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

Composition of trigonometric function and its inverse

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

Trigonometric and inverse trigonometric functions are inverse to each other. We can use them to compose new functions. In such composition, trigonometric function represents value of trigonometric ratio, whereas inverse trigonometric function represents angle. The composite function either evaluates to value or angle, depending on particular composition.

Composition representing value of trigonometric function

Sine inverse trigonometric function is given by :

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

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

The composition sin sin - 1 x sin sin - 1 x evaluates to a value. Clearly, x is a value of sine trigonometric function which falls within the range of sine function i.e x [ - 1,1 ] x [ - 1,1 ] . It is important to note that domain of inverse function is same as range of the corresponding trigonometric function. We write six compositions denoting value of trigonometric functions as :

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

cos cos - 1 x = x ; x [ - 1,1 ] cos cos - 1 x = x ; x [ - 1,1 ]

tan tan - 1 x = x ; x R tan tan - 1 x = x ; x R

cot cot - 1 x = x ; x R cot cot - 1 x = x ; x R

sec sec - 1 x = x ; x , 1 ] [ 1, sec sec - 1 x = x ; x , 1 ] [ 1,

cosec cosec - 1 x = x ; x , 1 ] [ 1, cosec cosec - 1 x = x ; x , 1 ] [ 1,

Composition representing angle

We shall discuss this composition with respect to individual inverse trigonometric ratio.

Composition with arcsine

Sine inverse trigonometric function is given by :

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

In order to maintain generality, we replace y by x as :

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

The composition sin - 1 sin x sin - 1 sin x evaluates to an angle. Clearly, x is angle value – not the value of trigonometric ratio. However, we know that we use a truncated domain of trigonometric function for defining range of inverse function. The values in the interval are selected such that all unique values of sine trigonometric function are represented. It means that expression on LHS of the equation i.e. sin - 1 sin x sin - 1 sin x evaluates to angle values lying in the interval [ - π / 2, π / 2 ] [ - π / 2, π / 2 ] .

sin - 1 sin x = x ; x [ - π 2 , π 2 ] sin - 1 sin x = x ; x [ - π 2 , π 2 ]

However, x as argument of sine function can assume angle values belonging to real number set. It means angles represented by LHS and RHS can be different if we consider angle values beyond principal set selected to render corresponding trigonometric function invertible.

Let us consider adjacent intervals such that all sine values are included once. Such intervals are [ π / 2, 3 π / 2 ] , [ 3 π / 2, 5 π / 2 ] [ π / 2, 3 π / 2 ] , [ 3 π / 2, 5 π / 2 ] etc on the right side and [ - 3 π / 2, - π / 2 ] , [ - 5 π / 2, - 3 π / 2 ] [ - 3 π / 2, - π / 2 ] , [ - 5 π / 2, - 3 π / 2 ] etc on the left side of the principal interval.

Our task now is to determine angles in any of these new intervals, say [ π / 2, 3 π / 2 ] [ π / 2, 3 π / 2 ] , corresponding to angles in the principal interval. We make use of value diagram which allows to determine angles having same trigonometric values. Let us consider a positive acute angle “θ” in the principal interval. This lies in the first quadrant. The new interval represents second and third quadrants. However, sine is positive in second quadrant and negative in third quadrant. Let the angle corresponding to positive acute angle in principal interval be x. Clearly, x corresponding to positive acute angle θ lies in second quadrant and is given by :

x = π θ x = π θ

θ = π x θ = π x

Hence,

sin - 1 sin x = π x ; x [ π 2 , 3 π 2 ] sin - 1 sin x = π x ; x [ π 2 , 3 π 2 ]

In order to find expression corresponding to negative angle interval [ - 3 π / 2, - π / 2 ] [ - 3 π / 2, - π / 2 ] , we need to construct negative value diagram. We know that equivalent negative angle is obtained by deducting “-2π” to the positive angle. Thus, corresponding to expression for positive angles in four quadrants, the expression in terms of negative angles are “-θ”,“-π+θ”,“-π-θ” and “-2π+θ” in four quadrants counted in clockwise direction in the value diagram. Now, we estimate from the sine plot that an angle, corresponding to a positive acute angle, θ, in the principal interval, lies in third negative quadrant. Therefore,

x = - π θ x = - π θ

θ = - π x θ = - π x

Hence,

sin - 1 sin x = - π x ; x [ - 3 π 2 , π 2 ] sin - 1 sin x = - π x ; x [ - 3 π 2 , π 2 ]

