Inverse Trigonometric Functions
Complete Class XII Mathematics chapter notes covering all six inverse trigonometric functions, principal values, domains, ranges, graphs, identities, transformations, solved examples and exam-oriented practice.
Chapter Foundation
Introduction
An inverse trigonometric function gives the angle corresponding to a given trigonometric ratio. Since trigonometric functions are periodic and therefore not one-to-one over their complete domains, we restrict their domains to suitable intervals before defining their inverses.
Inverse does not mean reciprocal
sin⁻¹x means the inverse sine of x, not 1/sin x . The reciprocal of sin x is cosec x .
Inverse vs Reciprocal
A useful distinction before beginning the chapter.
Core Concept
Why Do We Need Restricted Domains?
One-to-One Requirement
A function has an inverse function only when the function is one-to-one on the selected domain. Trigonometric functions repeat their values, so suitable domains are selected.
Standard Restricted Domains
Restricted Domain Visual
Complete Reference
All Six Inverse Trigonometric Functions
Inverse Sine
If y = sin⁻¹x, then sin y = x.
Range: [-π/2,π/2]
Inverse Cosine
If y = cos⁻¹x, then cos y = x.
Range: [0,π]
Inverse Tangent
If y = tan⁻¹x, then tan y = x.
Range: (-π/2,π/2)
Inverse Cotangent
If y = cot⁻¹x, then cot y = x.
Range: (0,π)
Inverse Secant
If y = sec⁻¹x, then sec y = x.
Range: [0,π] \ {π/2}
Inverse Cosecant
If y = cosec⁻¹x, then cosec y = x.
Range: [-π/2,π/2] \ {0}
Essential for CBSE
Principal Values
| Function | Domain | Principal Value 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 | (-∞,-1] ∪ [1,∞) | [0,π] \ {π/2} |
| cosec⁻¹x | (-∞,-1] ∪ [1,∞) | [-π/2,π/2] \ {0} |
Principal-Value Angle Map
Input & Output
Domain and Range
Domain
The domain tells us which values of x can be supplied to the inverse function.
Range
The range tells us the principal angle returned by the inverse function.
Remember
Inverse trigonometric answers must lie inside the corresponding principal-value range.
Visual Learning
Graphs of Inverse Trigonometric Functions
Graph of y = sin⁻¹x
Domain: [-1,1] | Range: [-π/2,π/2]
Graph of y = cos⁻¹x
Domain: [-1,1] | Range: [0,π]
Graph of y = tan⁻¹x
Domain: R | Range: (-π/2,π/2)
Inverse Functions and Reflection in y = x
The graph of an inverse function is the reflection of the original function's restricted graph in the line y = x.
Formula Bank
Fundamental Identities
Direct Composition
Reverse Composition
Complementary Identities
Negative-Argument Identities
Working With Expressions
Trigonometric Conversion Using Triangles
If θ = sin⁻¹x
Then sin θ = x .
For tan θ, require |x| < 1.
If θ = cos⁻¹x
Then cos θ = x .
Right-Triangle Method for θ = tan⁻¹x
Take tan θ = x = x/1 .
Addition and Subtraction of tan⁻¹
Principal-value caution
The addition formula for tan⁻¹ may require an adjustment by π depending on the signs of x, y and the resulting quadrant. Always check the principal-value range before writing the final answer.
Step-by-Step
Solved Examples
Evaluate sin⁻¹(1/2)
Therefore,
sin⁻¹(1/2) = π/6
Evaluate cos⁻¹(−1/2)
Since 2π/3 ∈ [0,π] ,
cos⁻¹(−1/2) = 2π/3
Evaluate tan⁻¹1
Hence,
tan⁻¹1 = π/4
Prove sin⁻¹x + cos⁻¹x = π/2
θ = sin⁻¹x
Then
sin θ = x
Therefore,
cos(π/2 − θ) = x
Hence,
cos⁻¹x = π/2 − θ
Therefore,
sin⁻¹x + cos⁻¹x = π/2
Simplify sin(cos⁻¹x)
θ = cos⁻¹x
Then
cos θ = x
Using
sin²θ + cos²θ = 1
sin²θ = 1 − x²
Since θ ∈ [0,π],
sin θ ≥ 0.
sin(cos⁻¹x) = √(1−x²)
Simplify cos(sin⁻¹x)
θ = sin⁻¹x
Therefore,
sin θ = x
cos²θ = 1 − x²
Since θ ∈ [-π/2,π/2], cos θ ≥ 0.
cos(sin⁻¹x) = √(1−x²)
Simplify tan(sin⁻¹x)
Then
sin θ = x
cos θ = √(1−x²)
Therefore,
tan θ = sin θ / cos θ
tan(sin⁻¹x) = x / √(1−x²)
If θ = tan⁻¹(3/4), find sinθ and cosθ.
Take opposite = 3 and adjacent = 4.
Hypotenuse = √(3²+4²) = 5
Therefore,
sinθ = 3/5
cosθ = 4/5
Simplify sin(tan⁻¹x)
Then tanθ = x = x/1.
Hypotenuse = √(1+x²).
