Inverse Trigonometric Functions
Inverse Trigonometry
Class 12 Math Complete Revision Kit
1. Concepts & Logic
The Golden Rule: Principal Value Branch
The inverse trigonometric functions are multivalued, but for functions to be defined, we restrict their domain and range. The restricted range is called the Principal Value Branch.
[Image of Unit Circle for Inverse Trigonometry]
Crucial Check: Always ensure your final answer $\theta$ lies strictly within the Principal Range defined below.
| Function Group | Range (Principal Value) | Memory Aid |
|---|---|---|
| $\sin^{-1}, \csc^{-1}, \tan^{-1}$ | $[-\frac{\pi}{2}, \frac{\pi}{2}]$ family | 1st & 4th Quadrant |
| $\cos^{-1}, \sec^{-1}, \cot^{-1}$ | $[0, \pi]$ family | 1st & 2nd Quadrant |
Handling Negative Inputs
How to solve $\text{fn}^{-1}(-x)$? There are only two rules.
Type A: The "Direct" Group
The minus sign simply comes outside.
$\sin^{-1}(-x) = -\sin^{-1}x$
$\tan^{-1}(-x) = -\tan^{-1}x$
$\csc^{-1}(-x) = -\csc^{-1}x$
$\tan^{-1}(-x) = -\tan^{-1}x$
$\csc^{-1}(-x) = -\csc^{-1}x$
Type B: The "Pi-Minus" Group
Subtract the value from $\pi$.
$\cos^{-1}(-x) = \pi - \cos^{-1}x$
$\sec^{-1}(-x) = \pi - \sec^{-1}x$
$\cot^{-1}(-x) = \pi - \cot^{-1}x$
$\sec^{-1}(-x) = \pi - \sec^{-1}x$
$\cot^{-1}(-x) = \pi - \cot^{-1}x$
Calculus Substitutions
Use these substitutions to simplify complex inverse expressions in differentiation or integration.
| Expression Form | Recommended Substitution ($x=$) |
|---|---|
| $\sqrt{a^2 - x^2}$ | $a\sin\theta$ or $a\cos\theta$ |
| $\sqrt{a^2 + x^2}$ | $a\tan\theta$ or $a\cot\theta$ |
| $\sqrt{x^2 - a^2}$ | $a\sec\theta$ or $a\csc\theta$ |
| $\sqrt{\frac{a-x}{a+x}}$ or $\sqrt{\frac{a+x}{a-x}}$ | $a\cos 2\theta$ |
2. Formula Sheet
1. Complete Domain & Range Table
| Function ($y$) | Domain ($x$) | Range (Principal Value Branch) |
|---|---|---|
| $y = \sin^{-1}x$ | $[-1, 1]$ | $[-\frac{\pi}{2}, \frac{\pi}{2}]$ |
| $y = \cos^{-1}x$ | $[-1, 1]$ | $[0, \pi]$ |
| $y = \tan^{-1}x$ | $\mathbb{R}$ | $(-\frac{\pi}{2}, \frac{\pi}{2})$ |
| $y = \cot^{-1}x$ | $\mathbb{R}$ | $(0, \pi)$ |
| $y = \sec^{-1}x$ | $\mathbb{R} - (-1, 1)$ | $[0, \pi] - \{\frac{\pi}{2}\}$ |
| $y = \csc^{-1}x$ | $\mathbb{R} - (-1, 1)$ | $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$ |
2. Key Identities & Properties
A. Reciprocal Properties
$\sin^{-1}\frac{1}{x} = \csc^{-1}x$
$\cos^{-1}\frac{1}{x} = \sec^{-1}x$
$\tan^{-1}\frac{1}{x} = \cot^{-1}x \quad (x>0)$
$\cos^{-1}\frac{1}{x} = \sec^{-1}x$
$\tan^{-1}\frac{1}{x} = \cot^{-1}x \quad (x>0)$
B. Co-function Properties
$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$
$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$
$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$
C. Sum & Difference Formulas (Tan Inverse)
$$ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad \text{if } xy < 1 $$
$$ \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \quad \text{if } xy > -1 $$
D. Conversion of $2\tan^{-1}x$
This is the "bridge" formula connecting tan inverse to sin and cos inverse.
$$ 2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) $$
3. Triple Angle Formulas
$$ 3\sin^{-1}x = \sin^{-1}(3x - 4x^3) $$
$$ 3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) $$
$$ 3\tan^{-1}x = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) $$
3. Solved Examples
Type 1: Basic Principal Values
1. Find the principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$
We know that $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Check: Does $\frac{\pi}{4}$ lie in $[-\frac{\pi}{2}, \frac{\pi}{2}]$? Yes.
Answer: $\frac{\pi}{4}$
2. Find the principal value of $\cot^{-1}(\sqrt{3})$
We know that $\cot(\frac{\pi}{6}) = \sqrt{3}$.
Check: Does $\frac{\pi}{6}$ lie in $(0, \pi)$? Yes.
Answer: $\frac{\pi}{6}$
Type 2: Negative Arguments (Most Important)
Remember: Sin/Tan/Csc output negative directly. Cos/Sec/Cot become $\pi - \theta$.
3. Find $\tan^{-1}(-1)$
Property: $\tan^{-1}(-x) = -\tan^{-1}x$
$\Rightarrow -\tan^{-1}(1)$
Since $\tan(\frac{\pi}{4}) = 1$, we get $-\frac{\pi}{4}$.
Answer: $-\frac{\pi}{4}$
4. Find $\cos^{-1}\left(-\frac{1}{2}\right)$
Property: $\cos^{-1}(-x) = \pi - \cos^{-1}x$
$\Rightarrow \pi - \cos^{-1}(\frac{1}{2})$
$\Rightarrow \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
Answer: $\frac{2\pi}{3}$
5. Find $\sec^{-1}\left(-\frac{2}{\sqrt{3}}\right)$
Property: $\sec^{-1}(-x) = \pi - \sec^{-1}x$
$\Rightarrow \pi - \sec^{-1}(\frac{2}{\sqrt{3}})$
$\Rightarrow \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
Answer: $\frac{5\pi}{6}$
Type 3: Complex Evaluation
6. Evaluate $\sin^{-1}\left(\sin \frac{2\pi}{3}\right)$
Warning $\frac{2\pi}{3}$ is outside the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Convert: $\sin \frac{2\pi}{3} = \sin(\pi - \frac{\pi}{3}) = \sin \frac{\pi}{3}$.
Solve: $\sin^{-1}(\sin \frac{\pi}{3}) = \frac{\pi}{3}$.
Answer: $\frac{\pi}{3}$
7. Evaluate $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$
Part A: $\tan^{-1}(1) = \frac{\pi}{4}$
Part B: $\cos^{-1}(-\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
Part C: $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$
Combine: $\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12}$
Answer: $\frac{3\pi}{4}$
4. Visual Graphs
Interactive Graphs
Hover over the lines to see exact coordinate values. Note how the graphs are restricted to their principal branches.