In Chapter 1, we studied that the inverse of a function $f$, denoted by $f^{-1}$, exists if and only if $f$ is one-one and onto (bijective). Trigonometric functions are periodic and thus not one-one and onto over their natural domains. However, by restricting their domains, we can define their inverse functions within specific intervals.
The following table summarizes the domains and ranges (principal value branches) of the inverse trigonometric functions:
| Function | Domain | Range (Principal Value Branch) | Key Properties & Limits |
|---|---|---|---|
| $y = \sin^{-1} x$ | $[-1, 1]$ | $[-\frac{\pi}{2}, \frac{\pi}{2}]$ | $\sin(y) = x$ where $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ |
| $y = \cos^{-1} x$ | $[-1, 1]$ | $[0, \pi]$ | $\cos(y) = x$ where $0 \le y \le \pi$ |
| $y = \csc^{-1} x$ | $\mathbb{R} - (-1, 1)$ | $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$ | $\csc(y) = x$ where $y \neq 0$ |
| $y = \sec^{-1} x$ | $\mathbb{R} - (-1, 1)$ | $[0, \pi] - \{\frac{\pi}{2}\}$ | $\sec(y) = x$ where $y \neq \frac{\pi}{2}$ |
| $y = \tan^{-1} x$ | $\mathbb{R}$ | $(-\frac{\pi}{2}, \frac{\pi}{2})$ | $\tan(y) = x$ where $-\frac{\pi}{2} < y < \frac{\pi}{2}$ |
| $y = \cot^{-1} x$ | $\mathbb{R}$ | $(0, \pi)$ | $\cot(y) = x$ where $0 < y < \pi$ |
The graph of an inverse function $y = f^{-1}(x)$ can be obtained from the graph of the original function $y = f(x)$ by interchanging the $x$ and $y$ axes. This is geometrically equivalent to reflecting the graph of $y = f(x)$ across the mirror line $y = x$.
The following properties and identities are valid within the principal value branches of the respective inverse trigonometric functions:
Mathematical Problem Statement: Find the principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$.
Step-by-Step Algorithmic Solution Trace:
- Let $y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$. This $\implies \sin y = \frac{1}{\sqrt{2}}$.
- We know that the principal value branch of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
- Since $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the principal value is $\frac{\pi}{4}$.
Final Evaluation Output: $\frac{\pi}{4}$
Mathematical Problem Statement: Find the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$.
Step-by-Step Algorithmic Solution Trace:
- Let $y = \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$. This $\implies \cot y = -\frac{1}{\sqrt{3}}$.
- We know that the principal value branch of $\cot^{-1}$ is $(0, \pi)$.
- Since $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$, we have $\cot\left(\pi - \frac{\pi}{3}\right) = \cot\left(\frac{2\pi}{3}\right) = -\frac{1}{\sqrt{3}}$.
- Since $\frac{2\pi}{3} \in (0, \pi)$, the principal value is $\frac{2\pi}{3}$.
Final Evaluation Output: $\frac{2\pi}{3}$
Mathematical Problem Statement: Show that $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1} x$ for $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.
Step-by-Step Algorithmic Solution Trace:
- Let $x = \sin \theta$. Then $\theta = \sin^{-1} x$.
- Substitute $x$ into the expression: $2x\sqrt{1-x^2} = 2\sin\theta\sqrt{1-\sin^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta$.
- Thus, $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1} x$.
Final Evaluation Output: $2\sin^{-1} x$
Mathematical Problem Statement: Express $\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$ in the simplest form, where $-\frac{3\pi}{2} < x < \frac{\pi}{2}$.
Step-by-Step Algorithmic Solution Trace:
- Express $\cos x$ as $\cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right)$ and $1 - \sin x$ as $\left(\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\right)^2$.
- The fraction becomes: $\frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2} = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$.
- Divide numerator and denominator by $\cos\frac{x}{2}$ to get: $\frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$.
- Therefore, $\tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) = \frac{\pi}{4} + \frac{x}{2}$.
Final Evaluation Output: $\frac{\pi}{4} + \frac{x}{2}$
Mathematical Problem Statement: Find the value of $\sin^{-1}\left(\sin \frac{3\pi}{5}\right)$.
Step-by-Step Algorithmic Solution Trace:
- Note that $\frac{3\pi}{5} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$, which is the principal value branch of $\sin^{-1}$.
- Rewrite the expression using the identity $\sin(\pi - \theta) = \sin\theta$: $\sin\left(\frac{3\pi}{5}\right) = \sin\left(\pi - \frac{2\pi}{5}\right) = \sin\left(\frac{2\pi}{5}\right)$.
- Since $\frac{2\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\sin^{-1}\left(\sin \frac{2\pi}{5}\right) = \frac{2\pi}{5}$.
Final Evaluation Output: $\frac{2\pi}{5}$
Find the principal values of the following:
Key problems include proving algebraic identities and simplifying expressions:
Historical Chronicle / Epigraph: Indian Trigonometry & Global Transmission