Inverse trigonometric functions, also called arc functions, reverse the operations of the basic six trigonometric functions – sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec), and cotangent (cot).
If y = g(x) and x = p(y) are two functions, then p = g-1 is the inverse function.
Thus, for the function y = g(x), g = y-1(x)
These functions are represented by adding the prefix ‘arc’ or -1 to the superscript of the basic functions.
For example, the inverse of the sine function (y = sin x) is sin-1 x or arcsin (x), where θ is the reference angle.
Here is the list of inverse trigonometric functions corresponding to the six trigonometric functions:
Trigonometric Functions
Inverse
x = sin y
y = arcsin x or sin-1 x, where ${-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}}$
x = cos y
y = arccos x or cos-1 x, where 0 ≤ y ≤ π
x = tan y
y = arctan x or tan-1 x, where ${-\dfrac{\pi }{2} <y <\dfrac{\pi }{2}}$
x = sec y
y = arcsec x or sec-1 x, where 0 ≤ y < π and y ≠ ${\dfrac{\pi }{2}}$
x = cosec y
y = arccosec x or cosec-1 x, where ${-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}}$ and y ≠ 0
x = cot y
y = arccot x or cot-1 x, where 0 < y < π
As we know, the basic trigonometric functions are used to determine the length of an unknown side in a right-angle triangle when given one side length and the measure of an angle. In contrast, inverse trigonometric functions help us to find an unknown angle of a right-angle triangle when one side and one angle are known.
Arcsine (arcsin or sin-1 x)
The arcsine function finds the angle y such that siny = x and ${-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}}$
Arccosine (arccos or cos-1 x)
The arccosine function finds the angle y such that cosy = x where 0 ≤ y ≤ π
Arctangent (arctan or tan-1 x)
The arctangent function finds the angle y such that tany = x where ${-\dfrac{\pi }{2} <y <\dfrac{\pi }{2}}$
Arcsecant (arcsec or sec-1 x)
The arcsecant function finds the angle y such that secy = x where 0 ≤ y < π and y ≠ ${\dfrac{\pi }{2}}$
Arccosecant (arccosec or cosec-1 x)
The arccosecant function finds the angle y such that cosec y = x where ${-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}}$ and y ≠ 0
Arccotangent (arccot or cot-1 x)
The arccotangent function finds the angle y such that coty = x where 0 < y < π
Identities
Arbitrary Values
For the inverse trigonometric functions sine, tangent, and cosecant, negative input values result in the negatives of their corresponding function outputs. On the other hand, for cosine, secant, and cotangent, negative input values are expressed as the function value subtracted from π.
sin-1 (-x) = -sin-1 x, x ∈ [-1, 1]
cos-1 (-x) = π – cos-1 x, x ∈ [-1, 1]
tan-1 (-x) = -tan-1 x, x ∈ ℝ
sec-1 (-x) = π – sec-1 x, x ∈ ℝ – (-1, 1)
cosec-1 (-x) = -cosec-1 x, x ∈ ℝ – (-1, 1)
cot-1 (-x) = π – cot-1 x, x ∈ ℝ
Reciprocal Functions
For reciprocal values of ${\dfrac{1}{x}}$, the inverse trigonometric function is converted into its corresponding reciprocal function.
Trigonometric functions are inherently periodic, which prevents their graphs from passing the Horizontal Line Test. As a result, they are not inherently one-to-one functions, and their inverses cannot exist without restricting their domains.
By carefully selecting specific domains, however, these functions can be made one-to-one, allowing the entire range to be utilized for their inverses.
Arcsine (y = sin-1 x)
The graph of arcsine is increasing and has symmetry about the origin (0, 0).
The unit circle is often used to define inverse trigonometric functions. It gives a geometric representation of angles and their corresponding trigonometric values, which help us to define the domains and ranges of these functions. In the unit circle, each point (x, y) corresponds to (cos θ, sin θ), where θ is the angle measured from the positive x-axis.
Given below is a unit circle:
Unit Circle Chart
Here,
The x-coordinate is for cosine (cos-1)
The y-coordinate is for sine (sin-1)
The ratios of the coordinates are for tangent (tan-1)
Solved Examples
Find the principal value of ${\sin ^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)}$
Solution:
Here, ${\sin ^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)}$ ⇒ sinx = ${-\dfrac{\sqrt{3}}{2}}$ The range of the principal value of sin-1x is ${\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]}$ Thus, the principal value of ${\sin ^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)}$ is ${-\dfrac{\pi }{3}}$
Find the value of ${\tan ^{-1}\left( 1\right) +\cos ^{-1}\left( -\dfrac{1}{2}\right)}$
Solution:
As we know, tanx = 1 ⇒ x = ${\dfrac{\pi }{4}}$ Since the range of cos-1x is [0, π] ⇒ cos(x) = -${\dfrac{1}{2}}$ ⇒ x = ${\dfrac{2\pi }{3}}$ Here, ${\tan ^{-1}\left( 1\right) +\cos ^{-1}\left( -\dfrac{1}{2}\right)}$ = ${\dfrac{\pi }{4}+\dfrac{2\pi }{3}}$ = ${\dfrac{11\pi }{12}}$
If tan(tan-1 x + cot-1 x) = 1, find the value of x
Solution:
As we know, tan-1 x + cot-1 x = ${\dfrac{\pi }{2}}$, x ∈ ℝ Here, tan(tan-1 x + cot-1 x) = 1 ⇒ tan(tan-1x + cot-1 x) = ${\tan \left( \dfrac{\pi }{2}\right)}$ Since the tangent of ${\dfrac{\pi }{2}}$ Thus, x is undefined.
Solve cos(tan-1 x)
Solution:
As we know, cos(tan-1 x) = cos θ Let tan-1 x = θ ⇒ tan θ = x ⇒ sec θ = ${\sqrt{1+\tan ^{2}\theta }}$ ⇒ sec θ = ${\sqrt{1+x^{2}}}$ ⇒ cos θ = ${\dfrac{1}{\sqrt{1+x^{2}}}}$ Thus, cos(tan-1 x) = ${\dfrac{1}{\sqrt{1+x^{2}}}}$
Evaluate sin-1(cosec 45°)
Solution:
As we know, cosec 45° = ${\sqrt{2}}$ Here, sin-1(cosec 45°) = ${\sin ^{-1}\left( \sqrt{2}\right)}$ since |x| ≤ 1 for sin-1 x, this is undefined. Thus, sin-1(cosec 45°) is undefined.
