Inverse Trigonometric Functions Class 12 Chapter 2 Revision Notes

Need for Restricting Domains

To define the inverse of trigonometric functions, we must restrict their domains because trigonometric functions are periodic and hence not one-to-one (injective) over their entire natural domains. By restricting their domains to specific intervals, we make them bijective (one-to-one and onto), allowing their unique inverses to exist. The branch of an inverse trigonometric function that corresponds to this restricted range is called the Principal Value Branch. For any given real value in the domain, the unique angle we obtain lying within this principal value branch is the Principal Value of that inverse trigonometric function. Mastery over these restricted intervals is crucial because writing an angle outside the principal branch in CBSE exams will lead to a complete loss of marks.

Domain and Range (Principal Value Branches)

Important Terminology

Bijective Function

A function that is both one-to-one (injective) and onto (surjective). An inverse function exists only if the function is bijective.

Inverse Trigonometric Function

The inverse of a restricted trigonometric function, written as sin⁻¹x, cos⁻¹x, etc., which yields an angle for a given ratio.

Principal Value Branch

The standard restricted range of an inverse trigonometric function where it yields unique, mathematically consistent values.

Principal Value

The unique value of an inverse trigonometric function which lies within its designated principal value branch.

Domain of Inverse Function

The set of all valid real numbers (inputs) for which the corresponding inverse trigonometric function is defined.

Key Properties & Formulas

• Negative Angle Property (Odd-like): sin⁻¹(-x) = -sin⁻¹(x), cosec⁻¹(-x) = -cosec⁻¹(x), tan⁻¹(-x) = -tan⁻¹(x).
• Negative Angle Property (Pi-subtraction): cos⁻¹(-x) = π - cos⁻¹(x), sec⁻¹(-x) = π - sec⁻¹(x), cot⁻¹(-x) = π - cot⁻¹(x).
• Reciprocal Identity: sin⁻¹(1/x) = cosec⁻¹(x) for |x| ≥ 1; and cos⁻¹(1/x) = sec⁻¹(x) for |x| ≥ 1.
• Complementary Angle Relations: sin⁻¹(x) + cos⁻¹(x) = π/2 for x ∈ [-1, 1].
• Complementary Angle Relations (tan & cot): tan⁻¹(x) + cot⁻¹(x) = π/2 for x ∈ R.
• Complementary Angle Relations (sec & cosec): sec⁻¹(x) + cosec⁻¹(x) = π/2 for |x| ≥ 1.
• Cancellation Rules: sin⁻¹(sin θ) = θ ONLY when θ ∈ [-π/2, π/2]. Similarly, cos⁻¹(cos θ) = θ ONLY when θ ∈ [0, π].

Step-by-Step Solved Examples

• {"title":"Example 1: Finding Principal Value of Negative Argument","description":"Find the principal value of sin⁻¹(-1/2).","steps":["Let y = sin⁻¹(-1/2). This implies sin(y) = -1/2.","We know that sin(π/6) = 1/2. Since sin(-θ) = -sin(θ), we get sin(-π/6) = -1/2.","The principal value branch of sin⁻¹ is [-π/2, π/2].","Since -π/6 lies within [-π/2, π/2], the principal value of sin⁻¹(-1/2) is -π/6."]}
• {"title":"Example 2: Angle Outside the Principal Range","description":"Evaluate cos⁻¹(cos (13π/6)).","steps":["We cannot directly write 13π/6 because 13π/6 does not lie in the principal value branch [0, π] of cos⁻¹.","Rewrite the angle: 13π/6 = 2π + π/6.","Since cos(2π + θ) = cos(θ), we have cos(13π/6) = cos(π/6).","Thus, cos⁻¹(cos(13π/6)) = cos⁻¹(cos(π/6)) = π/6, which lies within [0, π]."]}
• {"title":"Example 3: tan⁻¹ and cot⁻¹ Relation","description":"Evaluate tan⁻¹(tan (3π/4)).","steps":["The principal value branch of tan⁻¹ is (-π/2, π/2). The angle 3π/4 does not lie in this interval.","Rewrite the argument: tan(3π/4) = tan(π - π/4) = -tan(π/4) = tan(-π/4).","Therefore, tan⁻¹(tan(3π/4)) = tan⁻¹(tan(-π/4)) = -π/4, which lies within (-π/2, π/2)."]}

Board Exam Trap Alerts

1. Check the Domain First: Before applying the formula $sin^{-1}(sin(x)) = x$, always verify if $x$ belongs to $[-\pi/2, \pi/2]$. If it lies outside, you must use trigonometric reduction formulas (like $\pi - x$ or $2\pi - x$) to bring it into the principal branch.
2. Cos-Family Negatives: Never write $cos^{-1}(-x) = -cos^{-1}(x)$. Remember that for $cos$, $sec$, and $cot$, the negative argument requires subtraction from $\pi$: $cos^{-1}(-x) = \pi - cos^{-1}(x)$. This is a high-frequency source of errors in 2-mark questions.

Practice Questions with Solutions

• What is the principal value of cos⁻¹(-1/√2)? Let cos⁻¹(-1/√2) = y. Then cos(y) = -1/√2. Since cos(π/4) = 1/√2, cos⁻¹(-1/√2) = π - π/4 = 3π/4, which lies in [0, π].
• State the domain and range of sec⁻¹(x). The domain of sec⁻¹(x) is R - (-1, 1) or (-∞, -1] ∪ [1, ∞). Its range (principal value branch) is [0, π] - {π/2}.
• Evaluate: sin[π/3 - sin⁻¹(-1/2)] We know sin⁻¹(-1/2) = -π/6. Substituting this: sin[π/3 - (-π/6)] = sin[π/3 + π/6] = sin(π/2) = 1.
• Find the value of tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2). Using the properties: tan⁻¹(1) = π/4. Also, cos⁻¹(-1/2) + sin⁻¹(-1/2) = π/2 (since sin⁻¹x + cos⁻¹x = π/2). Thus, the sum is π/4 + π/2 = 3π/4.

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