Graphs of Inverse Trigonometric Functions-Formulas, Solved Examples, Class 12 Math Chapter 2 Notes Study Material Download free pdf

Inverse trigonometric functions are the inverse functions of the trigonometric ratios i.e. sin, cos, tan, cot, sec, cosec. These functions are widely used in fields like physics, mathematics, engineering and other research fields.

There are two popular notations used for inverse trigonometric functions:

Adding “arc” as a prefix.

Example: arcsin(x), arccos(x), arctan(x), …

Adding “-1” as superscript.

Example: sin^-1(x), cos^-1(x), tan^-1(x), …

In this article, we will learn about graphs and nature of various inverse functions.

Inverse of Sine Function, y = sin^-1(x)

sin^-1(x) is the inverse function of sin(x). Its domain is [−1, 1] and its range is [- π/2, π/2]. It intersects the coordinate axis at (0,0). It is an odd function and is strictly increasing in (-1, 1).

Graph of Function

Graph of Sin^-1(x)

Function Analysis

Domain	x∈[−1,1]x∈[−1,1]
Range	y∈[−π2,π2]y∈[2−π​,2π​]
X – Intercept	x=0x=0
Y – Intercept	y=0y=0
Minima	(−1,−π2)(−1,2−π​)
Maxima	(1,π2)(1,2π​)
Inflection Points	(0,0)(0,0)
Parity	Odd Function
Monotonicity	In (-1, 1) strictly increasing

Sample Problems on Inverse Sine Function

Problem 1: Find the principal value of the given equation:

y = sin^-1(1/√2)

Solution:

We are given that:

y = sin^-1(1/√2)

So we can say that,

sin(y) = (1/√2)

We know that the range of the principal value branch of sin^-1(x) is (−π/2, π/2) and sin(π/4) = 1/√2.

So, the principal value of sin^-1(1/√2) = π/4.

Problem 2: Find the principal value of the given equation:

y = sin^-1(1)

Solution:

We are given that:

y = sin^-1(1)

So we can say that,

sin(y) = 1

We know that the range of the principal value branch of sin^-1(x) is (−π/2, π/2) and sin(π/2) = 1.

So, the principal value of sin^-1(1) = π/2.

Inverse of Cosine Function, y = cos^-1(x)

cos^-1(x) is the inverse function of cos(x). Its domain is [−1, 1] and its range is [0, π]. It intersects the coordinate axis at (1, π/2). It is neither even nor an odd function and is strictly decreasing in (-1, 1).

Graph of Function

Graph of Cos^-1(x)

Function Analysis

Domain	x∈[−1,1]x∈[−1,1]
Range	y∈[0,π]y∈[0,π]
X – Intercept	x=1x=1
Y – Intercept	y=π2y=2π​
Minima	(1,0)(1,0)
Maxima	(−1,π)(−1,π)
Inflection Points	(0,π2)(0,2π​)
Parity	Neither Even Nor Odd
Monotonicity	In (-1, 1) strictly decreasing

Sample Problems on Inverse Cosine Function

Problem 1: Find the principal value of the given equation:

y = cos^-1(1/√2)

Solution:

We are given that:

y = cos^-1(1/√2)

So we can say that,

cos(y) = (1/√2)

We know that the range of the principal value branch of cos^-1(x) is (0, π) and cos(π/4) = 1/√2.

So, the principal value of cos^-1(1/√2) = π/4.

Problem 2: Find the principal value of the given equation:

y = cos^-1(1)

Solution:

We are given that:

y = cos^-1(1)

So we can say that,

cos(y) = 1

We know that the range of the principal value branch of cos^-1(x) is (0, π) and cos(0) = 1.

So, the principal value of cos^-1(1) = 0.

Inverse of Tangent Function, y = tan^-1(x)

tan^-1(x) is the inverse function of tan(x). Its domain is ℝ and its range is [-π/2, π/2]. It intersects the coordinate axis at (0, 0). It is an odd function which is strictly increasing in (-∞, ∞).

Graph of Function

Graph of tan^-1(x)

Function Analysis

Domain	x∈Rx∈R
Range	y∈(−π2,π2)y∈(2−π​,2π​)
X – Intercept	x=0x=0
Y – Intercept	y=0y=0
Minima	The function does not have any minima points.
Maxima	The function does not have any maxima points.
Inflection Points	(0,0)(0,0)
Parity	Odd Function
Monotonicity	In (−∞, ∞) strictly Increasing
Asymptotes	y=π2 and y=−π2y=2π​ and y=2−π​

Sample Problems on Inverse of Tangent Function

Problem 1: Find the principal value of the given equation:

y = tan^-1(1)

Solution:

We are given that:

y = tan^-1(1)

So we can say that,

tan(y) = (1)

We know that the range of the principal value branch of tan^-1(x) is (-π/2, π/2) and tan(π/4) = 1.

So, the principal value of tan^-1(1) = π/4.

Problem 2: Find the principal value of the given equation:

y = tan^-1(√3)

Solution:

We are given that:

y = tan^-1(√3)

So we can say that,

tan(y) = (√3)

We know that the range of the principal value branch of tan-1(x) is (-π/2, π/2) and tan(π/3) = √3.

