Inverse Trigonometric Functions Notes: Class 12 Mathematics Chapter 2
Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions covers the principal value branches of inverse trigonometric functions, their domains and ranges, and important identities used to simplify related expressions.
Overview of Topics & Subtopics (Click to Jump)
1. Principal Value Branches
Short Note:
• sin⁻¹x: domain [-1, 1], range [-π/2, π/2]; cos⁻¹x: domain [-1, 1], range [0, π]; tan⁻¹x: domain R, range (-π/2, π/2).
• cosec⁻¹x and sec⁻¹x are defined for |x| ≥ 1, while cot⁻¹x is defined for all real x, with range (0, π).
2. Important Identities
Short Note:
• sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; sec⁻¹x + cosec⁻¹x = π/2, each valid within its respective domain.
• sin⁻¹(-x) = -sin⁻¹x and tan⁻¹(-x) = -tan⁻¹x (odd functions), whereas cos⁻¹(-x) = π - cos⁻¹x.
3. Simplification and Graphs
Short Note:
• Inverse trigonometric functions are obtained by restricting trigonometric functions to intervals where they become one-one and onto, ensuring a well-defined inverse.
• Composite expressions such as sin(cos⁻¹x) or tan(sin⁻¹x) can be simplified using right-triangle relationships or the Pythagorean identity.
NCERT Textbook Exercise Questions
Why is the domain of sin⁻¹x restricted to [-1, 1]?
Answer: Since sin⁻¹x is the inverse of the sine function, and the range of sin x is [-1, 1], the domain of its inverse must correspondingly be restricted to [-1, 1]. Any value of x outside this range would not correspond to any real angle, since no angle exists whose sine falls outside this interval.
Prove that sin⁻¹x + cos⁻¹x = π/2 for x ∈ [-1, 1].
Answer: Let sin⁻¹x = θ, so sin θ = x, with θ lying in [-π/2, π/2]. Since sin θ = cos(π/2 - θ), we have x = cos(π/2 - θ), and π/2 - θ lies within [0, π], the principal range of cos⁻¹x. Hence cos⁻¹x = π/2 - θ = π/2 - sin⁻¹x, giving the identity sin⁻¹x + cos⁻¹x = π/2.
Find the principal value of tan⁻¹(√3).
Answer: We know that tan(π/3) = √3, and π/3 lies within the principal value branch (-π/2, π/2) of the tan⁻¹ function. Therefore, the principal value of tan⁻¹(√3) is π/3.
Why are the principal value branches of inverse trigonometric functions necessary?
Answer: Trigonometric functions like sine, cosine, and tangent are periodic and not one-one over their entire domain, meaning a single output corresponds to infinitely many input angles. To define a proper inverse function, the domain must be restricted to an interval where the function becomes one-one and onto; this restricted interval is called the principal value branch.
