Chapter 2: Inverse Trigonometric Functions

Math Part 1 • Class 12

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Chapter Analysis

Intermediate15 pages • English

Quick Summary

Chapter 2 of Class 12 Mathematics Part 1 focuses on Inverse Trigonometric Functions. It begins by reviewing the concept of inverse functions and the conditions under which they exist, particularly for trigonometric functions. The chapter discusses how to restrict the domains and ranges of trigonometric functions to allow their inverses to be defined. Examples are provided to demonstrate the calculation of inverse trigonometric values and to illustrate the use of inverse trigonometric functions in calculus, specifying their importance in integral evaluation. Additionally, the chapter includes exercises to reinforce the concepts.

Key Topics

  • Inverse Functions
  • Trigonometric Functions
  • Principal Value Branch
  • Domain Restriction
  • Graphical Representation
  • Properties of Inverse Trigonometric Functions
  • Applications in Calculus

Learning Objectives

  • Understand and define inverse trigonometric functions.
  • Apply domain and range restrictions to define inverse functions.
  • Calculate and interpret the principal values of inverse trigonometric functions.
  • Graph inverse trigonometric functions and understand their properties.
  • Apply inverse trigonometric functions in calculus, particularly for integration.

Questions in Chapter

Find the principal value of sin–1(1/2).

Page 26

Prove that sin–1(22x - 1) = 2 sin–1(x), for -1/2 <= x <= 1/2.

Page 29

Prove that 3cos–1(x) = cos–1(4x^3 - 3x), for x ∈ [-1, 1].

Page 29

Find the value of tan–1(2cos(2sin–1(x))) + cos–1(2sin(x)).

Page 31

Additional Practice Questions

What is the principal value of sec–1(2)?

medium

Answer: The principal value of sec–1(2) is pi/3, because secant is the reciprocal of cosine, and cos(pi/3) = 1/2.

Express cot–1(1) in terms of pi.

easy

Answer: cot–1(1) is equal to pi/4, because cotangent is the reciprocal of tangent, and tan(pi/4) = 1.

Evaluate tan–1(√3) + tan–1(1/√3).

medium

Answer: tan–1(√3) = pi/3 and tan–1(1/√3) = pi/6. Their sum equals pi/2.

If y = sin–1(x), find the derivative dy/dx.

hard

Answer: The derivative dy/dx = 1/√(1-x^2), where |x| < 1.

Show that sin–1(x) + cos–1(x) = pi/2 for all x in [-1, 1].

easy

Answer: For any x in [-1, 1], sin–1(x) and cos–1(x) are complementary angles, hence their sum is pi/2.