Class 12 Maths Chapter 2: Inverse Trigonometric Functions

Inverse Trigonometric Functions is Chapter 2 of NCERT Class 12 Maths. It explains why trigonometric functions need a restricted domain before they can be inverted, and then covers the domain, range, principal value branch and graph of each inverse trigonometric function. Use these notes, the summary of ranges and the solved examples to master principal values for your CBSE board exam.

What you will learn

Why sin x, cos x and the other trigonometric functions are not invertible on their full domain
The domain and principal value branch (range) of all six inverse trigonometric functions
How to find the principal value of expressions like sin⁻¹(−½) or sec⁻¹(−2)
How the graph of an inverse function is a reflection in the line y = x
How to simplify expressions such as sin⁻¹(sin θ) when θ lies outside the principal range

Key Concepts and Formulas

Why We Restrict the Domain

A function has an inverse only if it is one-one and onto. Trigonometric functions repeat their values, so they are many-one on R. By restricting each function to an interval where it is one-one and still takes all its values, we get an invertible function. The interval chosen in the CBSE syllabus is called the principal value branch.

sin⁻¹x = the angle whose sine is x
sin⁻¹x ≠ 1/sin x
(sin x)⁻¹ = 1/sin x = cosec x

The value of an inverse trigonometric function that lies in its principal branch is called its principal value.

Domain and Range of Each Function

Learn this list thoroughly; almost every question in this chapter depends on it.

Function   Domain        Range
sin⁻¹x     [−1, 1]       [−π/2, π/2]
cos⁻¹x     [−1, 1]       [0, π]
tan⁻¹x     R             (−π/2, π/2)
cot⁻¹x     R             (0, π)
sec⁻¹x     R − (−1, 1)   [0, π] − {π/2}
cosec⁻¹x   R − (−1, 1)   [−π/2, π/2] − {0}

Unit circle showing principal value branches of sin inverse and cos inverse for Class 12 Inverse Trigonometric Functions — Principal ranges of sin⁻¹x and cos⁻¹x on the unit circle

Finding Principal Values

To find the principal value of sin⁻¹a, ask: which angle in [−π/2, π/2] has sine equal to a? The same idea works for the others, using their own ranges.

sin(sin⁻¹x) = x,   for −1 ≤ x ≤ 1
sin⁻¹(sin θ) = θ,  only if −π/2 ≤ θ ≤ π/2
cos⁻¹(cos θ) = θ,  only if 0 ≤ θ ≤ π

For a negative input, sin⁻¹, tan⁻¹ and cosec⁻¹ give an angle in the fourth quadrant (a negative angle).
For a negative input, cos⁻¹, cot⁻¹ and sec⁻¹ give an angle in the second quadrant (between π/2 and π).

Graphs of Inverse Trigonometric Functions

The graph of y = sin⁻¹x is obtained by taking the graph of y = sin x on [−π/2, π/2] and reflecting it in the line y = x, that is, by interchanging the x and y axes.

Mirror line for all inverse graphs: y = x
Asymptotes of y = tan⁻¹x: y = π/2, y = −π/2

Graph of y equals sin inverse x as the reflection of the restricted sine curve in y equals x — y = sin⁻¹x is the mirror image of y = sin x

y = sin⁻¹x and y = tan⁻¹x are increasing functions.
y = cos⁻¹x and y = cot⁻¹x are decreasing functions.

Solved Examples

Example 1: Find the principal value of cos⁻¹(−√3/2).

Let θ = cos⁻¹(−√3/2), θ ∈ [0, π]
cos θ = −√3/2
cos(π/6) = √3/2
θ = π − π/6
∴ cos⁻¹(−√3/2) = 5π/6

Example 2: Find the principal value of tan⁻¹(−1/√3).

Let θ = tan⁻¹(−1/√3), θ ∈ (−π/2, π/2)
tan θ = −1/√3
tan(π/6) = 1/√3, so θ = −π/6
∴ tan⁻¹(−1/√3) = −π/6

Example 3: Find the principal value of cosec⁻¹(−√2).

Let θ = cosec⁻¹(−√2)
θ ∈ [−π/2, π/2], θ ≠ 0
cosec θ = −√2 ⇒ sin θ = −1/√2
θ = −π/4
∴ cosec⁻¹(−√2) = −π/4

Example 4: Find the principal value of sec⁻¹(−2).

Let θ = sec⁻¹(−2), θ ∈ [0, π], θ ≠ π/2
sec θ = −2 ⇒ cos θ = −½
cos(π/3) = ½
θ = π − π/3
∴ sec⁻¹(−2) = 2π/3

Example 5: Evaluate cot⁻¹(−√3) + tan⁻¹(1).

cot⁻¹(−√3): θ ∈ (0, π), cot θ = −√3
cot(π/6) = √3 ⇒ θ = π − π/6 = 5π/6
tan⁻¹(1) = π/4
Sum = 5π/6 + π/4
    = 10π/12 + 3π/12
∴ cot⁻¹(−√3) + tan⁻¹(1) = 13π/12

Example 6: Find the value of sin⁻¹(sin 4π/5).

4π/5 ∉ [−π/2, π/2], so answer ≠ 4π/5
sin(4π/5) = sin(π − 4π/5) = sin(π/5)
π/5 ∈ [−π/2, π/2]
sin⁻¹(sin 4π/5) = sin⁻¹(sin π/5)
∴ sin⁻¹(sin 4π/5) = π/5

Example 7: Find the domain of sin⁻¹(3x − 2).

−1 ≤ 3x − 2 ≤ 1
Add 2:        1 ≤ 3x ≤ 3
Divide by 3:  1/3 ≤ x ≤ 1
∴ Domain = [1/3, 1]

Common Mistakes to Avoid

Do not confuse sin⁻¹x with 1/sin x; the "−1" is not a power here.
Always check that your final angle lies in the principal range; writing 7π/6 for sin⁻¹(−½) is wrong, the correct answer is −π/6.
For cos⁻¹ of a negative number, students often write a negative angle. The range of cos⁻¹ is [0, π], so the answer must lie between 0 and π.
In questions like cos⁻¹(cos θ), first bring θ into [0, π] using the periodic and symmetry properties of cosine before removing the inverse.

Quick Revision

sin⁻¹:   [−1, 1] → [−π/2, π/2]
cos⁻¹:   [−1, 1] → [0, π]
tan⁻¹:   R → (−π/2, π/2)
cot⁻¹:   R → (0, π)
sec⁻¹:   R − (−1, 1) → [0, π] − {π/2}
cosec⁻¹: R − (−1, 1) → [−π/2, π/2] − {0}
sin⁻¹(sin θ) = θ only for θ ∈ [−π/2, π/2]