Combining three results,


|-π-x;   x∈ [-3π/2, -π/2]
sin⁻¹ sinx  = | x;     x∈ [-π/2, π/2]
| π- x;  x∈ [π/2, 3π/2]


We can similarly find expressions for more such intervals.

Graph of sin⁻¹sinx

Using three expressions obtained above, we can draw plot of the composition function. We extend the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :

y = x 2 π y = x 2 π

The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :

y = x + 2 π y = x + 2 π

We see that graph of composition is continuous. Its domain is R. Its range is [ - π / 2, π / 2 ] [ - π / 2, π / 2 ] . The function is periodic with period 2π.

Composition with arccosine

The composition cos - 1 cos x cos - 1 cos x evaluates to angle values lying in the interval [0, π].

cos - 1 cos x = x ; x [ 0, π ] cos - 1 cos x = x ; x [ 0, π ]

Let us consider adjacent intervals such that all cosine values are included once. Such intervals are [π, 2π], [2π, 3π] etc on the right side and [-π, 0], [-2π, -π] etc on the left side of the principal interval.

The new interval [ π , 2 π ] [ π , 2 π ] represents third and fourth quadrants. The angle x, corresponding to positive acute angle θ, lies in fourth quadrant. Then,

x = 2 π θ x = 2 π θ

θ = 2 π x θ = 2 π x

Hence,

cos - 1 cos x = 2 π x ; x [ π , 2 π ] cos - 1 cos x = 2 π x ; x [ π , 2 π ]

In order to find expression corresponding to negative angle interval [ - π , 0 ] [ - π , 0 ] , we estimate from the cosine plot that an angle corresponding to a positive acute angle, θ, in the principal interval lies in first negative quadrant. Therefore,

x = - θ x = - θ

θ = x θ = x

Hence,

cos - 1 cos x = - x ; x [ π , 0 ] cos - 1 cos x = - x ; x [ π , 0 ]

Combining three results,


|-x;     x∈ [-π, 0]
cos⁻¹ cosx  = | x;     x∈[0, π]
|2π- x;  x∈ [π, 2π]


We can similarly find expressions for other intervals.

Graph of cos⁻¹cosx

Using three expressions obtained above, we can draw plot of the composition function. We have extended the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :

y = x 2 π y = x 2 π

The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :

y = x + 2 π y = x + 2 π

We see that graph of composition is continuous. Its domain is R. Its range is [ 0, π ] [ 0, π ] . The function is periodic with period 2π.

Composition with arctangent

The composition tan - 1 tan x tan - 1 tan x evaluates to angle values lying in the interval - π / 2, π / 2 - π / 2, π / 2 .

tan - 1 tan x = x ; x [ π 2, π 2 ] tan - 1 tan x = x ; x [ π 2, π 2 ]

Let us consider adjacent intervals such that all tangent values are included once. Such intervals are (π/2, 3π/2), (3π/2, 5π/2) etc on the right side and (-3π/2, -π/2), (-5π/2, -3π/2) etc on the left side of the principal interval.

The new interval π / 2, 3 π / 2 π / 2, 3 π / 2 represents second and third quadrants. The angle x, corresponding to positive acute angle θ, lies in third quadrant. Then,

x = π + θ x = π + θ

θ = x π θ = x π

Hence,

tan - 1 tan x = x π ; x [ π 2 , 3 π 2 ] tan - 1 tan x = x π ; x [ π 2 , 3 π 2 ]

In order to find expression corresponding to negative angle interval - 3 π / 2, - π / 2 - 3 π / 2, - π / 2 , we estimate from the tangent plot that an angle corresponding to a positive acute angle, θ, in the principal interval lies in second negative quadrant. Therefore,

x = - π + θ x = - π + θ

θ = x + π θ = x + π

Hence,

tan - 1 tan x = x + π ; x [ 3 π 2, π 2 ] tan - 1 tan x = x + π ; x [ 3 π 2, π 2 ]

Combining three results,


| x+π;   x∈ (-3π/2, -π/2)
tan⁻¹ tanx  = | x;     x∈ (-π/2, π/2)
| x-π;   x∈ (π/2, 3π/2)


We can similarly find expressions for other intervals.

