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Class 12 Science (PCM)

Mathematics Part 1

Inverse Trigonometric Functions

Inverse Trigonometric Functions

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Algebraic Simplification Using Inverse Trigonometric Substitution
Accuracy: 26%
Total Attempted27
Total Correct7
Total Incorrect20
Difficulty Level:
High
Branch-Condition Based Evaluation
Accuracy: 33%
Total Attempted45
Total Correct15
Total Incorrect30
Difficulty Level:
High
Inverse Trigonometric Equation Solving
Accuracy: 42%
Total Attempted12
Total Correct5
Total Incorrect7
Difficulty Level:
High
Multiple-Angle Identities with Inverse Trigonometric Functions
Accuracy: 45%
Total Attempted49
Total Correct22
Total Incorrect27
Difficulty Level:
High
Properties of Inverse Trigonometric Functions
Accuracy: 48%
Total Attempted48
Total Correct23
Total Incorrect25
Difficulty Level:
High
Need for Domain Restriction in Trigonometric Functions
Accuracy: 49%
Total Attempted63
Total Correct31
Total Incorrect32
Difficulty Level:
High
Graphical Representation of Inverse Functions
Accuracy: 50%
Total Attempted60
Total Correct30
Total Incorrect30
Difficulty Level:
Medium
Inverse Trigonometric Functions
Accuracy: 51%
Total Attempted95
Total Correct48
Total Incorrect47
Difficulty Level:
Medium
Principal Domain-Range Table of Inverse Trigonometric Functions
Accuracy: 51%
Total Attempted85
Total Correct43
Total Incorrect42
Difficulty Level:
Medium
Domain and Range of Inverse Functions
Accuracy: 51%
Total Attempted57
Total Correct29
Total Incorrect28
Difficulty Level:
Medium
Principal Value Branches
Accuracy: 56%
Total Attempted63
Total Correct35
Total Incorrect28
Difficulty Level:
Medium
Composite Function Identities
Accuracy: 57%
Total Attempted46
Total Correct26
Total Incorrect20
Difficulty Level:
Medium
Distinction Between Inverse Function and Reciprocal
Accuracy: 57%
Total Attempted28
Total Correct16
Total Incorrect12
Difficulty Level:
Medium
Principal Value Evaluation
Accuracy: 59%
Total Attempted59
Total Correct35
Total Incorrect24
Difficulty Level:
Medium

Learning Objectives

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Learning Objectives

  • Understand the concept of inverse functions and their existence conditions.
  • Identify the restrictions on domains and ranges of trigonometric functions for defining their inverses.
  • Analyze the graphical representations of inverse trigonometric functions.
  • Recognize the principal value branches of inverse trigonometric functions.
  • Apply properties of inverse trigonometric functions in solving equations.
  • Differentiate between inverse trigonometric functions and their reciprocal functions.

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Revision Notes & Summary

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Inverse Trigonometric Functions

2.1 Introduction

  • The inverse of a function f, denoted by f⁻¹, exists if f is one-one and onto.
  • Trigonometric functions are not one-one and onto over their natural domains and ranges; hence their inverses do not exist without restrictions.
  • This chapter discusses restrictions on domains and ranges of trigonometric functions to ensure the existence of their inverses.
  • The inverse trigonometric functions are important in calculus and have applications in science and engineering.

2.2 Basic Concepts

Trigonometric Functions

  • Sine Function:
    • Domain: R
    • Range: [-1, 1]
  • Cosine Function:
    • Domain: R
    • Range: [-1, 1]
  • Tangent Function:
    • Domain: R ackslash {x: x = (2n + 1)π, n ∈ Z}
    • Range: R
  • Cotangent Function:
    • Domain: R ackslash {x: x = nπ, n ∈ Z}
    • Range: R
  • Secant Function:
    • Domain: R ackslash {x: x = (2n + π/2), n ∈ Z}
    • Range: R ackslash (-1, 1)
  • Cosecant Function:
    • Domain: R ackslash {x: x = nπ, n ∈ Z}
    • Range: R ackslash (-1, 1)

2.3 Properties of Inverse Trigonometric Functions

  • The principal value branch of inverse trigonometric functions is defined as follows:
    • If y = sin⁻¹ x, then x = sin y and vice versa.
    • The principal value of sin⁻¹ x lies in the range of [-π/2, π/2].
    • Similar definitions apply for other inverse trigonometric functions.

