Cliste School
Get started

Class 12 — Mathematics (NCERT)

Inverse Trigonometric Functions

Class 12

  • ✓By the end of this lesson students will be able to define inverse trigonometric functions and understand their necessity.
  • ✓By the end of this lesson students will be able to determine the domain and principal value range for each inverse trigonometric function.
  • ✓By the end of this lesson students will be able to apply various properties of inverse trigonometric functions to simplify expressions and solve problems.
  • ✓By the end of this lesson students will be able to prove identities involving inverse trigonometric functions.

Key concepts

Introduction to Inverse Trigonometric Functions

Recall that a function f: A → B has an inverse function f⁻¹: B → A if and only if f is both one-one (injective) and onto (surjective). Trigonometric functions (sine, cosine, tangent, etc.) are periodic, meaning they repeat their values over intervals. This periodicity implies they are not one-one over their natural domains. To define inverse trigonometric functions, we restrict the domain of each trigonometric function to an interval where it is one-one and onto. The range of the original function then becomes the domain of the inverse function, and the restricted domain of the original function becomes the principal value range of the inverse function. The notation sin⁻¹x, cos⁻¹x, etc., denotes the inverse trigonometric functions, not (sinx)⁻¹ or 1/sinx.

Domains and Principal Value Ranges

The principal value branch is the specific range chosen for each inverse trigonometric function to ensure it is well-defined and unique. It is crucial to remember these domains and ranges for solving problems.\n\nHere is a table summarising the domain and principal value range for each inverse trigonometric function:\n\n1. **sin⁻¹x**:\n * Domain: [-1, 1]\n * Principal Value Range: [-π/2, π/2]\n\n2. **cos⁻¹x**:\n * Domain: [-1, 1]\n * Principal Value Range: [0, π]\n\n3. **tan⁻¹x**:\n * Domain: R\n * Principal Value Range: (-π/2, π/2)\n\n4. **cosec⁻¹x**:\n * Domain: R - (-1, 1) or (-∞, -1] ∪ [1, ∞)\n * Principal Value Range: [-π/2, π/2] - {0}\n\n5. **sec⁻¹x**:\n * Domain: R - (-1, 1) or (-∞, -1] ∪ [1, ∞)\n * Principal Value Range: [0, π] - {π/2}\n\n6. **cot⁻¹x**:\n * Domain: R\n * Principal Value Range: (0, π)

Properties of Inverse Trigonometric Functions

Inverse trigonometric functions possess several important properties that are useful for simplifying expressions and solving equations. These properties are derived from the definitions and properties of standard trigonometric functions.

Property 1: Inverse Relations

These properties show the composition of a trigonometric function and its inverse. It is important to note the domain and range restrictions for these identities.

1. sin(sin⁻¹x) = x, for x ∈ [-1, 1]\n2. cos(cos⁻¹x) = x, for x ∈ [-1, 1]\n3. tan(tan⁻¹x) = x, for x ∈ R\n4. cosec(cosec⁻¹x) = x, for x ∈ R - (-1, 1)\n5. sec(sec⁻¹x) = x, for x ∈ R - (-1, 1)\n6. cot(cot⁻¹x) = x, for x ∈ R\n\n7. sin⁻¹(sinx) = x, for x ∈ [-π/2, π/2]\n8. cos⁻¹(cosx) = x, for x ∈ [0, π]\n9. tan⁻¹(tanx) = x, for x ∈ (-π/2, π/2)\n10. cosec⁻¹(cosecx) = x, for x ∈ [-π/2, π/2] - {0}\n11. sec⁻¹(secx) = x, for x ∈ [0, π] - {π/2}\n12. cot⁻¹(cotx) = x, for x ∈ (0, π)
Property 2: Reciprocal Relations

These properties relate an inverse trigonometric function to the inverse of its reciprocal function.

1. sin⁻¹(1/x) = cosec⁻¹x, for |x| ≥ 1\n2. cos⁻¹(1/x) = sec⁻¹x, for |x| ≥ 1\n3. tan⁻¹(1/x) = cot⁻¹x, for x > 0
Property 3: Negative Arguments

These properties show how to handle negative arguments within inverse trigonometric functions. Note the difference in behaviour for functions whose principal value range includes negative values (like sin⁻¹x) and those that don't (like cos⁻¹x).

