# 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 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 2a) 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 3a) 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 4a) \( \sin(\arcsin(-0.99)) = -0.99\) 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. - 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: \( [- \pi/2 , 0] \).
## More References and Links to Inverse Trigonometric FunctionsInverse Trigonometric FunctionsGraph, Domain and Range of Arcsin function Graph, Domain and Range of Arctan function Find Domain and Range of Arccosine Functions Find Domain and Range of Arcsine Functions Solve Inverse Trigonometric Functions Questions |