}{2}$
multiply all terms of the above inequality by -1 and change symbol of the double inequality

$-\dfrac{\pi}{2} \le -\arcsin(x) \le \dfrac{\pi}{2}$
add

$\dfrac{\pi}{2}$ to all terms of the inequality above and simplify

$0 \le - \arcsin(x) + \dfrac{\pi}{2} \le \pi$
the range of the given function

$- \arcsin(x) + \dfrac{\pi}{2}$ is given by the interval

$[ 0, \pi ]$ .

c)
The graph of the given function

$\arcsin(x-1)$ is the graph of

$\arcsin(x)$ 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

$\arcsin(x-1)$ is given by the interval

$[ -\dfrac{\pi}{2} , \dfrac{\pi}{2} ]$

Example 4
Evaluate if possible
a)

$\sin(\arcsin(-0.99))$ b)

$\arcsin(\sin(-\dfrac{\pi}{9}) )$ c)

$\sin(\arcsin(1.4))$ d)

$\arcsin(\sin(\dfrac{7 \pi}{6}) )$

Solution to Example 4
a)

$\sin(\arcsin(-0.99)) = -0.99$ using property 5 above
b)

$\arcsin(\sin(-\dfrac{\pi}{9}) ) = -\dfrac{\pi}{9}$ using property 6 above
c)
NOTE that we cannot use property 5 because 1.4 is not in the domain of

$\arcsin(x)$

$\sin(\arcsin(1.4))$ is undefined
d)
NOTE that we cannot use property 6 because

$\dfrac{7 \pi}{6}$ is not in the domain of that property. We will first evaluate

$\sin(\dfrac{7 \pi}{6}) = -1/2$
We now substitute

$\sin(\dfrac{7 \pi}{6})$ by

$-1/2$ in the given expression

$\arcsin(\sin(\dfrac{7 \pi}{6}) ) = \arcsin(-1/2) )$
We now use the definition of

$\arcsin(-1/2)$ to evaluate it

$\arcsin(-1/2) ) = -\dfrac{\pi}{6}$ because

$\sin(-\dfrac{\pi}{6}) = - 1/2$

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.

a =

-10+10

b =

-10+10

c =

-10+10

d =

-10+10

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 $a$ affects 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

Find the domain and range of $f(x) = \arcsin(x - 1) - \pi/2$ .
Find the domain and range of $g(x) = - \arcsin(2 x - 2) + \pi$ .
Find the domain and range of $h(x) = - \dfrac{1}{2} \arcsin(x) - \pi / 4$ .

Answers to Above Questions

Domain: $[0 , 2]$ , Range: $[- \pi , 0]$ .
Domain: $[1/2 , 3/2]$ , Range: $[ \pi / 2, 3\pi / 2]$ .
Domain: $[-1 , 1]$ , Range: $[- \pi/2 , 0]$ .

Graph, Domain and Range of arcsin(x) function

Definition of arcsin(x) Functions

Properties of y=arcsin(x) y = arcsin(x)

Interactive Tutorial to Explore the Transformed arcsin(x)

Exercises

More References and Links to Inverse Trigonometric Functions

Properties of $y = arcsin(x)$