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Simplifying and Solving Equations with Inverse Trigonometric Functions

Simplification and equation solving with inverse trigonometric functions means reducing expressions or finding variable values while preserving domain restrictions and principal value conditions.

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Student-friendly explanation

Inverse trigonometric equations often look algebraic after applying a trigonometric function to both sides, but this can introduce extra answers. Every candidate must be checked in the original equation and in the domain of each inverse function.

How to write this in exams

  1. 1

    Start with the exact idea

    Simplification and equation solving with inverse trigonometric functions means reducing expressions or finding variable values while preserving domain restrictions and principal value conditions.

  2. 2

    Then show how to use it

    List domain restrictions, simplify using valid identities, solve the resulting algebraic or trigonometric statement, substitute candidate values into the original equation, and reject any value outside the allowed domain or principal branch.

  3. 3

    Add one concrete example

    If sin⁻¹x = π/6, then x = sin(π/6) = 1/2. Since 1/2 ∈ [−1,1], it is valid.

  4. 4

    Avoid this incomplete answer

    Forgetting to check the original equation after squaring or applying a trigonometric function, leading to an extra value of x.

Definition

Simplification and equation solving with inverse trigonometric functions means reducing expressions or finding variable values while preserving domain restrictions and principal value conditions.

Example

If sin⁻¹x = π/6, then x = sin(π/6) = 1/2. Since 1/2 ∈ [−1,1], it is valid.

Rule to remember

Method rules: preserve domains such as x ∈ [−1,1] for sin⁻¹x and cos⁻¹x; use identities only under their conditions; when applying trig functions to both sides, check principal ranges; for equations involving tan⁻¹, verify denominator conditions in sum and difference formulae.

Memory hook

Solve first, but trust only the values that survive the domain and original-equation check.

Examples and method

Worked example

Solve 2sin⁻¹x = cos⁻¹x. Since sin⁻¹x + cos⁻¹x = π/2, write cos⁻¹x = π/2 − sin⁻¹x. Then 2sin⁻¹x = π/2 − sin⁻¹x, so 3sin⁻¹x = π/2. Hence sin⁻¹x = π/6 and x = sin(π/6) = 1/2. Check: 2sin⁻¹(1/2) = 2 × π/6 = π/3 and cos⁻¹(1/2) = π/3. Therefore x = 1/2.

Method to apply

List domain restrictions, simplify using valid identities, solve the resulting algebraic or trigonometric statement, substitute candidate values into the original equation, and reject any value outside the allowed domain or principal branch.

Diagram support

Usually not required unless the equation asks for graphical interpretation. A number line for domain restrictions is useful for filtering candidates.

How CBSE asks it

Questions appear as simplify, prove, evaluate, or solve for x, often combining identities, branch restrictions and exact angle values.

Avoid common mistakes

Common confusion

Taking sine, cosine or tangent on both sides and accepting every algebraic solution without checking the original inverse trigonometric equation.

Common wrong answer

Forgetting to check the original equation after squaring or applying a trigonometric function, leading to an extra value of x.

Exam tip

Write the allowed interval for x before solving. The final verification step is often the difference between a correct answer and an extraneous root.

Quick check

Solve sin⁻¹x = cos⁻¹x.

Using sin⁻¹x + cos⁻¹x = π/2, let both be equal. Then 2sin⁻¹x = π/2, so sin⁻¹x = π/4 and x = √2/2.

Answer writing and exam use

1-mark answer

Simplification and equation solving with inverse trigonometric functions means reducing expressions or finding variable values while preserving domain restrictions and principal value conditions.

2-mark answer

Simplification and equation solving with inverse trigonometric functions means reducing expressions or finding variable values while preserving domain restrictions and principal value conditions. Method rules: preserve domains such as x ∈ [−1,1] for sin⁻¹x and cos⁻¹x; use identities only under their conditions; when applying trig functions to both sides, check principal ranges; for equations involving tan⁻¹, verify denominator conditions in sum and difference formulae. If sin⁻¹x = π/6, then x = sin(π/6) = 1/2. Since 1/2 ∈ [−1,1], it is valid.

3-mark answer

Inverse trigonometric equations often look algebraic after applying a trigonometric function to both sides, but this can introduce extra answers. Every candidate must be checked in the original equation and in the domain of each inverse function. Method rules: preserve domains such as x ∈ [−1,1] for sin⁻¹x and cos⁻¹x; use identities only under their conditions; when applying trig functions to both sides, check principal ranges; for equations involving tan⁻¹, verify denominator conditions in sum and difference formulae. Solve 2sin⁻¹x = cos⁻¹x. Since sin⁻¹x + cos⁻¹x = π/2, write cos⁻¹x = π/2 − sin⁻¹x. Then 2sin⁻¹x = π/2 − sin⁻¹x, so 3sin⁻¹x = π/2. Hence sin⁻¹x = π/6 and x = sin(π/6) = 1/2. Check: 2sin⁻¹(1/2) = 2 × π/6 = π/3 and cos⁻¹(1/2) = π/3. Therefore x = 1/2. Questions appear as simplify, prove, evaluate, or solve for x, often combining identities, branch restrictions and exact angle values. Forgetting to check the original equation after squaring or applying a trigonometric function, leading to an extra value of x.
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