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# Inverse trigonometric functions

Module by: Sunil Kumar Singh. E-mail the author

Inverse trigonometric function returns an angle corresponding to a real number, following certain rule. They are inverse functions corresponding to trigonometric functions. The inverse function of sine , for example, is defined as :

f x = sin - 1 x ; x [ -1 , 1 ] f x = sin - 1 x ; x [ -1 , 1 ]

where “x” is a real number, "f(x)" is the angle. Clearly, "f(x)" is the angle, whose sine is “x”. Symbolically,

sin { f x } = sin { sin - 1 x } = x sin { f x } = sin { sin - 1 x } = x

In the representation of inverse function, we should treat “-1” as symbol – not as power. In particular,

sin - 1 x 1 sin x sin - 1 x 1 sin x

Inverse trigonometric functions are also called arc functions. This is an alternative notation. The corresponding functions are arcsine, arccosine, arctangent etc. For example,

f x = sin - 1 x = arcsin x f x = sin - 1 x = arcsin x

## Nature of trigonometric functions

Trigonometric functions are many-one relations. The trigonometric ratio of different angles evaluate to same value. If we draw a line parallel to x-axis such that 0 < y < 1, then it intersects sine plot for multiple times – ,in fact, infinite times. It follows, then, that we can associate many angles to the same sine value. The trigonometric functions are, therefore, not an injection and hence not a bijection. As such, we can not define an inverse of trigonometric function in the first place! We shall see that we need to redefine trigonometric functions in order to make them invertible.

In order to define, an inverse function, we require to have one-one relation in both directions between domain and range. The function needs to be a bijection. It emerges that we need to shorten the domain of trigonometric functions such that a distinct angle corresponds to a distinct real number. Similarly, a distinct real number corresponds to a distinct angle.

We can identify many such shortened intervals for a particular trigonometric function. For example, the shortened domain of sine function can be any one of the intervals defined by :

[ 2 n 1 π 2 , 2 n + 1 π 2 ] , n Z [ 2 n 1 π 2 , 2 n + 1 π 2 ] , n Z

The domain corresponding to n = 0 yields principal domain given by :

[ - π 2 , π 2 ] [ - π 2 , π 2 ]

The nature of trigonometric functions is periodic. Same values repeat after certain interval. Here, our main task is to identify an interval of “x” such that all possible values of a trigonometric function are included once. This will ensure one-one relation in both directions between domain and range of the function. This interval is easily visible on graphs of the corresponding trigonometric function.

## Inverse trigonometric functions

Every angle in the new domain (shortened) is related to a distinct real number in the range. Inversely, every real number in the range is related to a distinct angle in the domain of the trigonometric function. We are aware that the elements of the "ordered pair" in inverse relation exchanges their places. Therefore, it follows that domain and range of trigonometric function are exchanged for corresponding inverse function i.e. domain becomes range and range becomes domain.

### arcsine function

The arcsine function is inverse function of trigonometric sine function. From the plot of sine function, it is clear that an interval between - π / 2 - π / 2 and π / 2 π / 2 includes all possible values of sine function only once. Note that end points are included. The redefinition of domain of trigonometric function, however, does not change the range.

Domain of sine = [ - π 2 , π 2 ] Domain of sine = [ - π 2 , π 2 ]

Range of sine = [ - 1, 1 ] Range of sine = [ - 1, 1 ]

This redefinition renders sine function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :

Domain of arcsine = [ - 1, 1 ] Domain of arcsine = [ - 1, 1 ]

Range of arcsine = [ - π 2, π 2 ] Range of arcsine = [ - π 2, π 2 ]

Therefore, we define arcsine function as :

f : [ - 1,1 ] [ - π 2 , π 2 ] by f(x) = arcsin(x) f : [ - 1,1 ] [ - π 2 , π 2 ] by f(x) = arcsin(x)

The arcsin(x) .vs. x graph is shown here.

### arccosine function

The arccosine function is inverse function of trigonometric cosine function. From the plot of cosine function, it is clear that an interval between 0 and π π includes all possible values of cosine function only once. Note that end points are included. The redefinition of domain of trigonometric function, however, does not change the range.

Domain of cosine = [ 0, π ] Domain of cosine = [ 0, π ]

Range of cosine = [ - 1, 1 ] Range of cosine = [ - 1, 1 ]

This redefinition renders cosine function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :

Domain of arccosine = [ - 1,1 ] Domain of arccosine = [ - 1,1 ]

Range of arccosine = [ 0, π ] Range of arccosine = [ 0, π ]

Therefore, we define arccosine function as :

f : [ - 1,1 ] [ 0, π ] by f(x) = arccos(x) f : [ - 1,1 ] [ 0, π ] by f(x) = arccos(x)

The arccos (x) .vs. x graph is shown here.