So, the principal value of tan-1(√3) = π/3.

Inverse of Cosecant Function, y = cosec^-1(x)

cosec^-1(x) is the inverse function of cosec(x). Its domain is (-∞, -1] U [1, ∞) and its range is [-π/2, 0) U (0, π/2]. It doesn’t intercept the coordinate axis. It is an odd function that is strictly decreasing in its domain.

Graph of Function

Graph of Cosec^-1(x)

Function Analysis

Domain	x∈(−∞,−1]∪[1,∞)x∈(−∞,−1]∪[1,∞)
Range	y∈[−π2,0)∪(0,π2]y∈[2−π​,0)∪(0,2π​]
X – Intercept	ϕϕ
Y – Intercept	ϕϕ
Minima	(−1,−π2)(−1,2−π​)
Maxima	(1,π2)(1,2π​)
Inflection Points	The function does not have any inflection points.
Parity	Odd Function
Monotonicity	In (1, ∞) it is decreasing and in (-∞, -1) it is decreasing
Asymptotes	y = 0

Sample Problems on Inverse Cosecant Function

Problem 1: Find the principal value of the given equation:

y = cosec^-1(√2)

Solution:

We are given that:

y = cosec^-1(√2)

So we can say that,

cosec(y) = (√2)

We know that the range of the principal value branch of cosec^-1(x) is [-π/2, π/2] – {0} and cosec(π/4) = √2.

So, the principal value of cosec^-1(√2) = π/4.

Problem 2: Find the principal value of the given equation:

y = cosec^-1(1)

Solution:

We are given that:

y = cosec^-1(√2)

So we can say that,

cosec(y) = 1

We know that the range of the principal value branch of cosec^-1(x) is [-π/2, π/2] – {0} and cosec(π/2) = 1.

So, the principal value of cosec^-1(1) = π/2.

Inverse of Secant Function, y = sec^-1(x)

sec^-1(x) is the inverse function of sec(x). Its domain is (-∞, -1] U [1, ∞) and its range is [0, π/2) U (π/2, π]. It doesn’t intercept the coordinate axis as it is a discontinuous function. It is neither even nor odd function and is strictly increasing in its domain.

Graph of Function

Graph of Sec^-1(x)

Function Analysis

Domain	x∈(−∞,−1]∪[1,∞)x∈(−∞,−1]∪[1,∞)
Range	y∈[0,π2)∪(π2,π]y∈[0,2π​)∪(2π​,π]
X – Intercept	x=1x=1
Y – Intercept	ϕϕ
Minima	(1,0)(1,0)
Maxima	(−1,π)(−1,π)
Inflection Points	The function does not have any inflection points.
Parity	Neither Even Nor Odd
Monotonicity	In (1, ∞) it is increasing and in (-∞, -1) it is increasing
Asymptotes	y=π2y=2π​

Sample Problems on Inverse of Secant Function

Problem 1: Find the principal value of the given equation:

y = sec^-1(√2)

Solution:

We are given that:

y = sec^-1(√2)

So we can say that,

sec(y) = (√2)

We know that the range of the principal value branch of sec^-1(x) is [0, π] – {π/2} and sec(π/4) = √2.

So, the principal value of sec^-1(√2) = π/4.

Problem 2: Find the principal value of the given equation:

y = sec^-1(1)

Solution:

We are given that:

y = sec^-1(1)

So we can say that,

sec(y) = 1

We know that the range of the principal value branch of sec^-1(x) is [0, π] – {π/2} and sec(0) = 1.

So, the principal value of sec^-1(1) = 0.

Inverse of Cotangent Function, y = cot^-1(x)

cot^-1(x) is the inverse function of cot(x). Its domain is ℝ and its range is (0, π). It intersects the coordinate axis at (0, π/2). It is neither even nor odd function and is strictly decreasing in its domain.

Graph of Function

Graph of Cot^-1(x)

Function Analysis

Domain	x∈Rx∈R
Range	y∈(0,π)y∈(0,π)
X – Intercept	x = null
Y – Intercept	y=π2y=2π​
Minima	The function does not have any minima points.
Maxima	The function does not have any maxima points.
Inflection Points	The function does not have any inflection points.
Parity	Neither Even Nor Odd
Monotonicity	In (-∞, ∞) strictly decreasing
Asymptotes	y=0 and y=πy=0 and y=π

Sample Problems on Inverse of Cotangent Function

Problem 1: Find the principal value of the given equation:

y = cot^-1(1)

Solution:

We are given that:

y = cot^-1(1)

So we can say that,

cot(y) = 1

We know that the range of the principal value branch of cot^-1(x) is (-π/2, π/2) and cot(π/4) = 1.

So, the principal value of cot^-1(1) = π/4.

Problem 2: Find the principal value of the given equation:

y = cot^-1(1/√3)

Solution:

We are given that:

y = cot^-1(1/√3)

So we can say that,

cot(y) = (1/√3)

We know that the range of the principal value branch of cot^-1(x) is (-π/2, π/2) and cot(π/3) = 1/√3.

So, the principal value of cot^-1(1/√3) = π/3.