Inverse Trigonometric Functions
1. Introduction
Inverse trigonometric functions are the inverse functions of trigonometric functions restricted to principal value branches. They are widely used in calculus, integration, and solving differential equations.
2. Principal Values
A trigonometric function is not one-one on its natural domain. By restricting the domain, we obtain invertible branches called principal value branches.
2.1 Domain and Range
| Function | Domain | Range (Principal) |
|---|---|---|
| sin^{-1} x | [-1, 1] | [-π/2, π/2] |
| cos^{-1} x | [-1, 1] | [0, π] |
| tan^{-1} x | R | (-π/2, π/2) |
| cot^{-1} x | R | (0, π) |
| sec^{-1} x | (-∞, -1] ∪ [1, ∞) | [0, π] - {π/2} |
| cosec^{-1} x | (-∞, -1] ∪ [1, ∞) | [-π/2, π/2] - {0} |
2.2 Convention
'Always state the principal value. For example, sin^{-1}(1/2) = π/6 (not 5π/6).'
3. Graphs of Inverse Trigonometric Functions
The graph of sin^{-1} x is the reflection of sin x (restricted to [-π/2, π/2]) about the line y = x. Similar reflections give graphs for other functions.
3.1 Key Graph Features
- sin^{-1} x is increasing, odd, domain [-1, 1], range [-π/2, π/2].
- cos^{-1} x is decreasing, neither even nor odd, range [0, π].
- tan^{-1} x is increasing, odd, with asymptotes at y = ±π/2.
4. Properties and Identities
4.1 Inverse of Composites
sin(sin^{-1} x) = x, for x ∈ [-1, 1]. But sin^{-1}(sin θ) = θ only when θ ∈ [-π/2, π/2].
4.2 Complementary Relations
sin^{-1} x + cos^{-1} x = π/2 tan^{-1} x + cot^{-1} x = π/2 sec^{-1} x + cosec^{-1} x = π/2
4.3 Negative Arguments
sin^{-1}(-x) = - sin^{-1} x cos^{-1}(-x) = π - cos^{-1} x tan^{-1}(-x) = - tan^{-1} x
4.4 Sum and Difference Formulas
tan^{-1} x + tan^{-1} y = tan^{-1}((x + y)/(1 - xy)), provided xy < 1 tan^{-1} x - tan^{-1} y = tan^{-1}((x - y)/(1 + xy)), provided xy > -1
4.5 Conversion Formulas
sin^{-1} x = tan^{-1}(x/√(1 - x²)) cos^{-1} x = tan^{-1}(√(1 - x²)/x) tan^{-1} x = sin^{-1}(x/√(1 + x²))
5. Worked Problems
Problem 1: Find the principal value of tan^{-1}(-√3). Solution: tan^{-1}(-√3) = -tan^{-1}(√3) = -π/3.
Problem 2: Simplify sin^{-1}(√((1 - x)/2)). Solution: Let x = cos θ. Then √((1 - cos θ)/2) = √(sin²(θ/2)) = sin(θ/2). So sin^{-1}(sin(θ/2)) = θ/2 = (1/2)cos^{-1} x.
Problem 3: Solve tan^{-1}(x + 1) + tan^{-1}(x - 1) = tan^{-1}(8/31). Solution: Using sum formula: tan^{-1}((2x)/(2 - x²)) = tan^{-1}(8/31). So 2x/(2 - x²) = 8/31. Cross-multiplying: 62x = 16 - 8x² ⇒ 8x² + 62x - 16 = 0 ⇒ 4x² + 31x - 8 = 0 ⇒ (4x - 1)(x + 8) = 0. Hence x = 1/4 or x = -8 (reject as it doesn't satisfy domain condition).
6. Common Mistakes
'Students often forget to check domain conditions when applying sum formulas. Always verify that xy < 1 for the tan^{-1} sum formula.'
'Another frequent error: assuming sin^{-1}(sin θ) = θ for all θ. This holds only when θ is in the principal value range.'
7. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Principal values | 2 | 1 |
| Identities and simplification | 4 | 2 |
| Solving equations | 3 | 2 |
| Proofs using properties | 4 | - |
8. Self-Test Questions
- Find the principal value of sec^{-1}(-2).
- Prove that tan^{-1}(1) + tan^{-1}(2) + tan^{-1}(3) = π.
- Simplify: cos^{-1}(4x³ - 3x) for x ∈ [1/2, 1].
- Solve: sin^{-1}(x) + sin^{-1}(2x) = π/3.
- Prove: 2 tan^{-1}(1/5) + tan^{-1}(1/7) + 2 tan^{-1}(1/8) = π/4.
9. ISC Examination Tips
- Always state the principal value branch for inverse trigonometric functions.
- When using the sum formula for tan^{-1}, first verify the product condition.
- Remember to check domain restrictions when solving equations involving inverse trigonometric functions.
- For proving identities, substitution (such as x = sin θ, x = tan θ) is a powerful technique.
- Pay attention to the range of the function when simplifying compositions like sin^{-1}(sin θ).
10. Additional Worked Problems for ISC Practice
Problem A: Find the principal value of cosec^{-1}(-√2).
Solution: cosec^{-1}(-√2) = -cosec^{-1}(√2) = -π/4. The principal range of cosec^{-1} is [-π/2, π/2] - {0}.
Problem B: Prove that 2 tan^{-1}(1/3) + tan^{-1}(1/7) = π/4.
Solution: Let θ = tan^{-1}(1/3). Then tan 2θ = 2tanθ/(1-tan²θ) = (2/3)/(1-1/9) = (2/3)/(8/9) = 3/4. So 2 tan^{-1}(1/3) = tan^{-1}(3/4). Now tan^{-1}(3/4) + tan^{-1}(1/7) = tan^{-1}((3/4+1/7)/(1-3/28)) = tan^{-1}((25/28)/(25/28)) = tan^{-1}(1) = π/4.
Problem C: Simplify sin^{-1}(3x - 4x³) for x ∈ [-1/2, 1/2].
Solution: Let x = sin θ. Then sin^{-1}(3 sin θ - 4 sin³θ) = sin^{-1}(sin 3θ) = 3θ = 3 sin^{-1}x, since for x ∈ [-1/2, 1/2], θ ∈ [-π/6, π/6] and 3θ ∈ [-π/2, π/2] which is within the principal range.
