Inverse Trigonometric Functions

2.1 Core concepts

• A function f : X → Y has an inverse f⁻¹ : Y → X iff f is one-one and onto; trigonometric functions over their natural domains are NOT one-one (sine repeats with period 2π and oscillates between −1 and 1), so inverses are defined only after restricting domains (NCERT §2.1, p. 18). Without this restriction, "sin⁻¹(½)" would be ambiguous because infinitely many angles have sine ½.
• The six basic trig functions are: sin : R → [−1, 1], cos : R → [−1, 1], tan : R − {(2n+1)π/2} → R, cot : R − {nπ} → R, sec : R − {(2n+1)π/2} → R − (−1, 1), cosec : R − {nπ} → R − (−1, 1) (NCERT §2.2, p. 18). Each of these has a natural domain (the largest set of reals on which the function is defined) and a restricted domain on which it becomes invertible.
• Sine restricted to [−π/2, π/2] is bijective onto [−1, 1]; its inverse sin⁻¹ : [−1, 1] → [−π/2, π/2] is called the principal value branch (NCERT §2.2, p. 19). NCERT notes that other branches like [π/2, 3π/2] would also work, but [−π/2, π/2] is the conventional choice because it contains 0 and is symmetric about 0.
• Cosine restricted to [0, π] is bijective onto [−1, 1]; cos⁻¹ : [−1, 1] → [0, π] is the principal value branch of arc cosine (NCERT §2.2, p. 20–21). Note that the cos⁻¹ branch is NOT symmetric about 0; it starts at 0 and runs to π.
• Cosec restricted to [−π/2, π/2] − {0} is bijective onto R − (−1, 1); cosec⁻¹ : R − (−1, 1) → [−π/2, π/2] − {0} is the principal value branch (NCERT §2.2, p. 21–22). The point 0 is excluded because cosec is undefined at 0.
• Sec restricted to [0, π] − {π/2} is bijective onto R − (−1, 1); sec⁻¹ : R − (−1, 1) → [0, π] − {π/2} is the principal value branch (NCERT §2.2, p. 22–23). The point π/2 is excluded because sec is undefined there.
• Tan restricted to (−π/2, π/2) is bijective onto R; tan⁻¹ : R → (−π/2, π/2) is the principal value branch (NCERT §2.2, p. 23–24). The endpoints are excluded because tan diverges to ±∞ there.
• Cot restricted to (0, π) is bijective onto R; cot⁻¹ : R → (0, π) is the principal value branch (NCERT §2.2, p. 24–25). Again the endpoints are excluded because cot is undefined at 0 and π.
• The graph of any inverse trig function is obtained either by interchanging x and y axes of the original trig graph, or as the reflection of the original graph in the line y = x (NCERT §2.2, Remarks (i)–(ii), p. 19–20; Figs 2.1–2.6). Reflection in y = x is the standard "every inverse function" recipe and applies here too.
• A critical notational warning: sin⁻¹ x is NOT (sin x)⁻¹; in fact (sin x)⁻¹ = 1/sin x = cosec x (NCERT §2.2 Note (1), p. 26). The superscript −1 denotes "inverse function", not "reciprocal".
• Whenever no branch is mentioned, the principal value branch is assumed; the value of an inverse trig function that lies in this principal branch is called its principal value (NCERT §2.2 Notes (2)–(3), p. 26).
• For all suitable x: sin(sin⁻¹ x) = x for x ∈ [−1, 1], and sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2]; similar results hold for the other inverse trig functions in their principal ranges (NCERT §2.3, p. 27).