### arctangent function

The arctangent function is inverse function of trigonometric tangent function. From the plot of tangent function, it is clear that an interval between - π / 2 - π / 2 and π / 2 π / 2 includes all possible values of tangent function only once. Note that end points are excluded. The redefinition of domain of trigonometric function, however, does not change the range.

Domain of tangent = - π 2, π 2 Domain of tangent = - π 2, π 2

Range of tangent = R Range of tangent = R

This redefinition renders tangent function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :

Domain of arctangent = R Domain of arctangent = R

Range of arctangent = - π / 2, π / 2 Range of arctangent = - π / 2, π / 2

Therefore, we define arctangent function as :

f : R - π 2 , π 2 by f(x) = arctan (x) f : R - π 2 , π 2 by f(x) = arctan (x)

The arctan(x) .vs. x graph is shown here.

### arccosecant function

The arccosecant function is inverse function of trigonometric cosecant function. From the plot of cosecant function, it is clear that union of two disjointed intervals between “ - π / 2 - π / 2 and 0” and “0 and π / 2 π / 2 ” includes all possible values of cosecant function only once. Note that zero is excluded, but “ - π / 2 - π / 2 “ and “ π / 2 π / 2 ” are included . The redefinition of domain of trigonometric function, however, does not change the range.

Domain of cosecant = [ - π / 2, π / 2 ] { 0 } Domain of cosecant = [ - π / 2, π / 2 ] { 0 }

Range of cosecant = - , - 1 ] [ 1, = R - 1, 1 Range of cosecant = - , - 1 ] [ 1, = R - 1, 1

This redefinition renders cosecant function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :

Domain of arccosecant = R - 1, 1 Domain of arccosecant = R - 1, 1

Range of arccosecant = [ - π / 2, π / 2 ] { 0 } Range of arccosecant = [ - π / 2, π / 2 ] { 0 }

Therefore, we define arccosecant function as :

f : R - 1, 1 [ - π 2 , π 2 ] { 0 } by f(x) = arccosec (x) f : R - 1, 1 [ - π 2 , π 2 ] { 0 } by f(x) = arccosec (x)

The arccosec(x) .vs. x graph is shown here.

### arcsecant function

The arcsecant function is inverse function of trigonometric secant function. From the plot of secant function, it is clear that union of two disjointed intervals between “0 and π / 2 π / 2 ” and “ π / 2 π / 2 and π π ” includes all possible values of secant function only once. Note that “ π / 2 π / 2 ” is excluded. The redefinition of domain of trigonometric function, however, does not change the range.

Domain of secant = [ 0, π / 2 ) ( π / 2, π ] = [ 0, π ] { π / 2 } Domain of secant = [ 0, π / 2 ) ( π / 2, π ] = [ 0, π ] { π / 2 }

Range of secant = ( - , - 1 ] [ 1, ) = R - 1,1 Range of secant = ( - , - 1 ] [ 1, ) = R - 1,1

This redefinition renders secant function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :

Domain of arcsecant = R - 1, 1 Domain of arcsecant = R - 1, 1

Range of arcsecant = [ 0, π ] { π / 2 } Range of arcsecant = [ 0, π ] { π / 2 }

Therefore, we define arcsecant function as :

f : R - 1, 1 [ 0, π ] { π / 2 } by f(x) = arcsec (x) f : R - 1, 1 [ 0, π ] { π / 2 } by f(x) = arcsec (x)

The arcsec(x) .vs. x graph is shown here.

### arccotangent function

The arccotangent function is inverse function of trigonometric cotangent function. From the plot of cotangent function it is clear that an interval between 0 and π π includes all possible values of cotangent function only once. Note that end points are excluded. The redefinition of domain of trigonometric function, however, does not change the range.

Domain of cotangent = 0, π Domain of cotangent = 0, π

Range of cotangent = R Range of cotangent = R

This redefinition renders cotangent function invertible. Clearly, the domain and range are exchanged for the inverse function. Hence, domain and range of the inverse function are :

Domain of arccotangent = R Domain of arccotangent = R

Range of arccotangent = 0, π Range of arccotangent = 0, π

Therefore, we define arccotangent function as :

f : R 0, π by f(x) = arccot (x) f : R 0, π by f(x) = arccot (x)

The arccot(x) .vs. x graph is shown here.

