Inverse Trigonometric Functions – important questions for Boards 2027
CBSE Class 12 Maths
Which statement gives the ranges of sine and cosine on their natural domains? (A) Both have range R (B) Both have range [−1, 1] (C) Sine has range R and cosine has range [−1, 1] (D) Sine has range [−1, 1] and cosine has range R
Answer
(B) Both have range [−1, 1] — sine and cosine each have range [−1, 1] on their natural domains.
Let f : {1, 2} → {a, b} be given by f(1) = a and f(2) = b. Find f⁻¹, and verify that f⁻¹ is a bijection and that (f⁻¹)⁻¹ = f.
Answer
f⁻¹ : {a, b} → {1, 2} f⁻¹(a) = 1, f⁻¹(b) = 2 Range(f⁻¹) = {1, 2} f⁻¹(a) ≠ f⁻¹(b) ∴ f⁻¹ is one-one and onto, hence bijective. (f⁻¹)⁻¹(1) = a = f(1) (f⁻¹)⁻¹(2) = b = f(2) ∴ (f⁻¹)⁻¹ = f Hence proved.
Evaluate sin⁻¹(sin 2π/3) and tan⁻¹(tan 3π/4), using the principal-value ranges of sin⁻¹ and tan⁻¹.
Answer
The principal-value ranges are [−π/2, π/2] for sin⁻¹ and (−π/2, π/2) for tan⁻¹. sin⁻¹(sin 2π/3) = sin⁻¹(sin(π − π/3)) = sin⁻¹(sin π/3) = π/3 ∴ sin⁻¹(sin 2π/3) = π/3 tan⁻¹(tan 3π/4) = tan⁻¹(tan(−π/4)) = −π/4 ∴ tan⁻¹(tan 3π/4) = −π/4
For x ∈ [−1,1] and t ∈ [−π/2, π/2], establish the identities sin(sin⁻¹x) = x and sin⁻¹(sin t) = t. Explain how the graph of an inverse function is related to the graph of the original function.
Answer
The principal branch restricts sine to [−π/2, π/2], where it is one-one and has range [−1, 1]. 1. For x ∈ [−1, 1], let y = sin⁻¹x. Then y ∈ [−π/2, π/2] and sin y = x, by the definition of the inverse of the restricted sine function. Therefore, sin(sin⁻¹x) = sin y = x. 2. For t ∈ [−π/2, π/2], t is the unique angle in the principal range whose sine is sin t. Hence sin⁻¹(sin t) = t. 3. The graph of an inverse function is the reflection of the graph of the original function in the line y = x, because the coordinates of corresponding points are interchanged. Thus, both identities follow from the mutually inverse functions on their specified domains and ranges.
Find the range of f(x) = sin x when its domain is restricted to [−π/2, π/2].
Answer
f([−π/2, π/2]) = [−1, 1] ∴ Range of f is [−1, 1].
Evaluate sin⁻¹(−1/2), and write the principal range in which your answer lies.
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For −1 ≤ x ≤ 1, prove that sin⁻¹(2x√(1−x²)) = 2 sin⁻¹x, taking account of the principal-value range.
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Evaluate sin⁻¹(1/2) and 1/sin(1/2). Explain why these expressions are different, and state what is meant by the principal value of an inverse trigonometric function.
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For x > 1, let x = sec θ. Show that cot⁻¹√(x² − 1) = sec⁻¹x.
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For x ∈ [−1, 1] and θ ∈ [−π/2, π/2], evaluate sin(sin⁻¹x) and sin⁻¹(sin θ). Hence, state the range condition needed for sin⁻¹(sin θ) = θ.
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6 more probable questions in this chapter, every one with a full answer.
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