• For x outside the principal range one must convert: e.g. sin⁻¹(sin(3π/5)) = sin⁻¹(sin(π − 3π/5)) = sin⁻¹(sin(2π/5)) = 2π/5, because 2π/5 lies in [−π/2, π/2] (NCERT Misc. Example 6, p. 30). This "reduce to principal range first" routine appears on every CUET paper.
• Standard simplifications proved in NCERT include sin⁻¹(2x√(1−x²)) = 2 sin⁻¹ x (for −1/√2 ≤ x ≤ 1/√2) and sin⁻¹(2x√(1−x²)) = 2 cos⁻¹ x (for 1/√2 ≤ x ≤ 1) (NCERT Example 3, p. 28). The break-point x = 1/√2 (= sin(π/4)) is critical: the double-angle identity sin 2θ = 2 sin θ cos θ holds algebraically for all x, but the image of 2 sin⁻¹ x leaves the principal range of sin⁻¹ at x = 1/√2.
• Other simplifications: 3 sin⁻¹ x = sin⁻¹(3x − 4x³) on [−1/2, 1/2], and 3 cos⁻¹ x = cos⁻¹(4x³ − 3x) on [1/2, 1] (NCERT Exercise 2.2 Q1–Q2, p. 29). These echo the triple-angle identities sin 3θ = 3 sin θ − 4 sin³ θ and cos 3θ = 4 cos³ θ − 3 cos θ.
• The 2026-27 reprint drops most "Properties of Inverse Trigonometric Functions" (the long list of arc-tangent sum/difference identities); only the basic principal-branch identities and the few above remain in scope.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 2.1 (i), (ii), (iii), p. 20 — Graphs of y = sin x and y = sin⁻¹ x. The dark portion of y = sin x marks the principal branch [−π/2, π/2], and the y = sin⁻¹ x graph is its reflection in y = x. Frame (iii) overlays both to show the mirror symmetry, which is the visual definition of an inverse function.
• Fig 2.2 (i), (ii), p. 21 — Graphs of y = cos x (with [0, π] highlighted) and y = cos⁻¹ x. The cos⁻¹ graph is strictly decreasing, a fact many students forget.
• Fig 2.3 (i), (ii), p. 22 — Graphs of y = cosec x and y = cosec⁻¹ x. Note the hole at 0 in both the restricted domain of cosec and the range of cosec⁻¹.
• Fig 2.4 (i), (ii), p. 23 — Graphs of y = sec x and y = sec⁻¹ x. Note the hole at π/2 in both the restricted domain of sec and the range of sec⁻¹.
• Fig 2.5 (i), (ii), p. 24 — Graphs of y = tan x and y = tan⁻¹ x. The tan⁻¹ graph is strictly increasing, with horizontal asymptotes y = ±π/2.
• Fig 2.6 (i), (ii), p. 25 — Graphs of y = cot x and y = cot⁻¹ x. The cot⁻¹ graph is strictly decreasing, with horizontal asymptotes y = 0 and y = π.
• Summary table on p. 26 and p. 32 — Consolidated list of all six inverse trig functions with domains and principal-value ranges; this single table is the single highest-yield piece for CUET. Reproduce it cold every morning during revision.
• Process — principal value of sin⁻¹(k) when |k| ≤ 1: (i) find θ ∈ [0, π/2] with sin θ = |k|; (ii) take principal value = θ if k ≥ 0, else −θ. The answer is forced into [−π/2, π/2].
• Process — principal value of cos⁻¹(k): (i) find θ ∈ [0, π/2] with cos θ = |k|; (ii) answer = θ if k ≥ 0, else π − θ. The answer is forced into [0, π].
• Process — evaluation of sin⁻¹(sin α) for α outside [−π/2, π/2]: reduce α modulo 2π, then if the reduced value lies in (π/2, 3π/2) use sin α = sin(π − α). Otherwise use sin α = sin(α − 2π) to shift into [−π/2, π/2].