## Example

### Example 1

Problem : Find y when :

y = tan - 1 - 1 3 y = tan - 1 - 1 3

Solution : There are multiple angles for which :

tan y = x = - 1 3 tan y = x = - 1 3

However, range of sine function is [-π/2, π/2]. We need to find angle, which falls in this range. Now, acute angle corresponding to the value of 1/√3 is π/6. In accordance with sign diagram, tangent is negative in second and fourth quarters. But range is [-π/2, π/2]. Hence, we need to find angle in fourth quadrant. The angle in the fourth quadrant whose tangent has magnitude of 1/√3 is given by :

y = 2 π - π 6 = 11 π 6 y = 2 π - π 6 = 11 π 6

Corresponding negative angle is :

y = 11 π 6 - 2 π = - π 6 y = 11 π 6 - 2 π = - π 6

### Example 2

Problem : Find domain of the function given by :

f x = cos - 1 x [ x ] f x = cos - 1 x [ x ]

Solution : The given function is quotient of two functions having rational form :

f x = g x h x f x = g x h x

The domain of quotient is given by :

D = D 1 D 2 { x : x when h(x) = 0 } D = D 1 D 2 { x : x when h(x) = 0 }

Here, g x = cos - 1 x g x = cos - 1 x . The domain of arccosine is [-1,1]. Hence,

D 1 = Domain of “g” = [ - 1,1 ] D 1 = Domain of “g” = [ - 1,1 ]

The denominator function h(x) is greatest integer function. Its domain is “R”.

D 2 = Domain of “h” = R D 2 = Domain of “h” = R

The intersection of two domains is :

D 1 D 2 = [ - 1,1 ] R = [ - 1,1 ] D 1 D 2 = [ - 1,1 ] R = [ - 1,1 ]

Now, greatest integer function becomes zero for values of “x” in the interval [0,1). Hence, domain of given function is :

D = D 1 D 2 [ 0,1 ) D = D 1 D 2 [ 0,1 )

D = [ - 1,1 ] - [ 0,1 ) = - 1 x < 0 { 1 } D = [ - 1,1 ] - [ 0,1 ) = - 1 x < 0 { 1 }

## Summary

Redefined domains of trigonometric functions are tabulated here :

---------------------------------------------------------------------------------------------------
Trigonometric   Old				New			Old		New
Function        Domain 			        Domain                  Range           Range
---------------------------------------------------------------------------------------------------
sine		  R				[-π/2, π/2]		[-1,1]		[-1,1]
cosine		  R				[0, π]			[-1,1]		[-1,1]
tan 		  R – odd multiples of π/2      (-π/2, π/2)		R		R
cosecant	  R – integer multiple of π	[-π/2, π/2] – {0}	R – (-1,1)      R – (-1,1)
secant		  R - odd multiples of π/2      [0, π] – {π/2}		R – (-1,1)      R – (-1,1)
cotangent	  R – integer multiple of π	(0, π)			R		R
---------------------------------------------------------------------------------------------------

We observe that there is no change in the range – even though domains of the trigonometric functions have changed.

The corresponding domain and range of six inverse trigonometric functions are tabulated here.

--------------------------------------------------------------
Inverse  		Domain		     Range
Trigonometric
Function
--------------------------------------------------------------
arcsine		        [-1,1]		     [-π/2, π/2]
arccosine	        [-1,1]		     [0, π]
arctangent	        R 		     (-π/2, π/2)
arccosecant	        R – (-1,1)	     [-π/2, π/2] – {0}
arcsecant	        R – (-1,1)	     [0, π] – {π/2}
arccotangent	        R 		     (0, π)
--------------------------------------------------------------



## Exercise

### Exercise 1

Find the domain of the function given by :

f x = 2 sin - 1 x f x = 2 sin - 1 x

#### Solution

The exponent of the exponential function is inverse trigonometric function. Exponential function is real for all real values of exponent. We see here that given function is real for the values of “x” corresponding to which arcsine function is real. Now, domain of arcsine function is [-1,1]. This is the interval of "x" for which arcsine is real. Hence, domain of the given function, “f(x)” is :

Domain = [ - 1,1 ] Domain = [ - 1,1 ]

### Exercise 2

Problem : Find the domain of the function given by :

f x = cos - 1 3 3 + sin x f x = cos - 1 3 3 + sin x

#### Solution

Solution : The given function is an inverse cosine function whose argument is a rational function involving trigonometric function. The domain interval of inverse cosine function is [-1, 1]. Hence, value of argument to inverse cosine function should lie within this interval. It means that :