1. sin⁻¹(-x) = -sin⁻¹x, for x ∈ [-1, 1]\n2. tan⁻¹(-x) = -tan⁻¹x, for x ∈ R\n3. cosec⁻¹(-x) = -cosec⁻¹x, for x ∈ R - (-1, 1)\n4. cos⁻¹(-x) = π - cos⁻¹x, for x ∈ [-1, 1]\n5. sec⁻¹(-x) = π - sec⁻¹x, for x ∈ R - (-1, 1)\n6. cot⁻¹(-x) = π - cot⁻¹x, for x ∈ R
Property 4: Complementary Angle Identities

These identities relate pairs of inverse trigonometric functions whose sum is π/2, similar to complementary angles in trigonometry.

1. sin⁻¹x + cos⁻¹x = π/2, for x ∈ [-1, 1]\n2. tan⁻¹x + cot⁻¹x = π/2, for x ∈ R\n3. cosec⁻¹x + sec⁻¹x = π/2, for x ∈ R - (-1, 1)
Property 5: Sum and Difference Identities (tan⁻¹x)

These are crucial identities for combining or expanding inverse tangent expressions. Pay close attention to the conditions for their applicability.

1. tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)), if xy < 1\n2. tan⁻¹x - tan⁻¹y = tan⁻¹((x-y)/(1+xy)), if xy > -1\n3. 2tan⁻¹x = tan⁻¹(2x/(1-x²)), if |x| < 1\n4. 2tan⁻¹x = sin⁻¹(2x/(1+x²)), if |x| ≤ 1\n5. 2tan⁻¹x = cos⁻¹((1-x²)/(1+x²)), if x ≥ 0
Property 6: Conversion Identities

These identities allow conversion between different inverse trigonometric functions, often useful when simplifying expressions or proving identities. They can be derived using right-angled triangles. Note that the conditions for these identities are for positive x to keep the square roots real and positive, aligning with common NCERT problem types.

1. sin⁻¹x = cos⁻¹(√(1-x²)) = tan⁻¹(x/√(1-x²)), for x ∈ (0, 1]\n2. cos⁻¹x = sin⁻¹(√(1-x²)) = tan⁻¹(√(1-x²)/x), for x ∈ (0, 1]\n3. tan⁻¹x = sin⁻¹(x/√(1+x²)) = cos⁻¹(1/√(1+x²)), for x > 0

Key facts to remember

  • 1The domain and principal value range for each inverse trigonometric function are crucial and must be memorised.
  • 2sin⁻¹x and cosec⁻¹x have principal value range [-π/2, π/2] (with 0 excluded for cosec⁻¹x).
  • 3cos⁻¹x and sec⁻¹x have principal value range [0, π] (with π/2 excluded for sec⁻¹x).
  • 4tan⁻¹x has principal value range (-π/2, π/2) and cot⁻¹x has principal value range (0, π).
  • 5sin⁻¹(-x) = -sin⁻¹x, but cos⁻¹(-x) = π - cos⁻¹x.
  • 6sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2, cosec⁻¹x + sec⁻¹x = π/2.
  • 7The formula tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)) is valid only if xy < 1.
  • 8The formula sin⁻¹(sinx) = x is valid only if x ∈ [-π/2, π/2], and similarly for other inverse functions.

Worked examples

Example 1

Find the principal value of cos⁻¹(-1/2).

ILet cos⁻¹(-1/2) = y.
IIThen cos y = -1/2.
IIIWe know that the principal value range of cos⁻¹x is [0, π].
IVAlso, cos(π/3) = 1/2.
VSince cos y is negative, y must lie in the second quadrant.
VIWe can write cos y = -cos(π/3) = cos(π - π/3) = cos(2π/3).
VIISince 2π/3 ∈ [0, π], y = 2π/3.

Answer

The principal value of cos⁻¹(-1/2) is 2π/3.

Always ensure the value obtained lies within the principal value branch of the respective inverse trigonometric function.

Example 2

Prove that 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17).

