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Domain and Principal Value Branches of Inverse Trigonometric Functions

The principal value branch of an inverse trigonometric function is the chosen interval of angles on which the corresponding trigonometric function becomes one-one and onto its required range.

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Student-friendly explanation

For an inverse trigonometric function to exist as a function, each input must give exactly one output. For example, sin θ = 1/2 has many angle solutions, but sin⁻¹(1/2) means the principal angle in [−π/2, π/2], which is π/6. The domain is the allowed input set, and the range is the selected principal value interval.

How to write this in exams

  1. 1

    Start with the exact idea

    The principal value branch of an inverse trigonometric function is the chosen interval of angles on which the corresponding trigonometric function becomes one-one and onto its required range.

  2. 2

    Then show how to use it

    Identify the inverse function, write its domain and principal range, check that the input is allowed, find an angle whose trigonometric value matches the input, then select only the angle lying in the principal range.

  3. 3

    Add one concrete example

    sin⁻¹(−1/2) = −π/6 because −π/6 lies in [−π/2, π/2]. cos⁻¹(−1/2) = 2π/3 because 2π/3 lies in [0, π].

  4. 4

    Avoid this incomplete answer

    Using 3π/4 for tan⁻¹(−1) because tan(3π/4) = −1, even though 3π/4 is outside the range of tan⁻¹x.

Definition

The principal value branch of an inverse trigonometric function is the chosen interval of angles on which the corresponding trigonometric function becomes one-one and onto its required range.

Example

sin⁻¹(−1/2) = −π/6 because −π/6 lies in [−π/2, π/2]. cos⁻¹(−1/2) = 2π/3 because 2π/3 lies in [0, π].

Rule to remember

Domains and ranges: sin⁻¹x: domain [−1,1], range [−π/2,π/2]; cos⁻¹x: domain [−1,1], range [0,π]; tan⁻¹x: domain R, range (−π/2,π/2); cot⁻¹x: domain R, range (0,π); sec⁻¹x: domain (−∞,−1] ∪ [1,∞), range [0,π] excluding π/2; cosec⁻¹x: domain (−∞,−1] ∪ [1,∞), range [−π/2,π/2] excluding 0.

Memory hook

Inverse trig answers are not all angles; they are the one angle allowed by the function's selected branch.

Examples and method

Worked example

Evaluate tan⁻¹(−1). Since tan⁻¹x has range (−π/2, π/2), the required angle θ must lie in this interval. tan θ = −1 gives θ = −π/4 in the principal range. Therefore tan⁻¹(−1) = −π/4.

Method to apply

Identify the inverse function, write its domain and principal range, check that the input is allowed, find an angle whose trigonometric value matches the input, then select only the angle lying in the principal range.

Diagram support

A branch diagram is useful: show the restricted interval on the x-axis for each trigonometric function and indicate that only this branch is inverted.

How CBSE asks it

Questions usually ask for principal values, domains, ranges, or the validity of a simplification involving inverse functions.

Avoid common mistakes

Common confusion

Writing all possible angles instead of the principal value, such as giving sin⁻¹(1/2) = nπ + (−1)ⁿπ/6. That is a solution set for sin θ = 1/2, not the value of sin⁻¹(1/2).

Common wrong answer

Using 3π/4 for tan⁻¹(−1) because tan(3π/4) = −1, even though 3π/4 is outside the range of tan⁻¹x.

Exam tip

Before evaluating, write the range of the inverse function beside it. This prevents choosing an angle from the wrong quadrant.

Quick check

What is the principal value of cos⁻¹(−√3/2)?

5π/6, because cos⁻¹x has range [0, π] and cos(5π/6) = −√3/2.

Answer writing and exam use

1-mark answer

The principal value branch of an inverse trigonometric function is the chosen interval of angles on which the corresponding trigonometric function becomes one-one and onto its required range.

2-mark answer

The principal value branch of an inverse trigonometric function is the chosen interval of angles on which the corresponding trigonometric function becomes one-one and onto its required range. Domains and ranges: sin⁻¹x: domain [−1,1], range [−π/2,π/2]; cos⁻¹x: domain [−1,1], range [0,π]; tan⁻¹x: domain R, range (−π/2,π/2); cot⁻¹x: domain R, range (0,π); sec⁻¹x: domain (−∞,−1] ∪ [1,∞), range [0,π] excluding π/2; cosec⁻¹x: domain (−∞,−1] ∪ [1,∞), range [−π/2,π/2] excluding 0. sin⁻¹(−1/2) = −π/6 because −π/6 lies in [−π/2, π/2]. cos⁻¹(−1/2) = 2π/3 because 2π/3 lies in [0, π].

3-mark answer

For an inverse trigonometric function to exist as a function, each input must give exactly one output. For example, sin θ = 1/2 has many angle solutions, but sin⁻¹(1/2) means the principal angle in [−π/2, π/2], which is π/6. The domain is the allowed input set, and the range is the selected principal value interval. Domains and ranges: sin⁻¹x: domain [−1,1], range [−π/2,π/2]; cos⁻¹x: domain [−1,1], range [0,π]; tan⁻¹x: domain R, range (−π/2,π/2); cot⁻¹x: domain R, range (0,π); sec⁻¹x: domain (−∞,−1] ∪ [1,∞), range [0,π] excluding π/2; cosec⁻¹x: domain (−∞,−1] ∪ [1,∞), range [−π/2,π/2] excluding 0. Evaluate tan⁻¹(−1). Since tan⁻¹x has range (−π/2, π/2), the required angle θ must lie in this interval. tan θ = −1 gives θ = −π/4 in the principal range. Therefore tan⁻¹(−1) = −π/4. Questions usually ask for principal values, domains, ranges, or the validity of a simplification involving inverse functions. Using 3π/4 for tan⁻¹(−1) because tan(3π/4) = −1, even though 3π/4 is outside the range of tan⁻¹x.
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