Hence,
sin(tan⁻¹x) = x/√(1+x²)
Simplify cos(tan⁻¹x)
Adjacent = 1
Hypotenuse = √(1+x²)
Therefore,
cos(tan⁻¹x) = 1/√(1+x²)
Evaluate tan⁻¹(1/2) + tan⁻¹(1/3)
= tan⁻¹ [ ((1/2)+(1/3)) / (1−(1/6)) ]
= tan⁻¹ [ (5/6)/(5/6) ]
= tan⁻¹1
= π/4
Evaluate tan⁻¹2 + tan⁻¹3
= −1
Both A and B are positive, so A+B lies between π/2 and π.
The angle in (π/2,π) having tangent −1 is 3π/4.
tan⁻¹2 + tan⁻¹3 = 3π/4
Evaluate tan⁻¹3 − tan⁻¹(1/2)
= tan⁻¹ [ (5/2)/(5/2) ]
= tan⁻¹1
= π/4
Find sin(cos⁻¹(−3/5))
sin(cos⁻¹x) = √(1−x²)
Put x = −3/5.
= √[1−9/25]
= √(16/25)
= 4/5
Find cos(sin⁻¹(−5/13))
Therefore,
= √[1−25/169]
= √(144/169)
= 12/13
Simplify sin⁻¹(sin 5π/6)
Hence,
sin⁻¹(sin 5π/6) = sin⁻¹(1/2)
= π/6
It is not 5π/6 because 5π/6 is outside the principal range of sin⁻¹x.
Simplify cos⁻¹(cos 5π/3)
Therefore,
cos⁻¹(cos 5π/3) = cos⁻¹(1/2)
= π/3
Solve sin⁻¹x = π/6
Taking sine on both sides,
x = sin(π/6)
x = 1/2
Self Assessment
Practice Questions
1
State the domain and range of sin⁻¹x.
2
State the domain and range of cos⁻¹x.
3
State the domain and range of tan⁻¹x.
4
Evaluate sin⁻¹(√3/2).
5
Evaluate cos⁻¹(−√3/2).
6
Evaluate tan⁻¹(−1).
7
Prove sin⁻¹x + cos⁻¹x = π/2.
8
Simplify cos(sin⁻¹x).
9
Simplify sin(cos⁻¹x).
10
Simplify sin(tan⁻¹x).
11
Simplify cos(tan⁻¹x).
12
If θ = sin⁻¹(3/5), find cosθ.
13
If θ = tan⁻¹(5/12), find sinθ.
14
Evaluate tan⁻¹(1/2) + tan⁻¹(1/3).
15
Evaluate tan⁻¹2 + tan⁻¹3.
16
Evaluate tan⁻¹3 − tan⁻¹(1/2).
17
Simplify sin⁻¹(sin 3π/4).
18
Simplify cos⁻¹(cos 4π/3).
19
Why does sin⁻¹(sin x)=x need a restriction?
20
State the horizontal asymptotes of y = tan⁻¹x.
Higher Order Thinking
HOT Questions
Prove that sin⁻¹x + cos⁻¹x = π/2.
View Hint
Put θ = sin⁻¹x and use cos(π/2−θ)=sinθ.
Evaluate: tan⁻¹(1/2)+tan⁻¹(1/3)
Use the tan addition formula and then identify the principal angle.
Evaluate: tan⁻¹2 + tan⁻¹3
Since both angles are positive and their sum lies in the second quadrant, the principal result is:
Simplify: sin(cos⁻¹(−4/5))
Simplify: cos(sin⁻¹(−12/13))
If θ = sin⁻¹x, derive tanθ.
Explain geometrically why the graph of an inverse function is reflected in y = x.
The coordinates of a function and its inverse interchange: (x,y) becomes (y,x). Reflection in y=x performs exactly this interchange.
Find sin⁻¹(sin 7π/6).
Therefore answer = −π/6.
Find cos⁻¹(cos 5π/3).
Answer = π/3.
If tan⁻¹x + tan⁻¹y = π/4 for positive x,y, derive a relation between x and y.
x+y = 1−xy
Last-Minute Revision
Quick Revision Sheet
| 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 | |x| ≥ 1 | [0,π]\{π/2} |
| cosec⁻¹x | |x| ≥ 1 | [-π/2,π/2]\{0} |
Exam Strategy
Important Exam Tips
Never confuse inverse and reciprocal
sin⁻¹x is inverse sine. The reciprocal of sin x is cosec x.
Check principal values
Always make sure the final angle belongs to the correct principal-value interval.
Be careful with sin⁻¹(sin x)
It equals x directly only when x lies in [-π/2,π/2].
Use the correct range for cos⁻¹
The principal range of cos⁻¹x is [0,π].
Check the quadrant in tan⁻¹ addition
Do not stop after applying the tangent addition formula. Identify the correct principal angle.
Use radians
CBSE-style mathematical answers normally express angles in radians unless the question specifies otherwise.
Before submitting your answer
Master Inverse Trigonometric Functions
Revise all six inverse trigonometric functions, principal values, domains, ranges, graphs, identities and application-based problems before moving to the next chapter.