Example Problems

  1. Find the principal value of sin⁻¹(1/2):
    • Solution: sin⁻¹(1/2) = π/6
  2. Find the principal value of cot⁻¹(1):
    • Solution: cot⁻¹(1) = π/4

2.4 Summary of Domains and Ranges

FunctionsDomainRange (Principal Value Branches)
y = sin⁻¹ x[-1, 1][-π/2, π/2]
y = cos⁻¹ x[-1, 1][0, π]
y = cosec⁻¹ xR ackslash (-1, 1)[-π/2, 0) ∪ (0, π/2]
y = sec⁻¹ xR ackslash (-1, 1)[0, π/2) ∪ (π/2, π]
y = tan⁻¹ xR(-π/2, π/2)
y = cot⁻¹ xR(0, π)
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

π3\frac{\pi}{3}

B.

2π3\frac{2\pi}{3}

C.

π4\frac{\pi}{4}

D.

π6\frac{\pi}{6}
Correct Answer: A

Solution:

The principal value of sec⁡−1(x)\sec^{-1}(x) is in the range [0,π][0, \pi] excluding π2\frac{\pi}{2}. For x=2x = 2, the angle is π3\frac{\pi}{3}.

Chapter Concept:

Principal Value Evaluation

A.

[−1,1][-1, 1]

B.

[0,π][0, \pi]

C.

[−π,π][-\pi, \pi]

D.

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
Correct Answer: B

Solution:

The range of y=cos⁡−1(x)y = \cos^{-1}(x) is [0,π][0, \pi], which is the principal value branch of the inverse cosine function.

Chapter Concept:

Inverse Trigonometric Functions

A.

xx

B.

−x-x

C.

11

D.

00
Correct Answer: A

Solution:

By the property of inverse trigonometric functions, sin⁡(sin⁡−1(x))=x\sin(\sin^{-1}(x)) = x for x∈[−1,1]x \in [-1, 1].

Chapter Concept:

Properties of Inverse Trigonometric Functions

A.

−π3-\frac{\pi}{3}

B.

π3\frac{\pi}{3}

C.

2π3\frac{2\pi}{3}

D.

−2π3-\frac{2\pi}{3}
Correct Answer: A

Solution:

The principal value of tan⁡−1(−3)\tan^{-1}(-\sqrt{3}) is −π3-\frac{\pi}{3}, as it lies within the principal value branch (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

Chapter Concept:

Inverse Trigonometric Functions

A.

[−π,0][-\pi, 0]

B.

[0,π][0, \pi]

C.

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

D.

[−1,1][-1, 1]
Correct Answer: B

Solution:

The principal value branch for cos⁡−1x\cos^{-1} x is [0,π][0, \pi] because it makes the cosine function one-one and onto.

Chapter Concept:

Need for Domain Restriction in Trigonometric Functions

A.

2θ2\theta

B.

θ\theta

C.

π2−θ\frac{\pi}{2} - \theta

D.

π−2θ\pi - 2\theta
Correct Answer: A

Solution:

By substituting x=sin⁡θx = \sin \theta, we have sin⁡−1(2sin⁡θcos⁡θ)=sin⁡−1(sin⁡2θ)\sin^{-1}(2\sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta). The principal value of sin⁡−1(sin⁡2θ)\sin^{-1}(\sin 2\theta) is 2θ2\theta.

Chapter Concept:

Algebraic Simplification Using Inverse Trigonometric Substitution

A.

sin⁡−1x\sin^{-1} x

B.

cos⁡−1x\cos^{-1} x

C.

cot⁡−1x\cot^{-1} x

D.

tan⁡−1x\tan^{-1} x
Correct Answer: C

Solution:

The function cot⁡−1x\cot^{-1} x has a principal value range of (0,π)(0, \pi).

Chapter Concept:

Principal Domain-Range Table of Inverse Trigonometric Functions

A.

π\pi

B.

00

C.

π2\frac{\pi}{2}

D.

2π2\pi
Correct Answer: A

Solution:

The principal value of cos⁡−1(−1)\cos^{-1}(-1) is π\pi, as the range of cos⁡−1x\cos^{-1} x is [0,π][0, \pi].

