Inverse Trigonometric Functions: Class 12 Maths NCERT Guide

Understanding Inverse Trigonometric Functions

To understand inverse trigonometric functions (ITFs), let's first recall what an inverse function is. For a function f(x) to have an inverse, it must be both one-to-one (injective) and onto (surjective). However, standard trigonometric functions like sin(x), cos(x), etc., are periodic and thus many-to-one, meaning multiple x values map to the same y value. For instance, sin(0) = 0 and sin(π) = 0. This makes them not directly invertible over their entire domain.

To overcome this, we restrict the domain of each trigonometric function to an interval where it becomes one-to-one and onto. This restricted domain is called the principal value branch. Within this branch, the function is bijective, allowing us to define its inverse. The output of an inverse trigonometric function is always an angle, specifically the angle within its principal value branch. For example, sin⁻¹(x) (read as 'arcsin x' or 'sine inverse x') gives the angle y such that sin(y) = x and y lies within [-π/2, π/2]. It's crucial to remember that sin⁻¹(x) is not the same as (sin x)⁻¹ or 1/sin x. These distinctions are fundamental to correctly working with ITFs.

Definitions and Principal Value Branches

Sine Inverse (arcsin x)

If sin y = x, then y = sin⁻¹ x. The domain of sin⁻¹ x is [-1, 1] and its principal value branch (range) is [-π/2, π/2].

Cosine Inverse (arccos x)

If cos y = x, then y = cos⁻¹ x. The domain of cos⁻¹ x is [-1, 1] and its principal value branch (range) is [0, π].

Tangent Inverse (arctan x)

If tan y = x, then y = tan⁻¹ x. The domain of tan⁻¹ x is (-∞, ∞) (all real numbers) and its principal value branch (range) is (-π/2, π/2).

Cosecant Inverse (arccosec x)

If cosec y = x, then y = cosec⁻¹ x. The domain of cosec⁻¹ x is R - (-1, 1) and its principal value branch (range) is [-π/2, π/2] - {0}.

Secant Inverse (arcsec x)

If sec y = x, then y = sec⁻¹ x. The domain of sec⁻¹ x is R - (-1, 1) and its principal value branch (range) is [0, π] - {π/2}.

Cotangent Inverse (arccot x)

If cot y = x, then y = cot⁻¹ x. The domain of cot⁻¹ x is (-∞, ∞) and its principal value branch (range) is (0, π).

Worked Examples on Principal Values

• Example 1: Find the principal value of sin⁻¹(1/2). Step 1: Let y = sin⁻¹(1/2). By definition, sin y = 1/2. Step 2: We need to find an angle y such that sin y = 1/2 and y lies in the principal value branch of sin⁻¹ x, which is [-π/2, π/2]. Step 3: We know that sin(π/6) = 1/2. Step 4: Since π/6 is in [-π/2, π/2], the principal value of sin⁻¹(1/2) is π/6. Final answer: π/6
• Example 2: Find the principal value of cos⁻¹(-√3/2). Step 1: Let y = cos⁻¹(-√3/2). By definition, cos y = -√3/2. Step 2: We need to find an angle y such that cos y = -√3/2 and y lies in the principal value branch of cos⁻¹ x, which is [0, π]. Step 3: We know that cos(π/6) = √3/2. Since cos y is negative, y must be in the second quadrant. The angle in the second quadrant with a reference angle of π/6 is π - π/6 = 5π/6. Step 4: cos(5π/6) = -√3/2. Step 5: Since 5π/6 is in [0, π], the principal value of cos⁻¹(-√3/2) is 5π/6. Final answer: 5π/6
• Example 3: Evaluate tan⁻¹(1) + cos⁻¹(-1/2). Step 1: Find the principal value of tan⁻¹(1). Let A = tan⁻¹(1). Then tan A = 1. The principal value branch for tan⁻¹ x is (-π/2, π/2). We know tan(π/4) = 1, and π/4 is in (-π/2, π/2). So, A = π/4. Step 2: Find the principal value of cos⁻¹(-1/2). Let B = cos⁻¹(-1/2). Then cos B = -1/2. The principal value branch for cos⁻¹ x is [0, π]. We know cos(π/3) = 1/2. Since cos B is negative, B must be in the second quadrant. So, B = π - π/3 = 2π/3. Step 3: Add the principal values: tan⁻¹(1) + cos⁻¹(-1/2) = A + B = π/4 + 2π/3. Step 4: Find a common denominator: (3π + 8π) / 12 = 11π/12. Final answer: 11π/12

