Inverse Trigonometric Functions Class 12 Notes Maths Chapter 2 - CBSE

Chapter:2

What are Inverse Trigonometric Functions ?

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Inverse Function

Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f(a)=b. The inverse function of f is denoted by f ^-1. Hence, f ^-1(b)=a when f(a)=b. Inverse trigonometric functions are simply defined as the inverse functions of basic trigonometric functions.

Domains And Ranges Of Inverse Trigonometric Functions

The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:

Functions	Domain	Range (Principal Value Branches)
y = sin ^-1 x	[-1,1]	[—ℼ/2 , ℼ/2]
y = cos ^-1 x	[-1,1]	[0, ℼ]
y = cosec ^-1 x	R-(-1,1)	[—ℼ/2 , ℼ/2] - {0}
y = sec ^-1 x	R-(-1,1)	[0, ℼ] - {ℼ/2}
y = tan ^-1 x	R	(—ℼ/2 , ℼ/2)
y = cot ^-1 x	R	(0, ℼ)

Domain of inverse trigonometry function = Range of trigonometry function.

Graphs Of Inverse Trigonometric Functions

(i) Inverse sine function

In,[—ℼ/2 , ℼ/2], sinx is bijective hence its inverse is y = sin^-1 x, x∈[-1,1] and y ∈ [—ℼ/2 , ℼ/2].

(ii) Inverse cosine function

In, [0,ℼ] cosine function is bijective and hence its inverse is y = cos^-1 x, x ∈[-1,1] and y ∈ [0,∈].

(iii) Inverse tangent function

Tangent function is one-one and onto from (-ℼ/2 ,ℼ/2) to (-∞,∞) and so in this region the tangent function is invertible i.e., y = tan^-1 x, x ∈[-∞,∞] and y ∈ (-ℼ/2 ,ℼ/2)

(iv) Inverse cotangent function

Cotangent function is one-one and onto from (0,ℼ) to (-∞,∞) hence cot x is invertible in this region i.e., y = cot^-1 x; x ∈[-∞,∞] and y ∈ (0,ℼ)

(v) Inverse secant function

A function f : = [0, -ℼ/2) U (-ℼ/2, ℼ] → (-∞,-1] U [1,∞) defined by f(x) = sec x is one-one and onto and hence it is invertible in this region, hence y = sec^-1 x; x∈(-∞,-1] U [1,∞) and y ∈[0, ℼ/2) U (ℼ/2, ℼ]

(vi) Inverse cosecant function

A function f : = [- ℼ/2, 0) U (0, ℼ/2] → (-∞,-1] U [1,∞) defined by
f(x) = cosec x is one-one and onto and hence it is invertible in this
region, hence y = cosec-1 x; x∈(-∞,-1] U [1,∞) and y ∈[- ℼ/2, 0) U (0, ℼ/2]