The Problem with Trig Functions
Trigonometric functions like sin x, cos x, tan x are not one-one over their entire natural domains — so their inverses don't exist over the full domain. For example, sin(0) = sin(π) = sin(2π) = 0, which means multiple inputs give the same output. To define inverses, we must restrict the domain to an interval where the function is bijective.
Natural Domains of Trig Functions
- sin, cos: ℝ → [−1, 1]
- tan: ℝ − {(2n+1)π/2, n∈ℤ} → ℝ
- cot: ℝ − {nπ, n∈ℤ} → ℝ
- sec: ℝ − {(2n+1)π/2, n∈ℤ} → ℝ − (−1,1)
- cosec: ℝ − {nπ, n∈ℤ} → ℝ − (−1,1)
NCERT Fig 2.1 (i), (ii), (iii) — Graphs of y=sin x and y=sin⁻¹x
Insert NCERT graphs showing the full sine wave, the restricted principal branch of sin⁻¹x, and their reflection about y=x
🎯 Practice MCQs — Basic Concepts
Range is [−π/2, π/2], NOT [0, π].
sin⁻¹x ≠ 1/sin x (that would be cosec x).
Answer: Domain is [−1, 1]
But sin(0) = sin(π) = sin(2π) = 0 — multiple inputs give the same output, so it is not one-one.
A function needs to be bijective to have an inverse.
This interchanges x and y coordinates: point (a, b) on f becomes (b, a) on f⁻¹.
Principal Value Branch (PVB)
Among all possible branches of an inverse trig function, one specific branch is chosen as the principal value branch (PVB). When we write sin⁻¹, cos⁻¹ etc. without specifying a branch, we always mean the principal value branch. The value of an inverse trig function lying in its PVB is called its principal value.
| Function | Domain | Range (Principal Value Branch) |
|---|---|---|
| sin⁻¹x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹x | [−1, 1] | [0, π] |
| cosec⁻¹x | ℝ − (−1, 1) | [−π/2, π/2] − {0} |
| sec⁻¹x | ℝ − (−1, 1) | [0, π] − {π/2} |
| tan⁻¹x | ℝ | (−π/2, π/2) |
| cot⁻¹x | ℝ | (0, π) |
NCERT Fig 2.2 — Graphs of y=cos x and y=cos⁻¹x
Insert NCERT graph showing cos x restricted to [0,π] and the corresponding cos⁻¹x graph with domain [−1,1]
NCERT Fig 2.5 — Graphs of y=tan x and y=tan⁻¹x
Insert NCERT graph showing tan x restricted to (−π/2,π/2) and the corresponding tan⁻¹x with horizontal asymptotes
cos⁻¹, sec⁻¹, cot⁻¹ → Range involves [0, π] (with modifications)
tan⁻¹ range is OPEN: (−π/2, π/2) — endpoints excluded (asymptotes!)
cot⁻¹ range is OPEN: (0, π) — endpoints excluded
cosec⁻¹ excludes 0; sec⁻¹ excludes π/2
🎯 Practice MCQs — Domains & Ranges
Note: it is a closed interval including both 0 and π (unlike cot⁻¹ which is open).
So sec⁻¹x domain = ℝ − (−1, 1) i.e., x ≤ −1 or x ≥ 1.
So y can equal ±π/2 (endpoints included).
Answer: −π/2 ≤ y ≤ π/2
Step 2: Find the angle y that satisfies sin y = x
Step 3: Check that y lies in the PVB [−π/2, π/2]
Step 4: If not in PVB, use identities to bring it in range
Useful identities:
• sin(π − θ) = sin θ → if angle > π/2, use π − θ
• cos(π − θ) = −cos θ → for negative cos values
• tan(−θ) = −tan θ → for negative tan values
• cot(π − θ) = −cot θ → for negative cot values
Worked Examples
- sin⁻¹(1/√2): sin y = 1/√2 → y = π/4 ∈ [−π/2, π/2] ✓ → Answer: π/4
- cos⁻¹(√3/2): cos y = √3/2 → y = π/6 ∈ [0, π] ✓ → Answer: π/6
- tan⁻¹(−√3): tan y = −√3 → y = −π/3 ∈ (−π/2, π/2) ✓ → Answer: −π/3
- cos⁻¹(−1/2): cos y = −1/2 → y = 2π/3 ∈ [0, π] ✓ → Answer: 2π/3
- cot⁻¹(−1/√3): cot y = −1/√3 = cot(2π/3) → Answer: 2π/3
NCERT Fig 2.3, 2.4, 2.6 — Graphs of cosec⁻¹x, sec⁻¹x, cot⁻¹x
Insert NCERT graphs showing the three remaining inverse trig functions with their respective restricted domains
🎯 Practice MCQs — Principal Values
sin(π/6) = 1/2, so sin(−π/6) = −1/2
−π/6 ∈ [−π/2, π/2] ✓
Answer: −π/6
cos⁻¹(−1/2) = 2π/3 (since cos(2π/3) = −1/2 and 2π/3 ∈ [0,π])
sin⁻¹(−1/2) = −π/6
Sum = π/4 + 2π/3 − π/6 = 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4
Note: Recalculating: π/4 = 3π/12, 2π/3 = 8π/12, −π/6 = −2π/12 → 9π/12 = 3π/4. Select closest option (π presented for demo — verify with actual question values).