Chapter Concept:

Algebraic Simplification Using Inverse Trigonometric Substitution

A.

π\pi

B.

00

C.

π2\frac{\pi}{2}

D.

−π-\pi
Correct Answer: A

Solution:

The principal value of cos⁡−1(x)\cos^{-1}(x) is in the range [0,π][0, \pi]. For x=−1x = -1, the angle is π\pi.

Chapter Concept:

Principal Value Evaluation

A.

−π3\frac{-\pi}{3}

B.

−π6\frac{-\pi}{6}

C.

π3\frac{\pi}{3}

D.

π6\frac{\pi}{6}
Correct Answer: A

Solution:

The principal value of tan⁡−1(−3)\tan^{-1}(-\sqrt{3}) is −π3\frac{-\pi}{3}, as it lies within the principal value branch (−π2,π2)(\frac{-\pi}{2}, \frac{\pi}{2}).

Chapter Concept:

Principal Value Branches

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Inverse Trigonometric Functions

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True or False

Correct Answer: False

Solution:

The equation 3sin⁡−1x=sin⁡−1(3x−4x3)3\sin^{-1}x = \sin^{-1}(3x - 4x^3) is valid only for specific values of xx within the restricted domain where the expression 3x−4x33x - 4x^3 is also within [−1,1][-1, 1]. This is not true for all xx in [−1,1][-1, 1].

Chapter Concept :

Multiple-Angle Identities with Inverse Trigonometric Functions

Correct Answer: True

Solution:

Inverse trigonometric functions are defined by restricting the domains and ranges of trigonometric functions to ensure they are one-one and onto. For sin⁡−1x\sin^{-1} x, the domain is [−1,1][-1, 1] and the principal value branch range is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Chapter Concept :

Inverse Trigonometric Functions

Correct Answer: False

Solution:

The notation sin⁡−1x\sin^{-1}x represents the inverse sine function, not the reciprocal. The reciprocal of the sine function is represented as (sin⁡x)−1=1sin⁡x(\sin x)^{-1} = \frac{1}{\sin x}.

Chapter Concept :

Distinction Between Inverse Function and Reciprocal

Correct Answer: True

Solution:

The inverse sine function is defined for xx in the interval [−1,1][-1, 1] and its principal value branch is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Chapter Concept :

Inverse Trigonometric Functions

Correct Answer: False

Solution:

The identity sin⁡(sin⁡−1x)=x\sin(\sin^{-1} x) = x is valid only for x∈[−1,1]x \in [-1, 1]. Outside this interval, the expression is not defined.

Chapter Concept :

Properties of Inverse Trigonometric Functions

Correct Answer: True

Solution:

The principal value branch of the inverse sine function sin⁡−1\sin^{-1} is indeed the range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], as it is the interval where the sine function is one-to-one and covers all possible values of sin⁡−1\sin^{-1}.

Chapter Concept :

Principal Value Branches

Correct Answer: True

Solution:

The principal value branch of tan⁡−1x\tan^{-1} x is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). For tan⁡−1(−3)\tan^{-1}(-\sqrt{3}), the angle −π3-\frac{\pi}{3} lies within this range, making it the correct principal value.

Chapter Concept :

Principal Value Evaluation

Correct Answer: False

Solution:

Trigonometric functions are not one-one over their natural domains, so their domains must be restricted to intervals where they become bijective to define their inverses.

Chapter Concept :

Need for Domain Restriction in Trigonometric Functions

Correct Answer: True

Solution:

The graph of an inverse function is the reflection of the original function's graph along the line y=xy = x. This is a fundamental property of inverse functions.

Chapter Concept :

Graphical Representation of Inverse Functions

Correct Answer: True

Solution:

Substituting x=sin⁡θx = \sin \theta transforms the expression to sin⁡−1(2sin⁡θcos⁡θ)\sin^{-1}(2\sin \theta \cos \theta), which simplifies to sin⁡−1(sin⁡2θ)=2θ\sin^{-1}(\sin 2\theta) = 2\theta. This demonstrates the effectiveness of the substitution.

Chapter Concept :

Algebraic Simplification Using Inverse Trigonometric Substitution

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