# 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. 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 $y = arcsin(x)$

1. Domain: $[-1 , +1]$
2. Range: $[-\dfrac{\pi}{2} , \dfrac{\pi}{2}]$
3. $\arcsin(-x) = - \arcsin(x)$ , hence $\arcsin(x)$ is an odd function
4. $\arcsin(x)$ is a one to one function
5. $\sin(\arcsin(x)) = x$ , 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 $f^{-1}$
6. $\arcsin(\sin(x)) = x$ , 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 = 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 $f(x) = \arcsin(x)$ Check that the domain of $arcsin(x)$ is given by the interval $[-1 , 1]$ and the range is given by the interval $[-\dfrac{\pi}{2} , +\dfrac{\pi}{2} ]$ , $(\dfrac{\pi}{2} \approx 1.57)$
2. 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 $a \arcsin(x)$?
3. 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?
4. 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?
5. 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?
6. 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 $a \arcsin(x) + d$?
7. What is the domain and range of $a \arcsin(b x + c) + d$?

## Exercises

1. Find the domain and range of $f(x) = \arcsin(x - 1) - \pi/2$.
2. Find the domain and range of $g(x) = - \arcsin(2 x - 2) + \pi$.
3. Find the domain and range of $h(x) = - \dfrac{1}{2} \arcsin(x) - \pi / 4$.
1. Domain: $[0 , 2]$ , Range: $[- \pi , 0]$.
2. Domain: $[1/2 , 3/2]$ , Range: $[ \pi / 2, 3\pi / 2]$.
3. Domain: $[-1 , 1]$ , Range: $[- \pi/2 , 0]$.