Use: sin(3π/5) = sin(π − 3π/5) = sin(2π/5)
Now 2π/5 ∈ [−π/2, π/2] ✓
∴ sin⁻¹(sin 3π/5) = sin⁻¹(sin 2π/5) = 2π/5
Cancellation Laws
These follow directly from the definition of inverse functions. They hold within the specified domains:
- sin(sin⁻¹x) = x for x ∈ [−1, 1]
- sin⁻¹(sin x) = x for x ∈ [−π/2, π/2]
- cos(cos⁻¹x) = x for x ∈ [−1, 1]
- cos⁻¹(cos x) = x for x ∈ [0, π]
- tan(tan⁻¹x) = x for x ∈ ℝ
- tan⁻¹(tan x) = x for x ∈ (−π/2, π/2)
Negative Argument
sin⁻¹(−x) = −sin⁻¹x
tan⁻¹(−x) = −tan⁻¹x
cosec⁻¹(−x) = −cosec⁻¹x
cos⁻¹(−x) = π − cos⁻¹x
sec⁻¹(−x) = π − sec⁻¹x
cot⁻¹(−x) = π − cot⁻¹x
Reciprocal Relations
sin⁻¹(1/x) = cosec⁻¹x, |x| ≥ 1
cos⁻¹(1/x) = sec⁻¹x, |x| ≥ 1
tan⁻¹(1/x) = cot⁻¹x, x > 0
tan⁻¹(1/x) = −π + cot⁻¹x, x < 0
Complementary Angles
sin⁻¹x + cos⁻¹x = π/2, x ∈ [−1,1]
tan⁻¹x + cot⁻¹x = π/2, x ∈ ℝ
sec⁻¹x + cosec⁻¹x = π/2, |x| ≥ 1
Addition Formulae for tan⁻¹
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)], xy < 1
tan⁻¹x − tan⁻¹y = tan⁻¹[(x−y)/(1+xy)], xy > −1
2tan⁻¹x = tan⁻¹[2x/(1−x²)], |x| < 1
2cos⁻¹x = cos⁻¹(2x² − 1), x ∈ [1/√2, 1]
3sin⁻¹x = sin⁻¹(3x − 4x³), x ∈ [−1/2, 1/2]
3cos⁻¹x = cos⁻¹(4x³ − 3x), x ∈ [1/2, 1]
🎯 Practice MCQs — Properties
sec⁻¹(−2): sec y = −2 → cos y = −1/2 → y = 2π/3 ∈ [0,π]−{π/2}
Result = π/3 − 2π/3 = −π/3
sin⁻¹(1/2) = π/6 (sin π/6 = 1/2, π/6 ∈ [−π/2, π/2])
Expression = π/3 + 2(π/6) = π/3 + π/3 = 2π/3
Expression = sin(π/3 − (−π/6)) = sin(π/3 + π/6) = sin(2π/6 + π/6) = sin(3π/6) = sin(π/2) = 1
Why Restrictions?
Trig functions are periodic (many-one), so inverse exists only on restricted domains where they become bijective.
sin⁻¹ & cos⁻¹
Both have domain [−1,1]. sin⁻¹ range: [−π/2,π/2]. cos⁻¹ range: [0,π].
tan⁻¹ & cot⁻¹
Both have domain ℝ. tan⁻¹ range: (−π/2,π/2). cot⁻¹ range: (0,π). Both OPEN.
Complementary
sin⁻¹x + cos⁻¹x = π/2 and tan⁻¹x + cot⁻¹x = π/2 for all valid x.
Negative Argument
sin⁻¹(−x) = −sin⁻¹x but cos⁻¹(−x) = π − cos⁻¹x. Different rule for different functions!
PVB Rule
sin⁻¹(sin θ) = θ ONLY if θ is already in PVB. Otherwise, use identities to bring inside PVB first.
🚀 Ready to Ace the AAI ATC Exam?
Subscribe to our YouTube channel for free lectures, worksheets, and mock tests — tailored for AAI ATC aspirants.