Graph of tan⁻¹tanx

Using three expressions obtained above, we can draw plot of the composition function. We have extended the plot, using the fact that composition is a periodic function with a period of π. The equation of plot, which is equivalent to plot y=x shifted by π towards right is :

y = x π y = x π

The equation of plot, which is equivalent to plot y=x shifted by π towards left is :

y = x + π y = x + π

These results are same as obtained earlier. It means that nature of plot is same in the adjacent intervals.

We see that graph of composition is discontinuous. Its domain is R { 2 n + 1 π / 2 ; n Z } R { 2 n + 1 π / 2 ; n Z } . Its range is [ - π / 2, π / 2 ] [ - π / 2, π / 2 ] . The function is periodic with period π.

Composition with arccosecant

The composition cosec - 1 cosec x cosec - 1 cosec x evaluates to angle values lying in the interval [ - π / 2, π / 2 ] { 0 } [ - π / 2, π / 2 ] { 0 } .

cosec - 1 cosec x = x ; x [ π 2, π 2 ] { 0 } cosec - 1 cosec x = x ; x [ π 2, π 2 ] { 0 }

Let us consider adjacent intervals such that all cosine values are included once. Such intervals are [π/2, 3π/2] – {π}, [3π/2, 5π/2] – {2π} etc on the right side and [-3π/2, -π/2] – {-π}, [-5π/2, -3π/2] – {-2π} etc on the left side of the principal interval.

The new interval [ π / 2, 3 π / 2 ] { π } [ π / 2, 3 π / 2 ] { π } lies in second and third quadrants. The angle x corresponding to positive acute angle θ, lies in second quadrant. Then,

x = π θ x = π θ

θ = π x θ = π x

Hence,

cosec - 1 cosec x = x ; π x [ π 2 , 3 π 2 ] { π } cosec - 1 cosec x = x ; π x [ π 2 , 3 π 2 ] { π }

In order to find expression corresponding to negative angle interval [ - 3 π / 2, - π / 2 ] { - π } [ - 3 π / 2, - π / 2 ] { - π } , we estimate from the cosecant plot that an angle corresponding to a positive acute angle, θ, in the principal interval lies in third negative quadrant. Therefore,

x = - π - θ x = - π - θ

θ = - π - x θ = - π - x

Hence,

cosec - 1 cosec x = π - x ; x [ 3 π 2 , π 2 ] { π } cosec - 1 cosec x = π - x ; x [ 3 π 2 , π 2 ] { π }

Combining three results,


|- π-x;   x∈[-3π/2, -π/2] – {-π}
cosec⁻¹ cosecx  = | x;      x∈[-/2, π/2]-{0}
| π- x;   x∈[π/2, 3π/2] – {π}


We can similarly find expressions for other intervals.

Graph of cosec⁻¹cosecx

Using three expressions obtained above, we can draw plot of the composition function. We have extended the plot, using the fact that composition is a periodic function with a period of 2π. The equation of plot, which is equivalent to plot y=x shifted by 2π towards right, is :

y = x - 2 π y = x - 2 π

The equation of plot, which is equivalent to plot y=x shifted by 2π towards left, is :

y = x + 2 π y = x + 2 π

We see that graph of composition is discontinuous. Its domain is R { n π ; n Z } R { n π ; n Z } . Its range is [ - π / 2, π / 2 ] { 0 } [ - π / 2, π / 2 ] { 0 } . The function is periodic with period 2π.

Note:

We can similarly find out expressions for different intervals for arcsecant and arccotangent compositions. We have left out discussion of these two functions as exercise.

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