Important Properties and Exam Tips

• Key Properties to Remember: sin⁻¹(x) + cos⁻¹(x) = π/2, for x ∈ [-1, 1] tan⁻¹(x) + cot⁻¹(x) = π/2, for x ∈ R sec⁻¹(x) + cosec⁻¹(x) = π/2, for x ∈ R - (-1, 1) tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)), if xy < 1 2tan⁻¹(x) = sin⁻¹(2x/(1+x²)), if |x| ≤ 1 2tan⁻¹(x) = cos⁻¹((1-x²)/(1+x²)), if x ≥ 0 * 2tan⁻¹(x) = tan⁻¹(2x/(1-x²)), if -1 < x < 1
• Principal Value Branch is Crucial: Always ensure your final answer for an inverse trigonometric function lies within its specified principal value range. Failing to do so is a very common mistake in exams.
• Domain Restrictions: Be mindful of the domain of each inverse trigonometric function. For example, sin⁻¹(2) is undefined because 2 is outside the domain [-1, 1]. Questions often test this understanding.
• Signs Matter: When evaluating cos⁻¹(-x), sec⁻¹(-x), or cot⁻¹(-x), remember the property: cos⁻¹(-x) = π - cos⁻¹(x). For sin⁻¹(-x), tan⁻¹(-x), cosec⁻¹(-x), the property is f⁻¹(-x) = -f⁻¹(x). This is essential for getting the correct principal value.
• Simplify Before Evaluating: If an expression involves nested functions like sin(tan⁻¹(x)), it's often helpful to form a right-angled triangle using the inner inverse function to find the values of other trigonometric ratios.

Practice Questions with Solutions

• Q: Find the principal value of cosec⁻¹(-√2). A: Step 1: Let y = cosec⁻¹(-√2). Then cosec y = -√2. Step 2: The principal value branch for cosec⁻¹ x is [-π/2, π/2] - {0}. Step 3: We know cosec(π/4) = √2. Since cosec y is negative, y must be in the fourth quadrant. The angle is -π/4. Step 4: Since -π/4 is in [-π/2, π/2] - {0}, the principal value is -π/4. Final answer: -π/4
• Q: Prove that 3sin⁻¹x = sin⁻¹(3x - 4x³) for x ∈ [-1/2, 1/2]. A: Step 1: Let x = sinθ. Then θ = sin⁻¹x. The condition x ∈ [-1/2, 1/2] implies θ ∈ [-π/6, π/6]. Step 2: Substitute x = sinθ into the RHS: sin⁻¹(3sinθ - 4sin³θ). Step 3: Recognize the trigonometric identity: sin(3θ) = 3sinθ - 4sin³θ. Step 4: The expression becomes sin⁻¹(sin(3θ)). Step 5: Since θ ∈ [-π/6, π/6], then 3θ ∈ [-π/2, π/2]. This interval is the principal value branch of sin⁻¹. Therefore, sin⁻¹(sin(3θ)) = 3θ. Step 6: Substitute back θ = sin⁻¹x. So, RHS = 3sin⁻¹x. Step 7: LHS = RHS. Hence proved. Final answer: Proof shown above.
• Q: Evaluate tan⁻¹(√3) - sec⁻¹(-2). A: Step 1: Find the principal value of tan⁻¹(√3). Let A = tan⁻¹(√3). Then tan A = √3. The principal value branch for tan⁻¹ x is (-π/2, π/2). We know tan(π/3) = √3, and π/3 is in (-π/2, π/2). So, A = π/3. Step 2: Find the principal value of sec⁻¹(-2). Let B = sec⁻¹(-2). Then sec B = -2. The principal value branch for sec⁻¹ x is [0, π] - {π/2}. We know sec(π/3) = 2. Since sec B is negative, B must be in the second quadrant. So, B = π - π/3 = 2π/3. Step 3: Subtract the values: tan⁻¹(√3) - sec⁻¹(-2) = A - B = π/3 - 2π/3. Step 4: π/3 - 2π/3 = -π/3. Final answer: -π/3
• Q: Simplify tan⁻¹( (cos x - sin x) / (cos x + sin x) ), where x ∈ (-π/4, π/4). A: Step 1: Divide the numerator and denominator by cos x (since cos x ≠ 0 in the given interval (-π/4, π/4)). tan⁻¹( (1 - tan x) / (1 + tan x) ) Step 2: Recall the formula for tan(A - B) = (tan A - tan B) / (1 + tan A tan B). We know tan(π/4) = 1. Step 3: Substitute 1 = tan(π/4) into the expression: tan⁻¹( (tan(π/4) - tan x) / (1 + tan(π/4)tan x) ) Step 4: This simplifies to tan⁻¹(tan(π/4 - x)). Step 5: Given x ∈ (-π/4, π/4), we have -π/4 < x < π/4. Multiplying by -1 and reversing inequality signs, -π/4 < -x < π/4. Adding π/4 to all parts: 0 < π/4 - x < π/2. Step 6: Since π/4 - x lies in (0, π/2), which is within the principal value branch (-π/2, π/2) of tan⁻¹, we have tan⁻¹(tan(π/4 - x)) = π/4 - x. Final answer: π/4 - x

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