- 1 3 3 + sin x 1 - 1 3 3 + sin x 1

Comparing with the form of modulus, | x | 1 - 1 x 1 | x | 1 - 1 x 1 , we conclude :

| 3 | | 3 + sin x | 1 | 3 | | 3 + sin x | 1

Since, modulus is a non-negative number, the inequality sign remains same after simplification :

| 3 | | 3 + sin x | | 3 | | 3 + sin x |

Again 3 > 0 and 3+sin x > 0, we can open up the expression within the modulus operator without any change in inequality sign :

3 3 + sin x 0 sin x sin x 0 3 3 + sin x 0 sin x sin x 0

The solution of sine function is the domain of the given function :

Domain = 2 n π x 2 n + 1 π , x Z Domain = 2 n π x 2 n + 1 π , x Z

### Exercise 3

Find range of the function :

f x = 1 2 sin 2 x f x = 1 2 sin 2 x

#### Solution

We have already solved this problem by building up interval in earlier module. Here, we shall find domain conventionally by solving for x. The denominator of given function is non-negative as value of sin2x can not exceed 1. Hence, domain of function is real number set R. Further, maximum value of sin2x is 1. Hence,

y = f x = 1 2 - sin 2 x > 1 y = f x = 1 2 - sin 2 x > 1

This means given function is positive for all real x. Now, solving for x,

2 y y sin 2 x = 1 2 y y sin 2 x = 1 sin 2 x = 2 y 1 y sin 2 x = 2 y 1 y x = 1 2 sin - 1 2 y - 1 y x = 1 2 sin - 1 2 y - 1 y

We know that domain of sine inverse function is [-1,1]. Hence,

- 1 2 y - 1 y 1 - 1 2 y - 1 y 1

Since y>0, we can simplify this inequality as :

- y 2 y - 1 y - y 2 y - 1 y

Either,

2 y - 1 - y 2 y - 1 - y y 1 3 y 1 3

Or,

2 y - 1 y 2 y - 1 y y 1 y 1

Range = [ 1 3 , 1 ] Range = [ 1 3 , 1 ]

### Exercise 4

Find domain of function

f x = sin - 1 { log 2 x 2 + 3 x + 4 } f x = sin - 1 { log 2 x 2 + 3 x + 4 }

#### Solution

This is a composite function in which quadratic function is argument of logarithmic function. The logarithmic function is, in turn, argument of inverse sine function. In such case, it is advantageous to evaluate from outer to inner part. The domain of outermost inverse trigonometric function is [-1,1].

- 1 { log 2 x 2 + 3 x + 4 } 1 - 1 { log 2 x 2 + 3 x + 4 } 1 log 2 2 - 1 { log 2 x 2 + 3 x + 4 } log 2 2 log 2 2 - 1 { log 2 x 2 + 3 x + 4 } log 2 2 1 2 x 2 + 3 x + 4 2 1 2 x 2 + 3 x + 4 2

For the first inequality,

2 x 2 + 6 x + 8 1 2 x 2 + 6 x + 8 1 2 x 2 + 6 x + 7 0 2 x 2 + 6 x + 7 0

This quadratic function is positive for all value of x. For the second inequality,

x 2 + 3 x + 4 2 x 2 + 3 x + 4 2 x 2 + 3 x + 2 0 x 2 + 3 x + 2 0

The solution of this inequality is [1,2]. The intersection of R and [1,2] is [1,2]. Hence, domain of given function is [1,2].

### Exercise 5

Find the range of the function

f x = cos - 1 x 2 1 + x 2 f x = cos - 1 x 2 1 + x 2

#### Solution

Hint : The range of rational expression as argument of inverse trigonometric function is [0,1]. But, domain of arccosine is [-1,1] and range is [0,π]. The function is continuously decreasing. The maximum and minimum values are 0 and1 (see arccosine graph). Hence, range of given function is [0, π/2].

### Exercise 6

Find the range of the function

f x = cosec - 1 [ 1 + sin 2 x ] f x = cosec - 1 [ 1 + sin 2 x ]

where [.] denotes greatest integer function.

#### Solution

The minimum and maximum value of sin 2 x sin 2 x is 0 and 1. Hence, range of 1 + sin 2 x 1 + sin 2 x is defined in the interval given by :

1 1 + sin 2 x 2 1 1 + sin 2 x 2

The corresponding values returned by GIF are 1 and 2. It means :

[ 1 + sin 2 x ] = { 1,2 } [ 1 + sin 2 x ] = { 1,2 }

But domain of arccosecant is [-π/2, π/2] – {0}. Refer graph of arccosecant. Thus, arccosecant can take only 1 as its argument, which falls within the domain of arccosecant. Hence, range of given function is a singleton :

Range = { cosec - 1 1 } Range = { cosec - 1 1 }

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