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 \( y = arcsin(x) \)
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.
Exercises
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 |