IL.H.S. = 2tan⁻¹(1/2) + tan⁻¹(1/7).
IIFirst, apply the formula 2tan⁻¹x = tan⁻¹(2x/(1-x²)) for x = 1/2.
IIIHere, x = 1/2, so |x| < 1. The formula is applicable.
IV2tan⁻¹(1/2) = tan⁻¹( (2 * 1/2) / (1 - (1/2)²) )
V= tan⁻¹( 1 / (1 - 1/4) )
VI= tan⁻¹( 1 / (3/4) )
VII= tan⁻¹(4/3).
VIIINow, substitute this back into the L.H.S.:
9L.H.S. = tan⁻¹(4/3) + tan⁻¹(1/7).
10Apply the formula tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)).
11Here, x = 4/3 and y = 1/7. Check condition xy < 1:
12xy = (4/3) * (1/7) = 4/21, which is < 1. The formula is applicable.
13L.H.S. = tan⁻¹( (4/3 + 1/7) / (1 - (4/3)*(1/7)) )
14= tan⁻¹( ( (28+3)/21 ) / (1 - 4/21) )
15= tan⁻¹( (31/21) / ( (21-4)/21 ) )
16= tan⁻¹( (31/21) / (17/21) )
17= tan⁻¹(31/17).
18This is equal to R.H.S.

Answer

Hence proved: 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17).

Always verify the conditions (e.g., xy < 1 or |x| < 1) before applying the sum/difference formulas for inverse tangent functions.

Example 3

Simplify: tan⁻¹( (cos x - sin x) / (cos x + sin x) ), where -π/4 < x < π/4.

ILet the given expression be y = tan⁻¹( (cos x - sin x) / (cos x + sin x) ).
IIDivide the numerator and denominator by cos x (since cos x ≠ 0 for -π/4 < x < π/4).
IIIy = tan⁻¹( ( (cos x / cos x) - (sin x / cos x) ) / ( (cos x / cos x) + (sin x / cos x) ) )
IVy = tan⁻¹( (1 - tan x) / (1 + tan x) ).
VRecall the trigonometric identity: tan(π/4 - x) = (tan(π/4) - tan x) / (1 + tan(π/4) * tan x) = (1 - tan x) / (1 + tan x).
VISo, y = tan⁻¹(tan(π/4 - x)).
VIIFor tan⁻¹(tanθ) = θ, we need θ to be in the principal value range of tan⁻¹x, i.e., (-π/2, π/2).
VIIIGiven -π/4 < x < π/4.
9Multiply by -1: -π/4 < -x < π/4.
10Add π/4 to all parts: π/4 - π/4 < π/4 - x < π/4 + π/4.
110 < π/4 - x < π/2.
12Since (π/4 - x) ∈ (0, π/2), which is a subset of (-π/2, π/2), the condition for tan⁻¹(tanθ) = θ is satisfied.
13Therefore, y = π/4 - x.

Answer

The simplified expression is π/4 - x.

When simplifying expressions of the form tan⁻¹(tanθ), always check if θ lies within the principal value branch (-π/2, π/2). If not, adjust it using periodicity.

Common mistakes

  • ✗Confusing sin⁻¹x with (sinx)⁻¹ or 1/sinx. They are distinct concepts.
  • ✗Not considering the principal value branch when finding the value of an inverse trigonometric function (e.g., giving 7π/6 for sin⁻¹(1/2)).
  • ✗Incorrectly applying the properties for negative arguments, especially for cos⁻¹(-x), sec⁻¹(-x), and cot⁻¹(-x).
  • ✗Using the sum/difference formulas for tan⁻¹x and tan⁻¹y without checking the conditions (e.g., xy < 1).
  • ✗Assuming sin⁻¹(sinx) = x for any value of x, without restricting x to the principal value range of sin⁻¹x.
  • ✗Errors in converting between different inverse trigonometric functions, often due to incorrect sign conventions or triangle constructions.

Exam tips

  • ★Memorise the domains and principal value ranges of all six inverse trigonometric functions thoroughly. This is fundamental for solving problems correctly.
  • ★Learn all the properties of inverse trigonometric functions. Practice deriving them to understand their origins, which helps in recall and application.
  • ★When solving problems involving tan⁻¹x + tan⁻¹y, always check the condition xy < 1. If xy > 1, the formula changes (e.g., π + tan⁻¹((x+y)/(1-xy)) for x,y > 0). NCERT usually focuses on xy < 1 cases.
  • ★For expressions like sin⁻¹(sinx) or cos⁻¹(cosx), ensure the argument 'x' lies within the principal value branch. If not, adjust it using the periodicity of trigonometric functions.
  • ★Practice converting between different inverse trigonometric functions (e.g., sin⁻¹x to tan⁻¹y) using right-angled triangles. This is a common technique in proofs and simplifications.
  • ★Show all steps clearly and logically, especially when proving identities. State the property being used at each step.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.

Solve a problem →← Maths curriculum