2.4 Common confusions / NTA trap points

• Confusing sin⁻¹ x with (sin x)⁻¹ = 1/sin x = cosec x (NCERT §2.2 Note 1, p. 26). NTA loves planting this as a distractor.
• For sin⁻¹(sin x), forgetting that the answer is x ONLY if x ∈ [−π/2, π/2]; for other x you must use sin x = sin(π − x) or sin(x − 2π) etc. to land inside the principal branch (NCERT Misc. Example 6, p. 30).
• Mis-identifying the principal range of sec⁻¹ as [−π/2, π/2] − {0} (that's cosec⁻¹) instead of [0, π] − {π/2} (NCERT summary table, p. 26, 32).
• For cot⁻¹ of a negative number, students wrongly answer a negative angle; the principal range is (0, π), so cot⁻¹(−1/√3) = 2π/3, not −π/3 (NCERT Example 2, p. 26).
• Forgetting the identity sin⁻¹(2x√(1−x²)) = 2 sin⁻¹ x is only valid for −1/√2 ≤ x ≤ 1/√2; outside this interval the simpler RHS is no longer in the principal range (NCERT Example 3 (i), p. 28).
• Assuming cos⁻¹(−x) = −cos⁻¹ x. Actually cos⁻¹(−x) = π − cos⁻¹ x because cos⁻¹ has range [0, π], which forbids negative values.
• Reading "tan⁻¹ x" as having range [−π/2, π/2]; the correct range is the open interval (−π/2, π/2). The endpoints are never attained.
• Writing sin⁻¹(2) as a real number; the domain of sin⁻¹ is [−1, 1], so sin⁻¹(2) is undefined in the real-valued setting.
• Forgetting that cot⁻¹ has the same domain R as tan⁻¹, not the restricted domain of cot.
• Writing the graph of y = sin⁻¹ x as a reflection in the x-axis (which gives y = −sin x) instead of in y = x.
• Misreading sec⁻¹ x's range as (0, π); the correct range excludes π/2.
• Setting sin⁻¹(sin(3π/5)) = 3π/5; the actual value is 2π/5 since 3π/5 > π/2 is outside the principal range.
• Mis-computing tan⁻¹(−√3) = −2π/3 instead of −π/3; −2π/3 is outside (−π/2, π/2).
• Treating the inverse trig functions as "multivalued": in the NCERT framework they are single-valued functions on the principal branch. Treating sin⁻¹(½) as both π/6 and 5π/6 is a leftover bad habit from school algebra and produces wrong CUET answers.
• Confusing the convex/concave shape of inverse graphs: y = sin⁻¹ x is increasing, y = cos⁻¹ x is decreasing, y = tan⁻¹ x is increasing, y = cot⁻¹ x is decreasing. Mis-sketching produces wrong sign of derivative in Class XII calculus problems.
• Forgetting that the inverse function exists only because we restricted the domain — the inverse of the unrestricted sine function is not a function at all. CUET sometimes phrases this in conceptual MCQs.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 1, p. 26). Find principal value of sin⁻¹(1/2).

Step 1 — identify reference angle: sin θ = 1/2 ⇒ θ = π/6 is the standard reference. Step 2 — check principal range: π/6 ∈ [−π/2, π/2]. ✓ Step 3 — conclude: sin⁻¹(1/2) = π/6. Answer: π/6.

Example B (NCERT Example 2, p. 26). Find principal value of cot⁻¹(−1/√3).

Step 1 — find reference angle: cot θ = 1/√3 ⇒ θ = π/3. Step 2 — adjust for negative input: cot⁻¹ range is (0, π); cot(π − π/3) = −cot(π/3) = −1/√3, so principal value = π − π/3 = 2π/3. Step 3 — verify range: 2π/3 ∈ (0, π). ✓ Answer: 2π/3.

Example C (NCERT Misc. Example 6, p. 30). Evaluate sin⁻¹(sin(3π/5)).

Step 1 — check range: 3π/5 ≈ 0.6π is outside [−π/2, π/2]. Step 2 — use sin α = sin(π − α): sin(3π/5) = sin(π − 3π/5) = sin(2π/5). Step 3 — check 2π/5 ∈ [−π/2, π/2]: Yes, 2π/5 = 0.4π < π/2. So answer = 2π/5. Answer: 2π/5.

Example D (NCERT Example 3 (i), p. 28). Show sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹ x for x ∈ [−1/√2, 1/√2].

Step 1 — substitute x = sin θ: with θ ∈ [−π/4, π/4] so 2θ ∈ [−π/2, π/2]. Then √(1 − x²) = cos θ. Step 2 — apply double-angle: 2x√(1 − x²) = 2 sin θ cos θ = sin 2θ. Step 3 — take sin⁻¹: sin⁻¹(sin 2θ) = 2θ (since 2θ ∈ [−π/2, π/2]) = 2 sin⁻¹ x. QED.

Example E (NCERT Exercise 2.1, Q14, p. 27). Find value of tan⁻¹(√3) − sec⁻¹(−2).

Step 1 — evaluate tan⁻¹(√3): tan(π/3) = √3 and π/3 ∈ (−π/2, π/2) ⇒ tan⁻¹(√3) = π/3. Step 2 — evaluate sec⁻¹(−2): sec θ = −2 ⇒ cos θ = −1/2 ⇒ θ = 2π/3 ∈ [0, π] − {π/2}. So sec⁻¹(−2) = 2π/3. Step 3 — subtract: π/3 − 2π/3 = −π/3. Answer: −π/3.