Class 12 - Inverse Trigonometric Functions

1. Introduction

  • Definition: Recall that a function ff has an inverse f−1f^{-1} if and only if it is one-one and onto (bijective).

  • The Problem with Trig Functions: Trigonometric functions are periodic and thus not one-one over their entire natural domains.

  • The Solution: By restricting the domains of trigonometric functions, we can make them one-one and onto, allowing us to define their inverses.

  • Applications: Used extensively in Calculus (integration), science, and engineering.

2. Basic Concepts

  • Inversion of Functions:

    • If y=sin⁡xy = \sin x, then x=sin⁡−1yx = \sin^{-1} y (provided xx is in the restricted domain).

    • Note: sin⁡−1x\sin^{-1} x is an angle, not to be confused with (sin⁡x)−1(\sin x)^{-1} or 1sin⁡x\frac{1}{\sin x}.

  • Principal Value Branch: The specific range of an inverse trigonometric function where it is defined as a unique function.

  • Domain and Range Table: y=sin⁡−1xy = \sin^{-1} x | [−1,1][-1, 1] | [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] | y=cos⁡−1xy = \cos^{-1} x | [−1,1][-1, 1] | [0,π][0, \pi] | y=tan⁡−1xy = \tan^{-1} x | R\mathbb{R} | (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) | y=cosec−1xy = \text{cosec}^{-1} x | R−(−1,1)\mathbb{R} - (-1, 1) | [−π2,π2]−{0}[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\} | y=sec⁡−1xy = \sec^{-1} x | R−(−1,1)\mathbb{R} - (-1, 1) | [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\} | y=cot⁡−1xy = \cot^{-1} x | R\mathbb{R} | (0,π)(0, \pi) |

  • Principal Value: The value of an inverse trigonometric function that lies within its principal value branch.

3. Properties of Inverse Trigonometric Functions

  • Self-Inverse Properties:

    • sin⁡(sin⁡−1x)=x, for x∈[−1,1]\sin(\sin^{-1} x) = x, \text{ for } x \in [-1, 1]

    • sin⁡−1(sin⁡x)=x, for x∈[−π2,π2]\sin^{-1}(\sin x) = x, \text{ for } x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

  • Reciprocal Identities:

    • sin⁡−1(1x)=cosec−1x,x≥1\sin^{-1}(\frac{1}{x}) = \text{cosec}^{-1} x, x \geq 1 or x≤−1x \leq -1

    • cos⁡−1(1x)=sec⁡−1x,x≥1\cos^{-1}(\frac{1}{x}) = \sec^{-1} x, x \geq 1 or x≤−1x \leq -1

    • tan⁡−1(1x)=cot⁡−1x,x>0\tan^{-1}(\frac{1}{x}) = \cot^{-1} x, x > 0

  • Negative Argument Identities:

    • sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1} x

    • tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1} x

    • cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1} x

  • Complementary Angle Identities:

    • sin⁡−1x+cos⁡−1x=π2,x∈[−1,1]\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, x \in [-1, 1]

    • tan⁡−1x+cot⁡−1x=π2,x∈R\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}, x \in \mathbb{R}

    • cosec−1x+sec⁡−1x=π2,∣x∣≥1\text{cosec}^{-1} x + \sec^{-1} x = \frac{\pi}{2}, |x| \geq 1

  • Sum and Difference of tan⁡−1\tan^{-1}:

    • \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), xy < 1

    • tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy),xy>−1\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right), xy > -1

  • Double Angle Conversions:

    • 2tan⁡−1x=sin⁡−1(2x1+x2)=cos⁡−1(1−x21+x2)=tan⁡−1(2x1−x2)2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)

4. Graphical Representations

  • Understanding the visual behavior of inverse functions (e.g., how the graph of y=sin⁡−1xy = \sin^{-1} x is the reflection of y=sin⁡xy = \sin x over the line y=xy = x within the restricted interval).

5. Historical Note

  • The study of trigonometry originated in India with ancient mathematicians like Aryabhata (476 A.D.), Brahmagupta, and Bhaskara I.

Quick Summary for Revision

  1. Always check the range: When calculating sin⁡−1(x)\sin^{-1}(x), ensure your answer is within [−π/2,π/2][-\pi/2, \pi/2].

  2. Conversion: You can convert any inverse trig function to another using right-angled triangle properties (Pythagoras theorem).

  3. Simplification: Use the tan⁡−1\tan^{-1} properties for most proof-based problems as they are often the most straightforward to apply.

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