Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angle when the value of a trigonometric ratio is known. These functions are the inverses of the standard trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent.

Concept Explanation

For a function \( f: X \to Y \) that is one-to-one and onto, the inverse function \( g: Y \to X \) satisfies \( g = f^{-1} \), with domain and range interchanged:

\[ \text{Domain of } g = \text{Range of } f, \quad \text{Range of } g = \text{Domain of } f \]

Inverse trigonometric functions are defined by restricting the domain of the original trigonometric functions to make them one-to-one and onto, enabling the existence of inverses.

Formula Derivation

For example, the inverse sine function \( y = \sin^{-1} x \) is defined as the inverse of \( y = \sin x \) restricted to \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Thus,

\[ \sin y = x, \quad y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \quad x \in [-1,1] \]

Similarly, inverse cosine and inverse tangent functions are defined with appropriate domain restrictions to ensure invertibility.

Worked Illustrations

Example 1: Find the principal value of \( \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) \).

Solution:

Let \( y = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) \). Then,

\[ \sin y = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \implies y = \frac{\pi}{4} \]

Since \( y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), the principal value is \( \frac{\pi}{4} \).

Solved Examples

Example 2: Show that \( \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11} = \tan^{-1} \frac{3}{4} \).

Solution:

Using the formula for sum of inverse tangents:

\[ \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A + B}{1 - AB} \right) \]

Let \( A = \frac{1}{2} \), \( B = \frac{2}{11} \), then

\[ \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{2}{11} = \tan^{-1} \left( \frac{\frac{1}{2} + \frac{2}{11}}{1 - \frac{1}{2} \times \frac{2}{11}} \right) = \tan^{-1} \left( \frac{\frac{15}{22}}{1 - \frac{1}{11}} \right) = \tan^{-1} \left( \frac{\frac{15}{22}}{\frac{10}{11}} \right) = \tan^{-1} \frac{15}{20} = \tan^{-1} \frac{3}{4} \]

Hence proved.

Practice Set

Level 1 – Easy

• Find \( \sin^{-1} 0 \).
• Evaluate \( \cos^{-1} 1 \).
• Find the principal value of \( \tan^{-1} 1 \).

Level 2 – Moderate

• Prove that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) for \( x \in [-1,1] \).
• Find \( \tan^{-1} 2 + \tan^{-1} 3 \).
• Evaluate \( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \).

Level 3 – Challenging

• Show that \( \tan^{-1} x + \tan^{-1} \frac{1}{x} = \frac{\pi}{2} \) for \( x > 0 \).
• Find the exact value of \( \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{4}{5} \).
• Prove that \( \cos^{-1} x + \cos^{-1} y = \pi \) if \( x^2 + y^2 = 1 \) and \( x,y \in [0,1] \).

Answer Key

• \( \sin^{-1} 0 = 0 \)
• \( \cos^{-1} 1 = 0 \)
• \( \tan^{-1} 1 = \frac{\pi}{4} \)
• \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) (proved by definition)
• \( \tan^{-1} 2 + \tan^{-1} 3 = \tan^{-1} \left( \frac{2+3}{1-2\times3} \right) = \tan^{-1} (-1) = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} \)
• \( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \)
• \( \tan^{-1} x + \tan^{-1} \frac{1}{x} = \frac{\pi}{2} \) for \( x > 0 \) (using tangent addition formula)
• \( \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{4}{5} = \frac{\pi}{2} \)
• \( \cos^{-1} x + \cos^{-1} y = \pi \) if \( x^2 + y^2 = 1 \) (using trigonometric identities)

Quick Reference

Glossary

• Domain: Set of all possible input values for a function.
• Range: Set of all possible output values of a function.
• Principal Value: The unique value of an inverse trigonometric function chosen from its range to make it a function.
• One-to-One Function: A function where each input corresponds to a unique output.
• Inverse Function: A function that reverses the effect of the original function.