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MathsMediumClass 11

Composition — sin(sin⁻¹ x), sin⁻¹(sin x)

Inverse Trigonometric Functions

3

JEE Qs

8%

Hard

75

min

Always prioritize the domain and range restrictions for both the inner and outer functions when dealing with compositions of trigonometric and inverse trigonometric functions.

🧮 Key Formulas

sin(sin⁻¹ x) = x, for x ∈ [-1, 1]
sin⁻¹(sin x) = x, for x ∈ [-pi/2, pi/2]

✅ Key Points for JEE

  • 1For `sin(sin⁻¹ x)`, the expression is defined as `x` ONLY if `x` is within the domain of `sin⁻¹ x`, i.e., `x ∈ [-1, 1]`. Outside this interval, `sin⁻¹ x` is undefined, and hence the composite function is undefined.
  • 2For `sin⁻¹(sin x)`, the output of the inverse sine function must ALWAYS lie within its principal value branch, which is `[-pi/2, pi/2]`. The expression equals `x` only if `x` itself is already in this range.
  • 3When evaluating `sin⁻¹(sin x)` for `x` outside `[-pi/2, pi/2]`, use trigonometric identities (e.g., `sin(π - θ) = sin θ`, `sin(2π + θ) = sin θ`, `sin(-θ) = -sin θ`) to find an equivalent angle `y` such that `sin y = sin x` and `y ∈ [-pi/2, pi/2]`. Then `sin⁻¹(sin x) = y`.
  • 4The graph of `y = sin⁻¹(sin x)` is a periodic, piecewise linear function with a period of `2π`, bounded between `[-pi/2, pi/2]`. Understanding its graph is crucial for solving equations and inequalities.

⚠️ Common Mistakes

  • Blindly assuming `sin(sin⁻¹ x) = x` for any `x` without checking the domain `[-1, 1]`. For `x` outside this domain, `sin(sin⁻¹ x)` is undefined.
  • Blindly assuming `sin⁻¹(sin x) = x` for any `x` without considering the principal value branch `[-pi/2, pi/2]`. This is the most common mistake for this topic.
  • Incorrectly simplifying `sin⁻¹(sin x)` by misapplying trigonometric identities or failing to bring the argument into the principal value branch.

NCERT Chapters

  • Class 11 Maths Ch 3: Trigonometric Functions
  • Class 12 Maths Ch 2: Inverse Trigonometric Functions