Inverse Trigonometric Functions revision notes

Key Concepts & Definitions

Inverse of a Function: The inverse of a function $f$, denoted by $f^{-1}$, exists only if $f$ is a bijective function (both one-one and onto). If $f: X \rightarrow Y$ such that $f(x) = y$ is one-one and onto, we can define a unique function $g: Y \rightarrow X$ such that $g(y) = x$. Here, the domain of $g$ is the range of $f$, and the range of $g$ is the domain of $f$. Also, $g^{-1} = (f^{-1})^{-1} = f$.

Trigonometric Functions and Bijections: Standard trigonometric functions are not one-one and onto over their natural domains. To ensure the existence of their inverses, their domains and ranges must be restricted so that they become bijective.

Principal Value Branch: When restricting the domains of trigonometric functions to make them bijective, infinitely many intervals are possible. The standard, universally accepted restricted interval is called the principal value branch. The value of an inverse trigonometric function that lies in its principal value branch is known as its principal value.

Notation Convention: $\sin^{-1}x$ denotes the inverse sine function (arc sine function). [JEE TIP] Do not confuse $\sin^{-1} x$ with $(\sin x)^{-1}$. The latter means $\frac{1}{\sin x}$, which is $\text{cosec } x$. The notation using $-1$ as a superscript for inverse was suggested by astronomer Sir John F.W. Herschel in 1813.

Historical Context: The study of trigonometry originated in India with mathematicians like Aryabhata, Brahmagupta, Bhaskara I, and Bhaskara II. Thales is credited with early height and distance calculations using shadows.

Domain, Range & Principal Value Branches

The following table dictates the strictly defined domains and ranges (principal value branches) of inverse trigonometric functions. [JEE TIP] Memorize this table perfectly; nearly all JEE Advanced range and domain restriction questions stem from here.

Important Graphs & Graphical Transformations

General Transformation: 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. Visually, this is the mirror image (reflection) of the original graph along the line $y = x$.

Graphs of Inverse Trigonometric Functions:

• $y = \sin^{-1} x$: Domain $[-1, 1]$, strictly increasing from $-\pi/2$ to $\pi/2$. Point of inflection at origin.
• $y = \cos^{-1} x$: Domain $[-1, 1]$, strictly decreasing from $\pi$ to $0$. Crosses y-axis at $\pi/2$.
• $y = \tan^{-1} x$: Domain $R$, strictly increasing. Horizontal asymptotes at $y = \pi/2$ and $y = -\pi/2$. Passes through origin.
• $y = \cot^{-1} x$: Domain $R$, strictly decreasing. Horizontal asymptotes at $y = \pi$ and $y = 0$. Crosses y-axis at $\pi/2$.
• $y = \sec^{-1} x$: Domain $R - (-1, 1)$. Increasing in $(-\infty, -1]$ and $[1, \infty)$. Horizontal asymptote at $y = \pi/2$.
• $y = \text{cosec}^{-1} x$: Domain $R - (-1, 1)$. Decreasing in $(-\infty, -1]$ and $[1, \infty)$. Horizontal asymptote at $y = 0$.

[JEE TIP] Graphs of Self-Inverse Compositions (Sawtooth & Triangle Waves): These graphs are paramount for JEE Advanced area under curve and continuity/differentiability questions:

• $y = \sin^{-1}(\sin x)$: A continuous zig-zag (triangle wave) passing through the origin. Domain $R$, Range $[-\pi/2, \pi/2]$. Period is $2\pi$. Slope is alternately $+1$ and $-1$.

• $y = \cos^{-1}(\cos x)$: A continuous triangular wave starting at $(0,0)$ and peaking at $(\pi, \pi)$. Domain $R$, Range $[0, \pi]$. Period is $2\pi$.

• $y = \tan^{-1}(\tan x)$: Parallel line segments of slope $+1$ with points of discontinuity (open circles) at odd multiples of $\pi/2$. Domain $R - \{(2n+1)\pi/2\}$, Range $(-\pi/2, \pi/2)$. Period is $\pi$.

Formulae, Equations & Properties of ITF

All properties of inverse trigonometric functions are strictly valid only within their defined domains and principal value branches.

1. Self-Cancelling Properties

• $f(f^{-1}(x)) = x$:
  • $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$
  • $\cos(\cos^{-1} x) = x$ for $x \in [-1, 1]$
  • $\tan(\tan^{-1} x) = x$ for $x \in R$
• $f^{-1}(f(x)) = x$:
  • $\sin^{-1}(\sin x) = x$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
  • $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$
  • $\tan^{-1}(\tan x) = x$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$JEE TIPIf $x$ is outside these principal intervals, use the periodic and symmetric properties of trigonometric functions to reduce the angle into the principal branch before cancelling.

2. Negative Argument Properties (Odd/Even Analogs)

• $\sin^{-1}(-x) = -\sin^{-1} x$, for $x \in [-1, 1]$
• $\tan^{-1}(-x) = -\tan^{-1} x$, for $x \in R$
• $\text{cosec}^{-1}(-x) = -\text{cosec}^{-1} x$, for $|x| \ge 1$
• $\cos^{-1}(-x) = \pi - \cos^{-1} x$, for $x \in [-1, 1]$
• $\cot^{-1}(-x) = \pi - \cot^{-1} x$, for $x \in R$
• $\sec^{-1}(-x) = \pi - \sec^{-1} x$, for $|x| \ge 1$JEE TIPThe $\pi -$ adjustment for $\cos^{-1}, \cot^{-1}$, and $\sec^{-1}$ is heavily tested. Forgetting the $\pi$ leads to answers in the wrong quadrant.

3. Reciprocal Properties

• $\sin^{-1}\left(\frac{1}{x}\right) = \text{cosec}^{-1} x$, for $|x| \ge 1$
• $\cos^{-1}\left(\frac{1}{x}\right) = \sec^{-1} x$, for $|x| \ge 1$
• $\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x$, for $x > 0$
• $\tan^{-1}\left(\frac{1}{x}\right) = -\pi + \cot^{-1} x$, for $x < 0$JEE TIPThis split condition for $\tan^{-1}(1/x)$ based on the sign of $x$ is a notorious JEE trap!

4. Complementary Angles Properties

• $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, for $x \in [-1, 1]$
• $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, for $x \in R$
• $\sec^{-1} x + \text{cosec}^{-1} x = \frac{\pi}{2}$, for $|x| \ge 1$

5. Sum and Difference Formulas

[JEE TIP] Always evaluate the product $xy$ before applying these.

• $\tan^{-1} x + \tan^{-1} y$:
  • $= \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $xy < 1$
  • $= \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $x > 0, y > 0, xy > 1$
  • $= -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $x < 0, y < 0, xy > 1$
  • $= \frac{\pi}{2}$ if $x>0, y>0, xy=1$
• $\tan^{-1} x - \tan^{-1} y$:
  • $= \tan^{-1}\left(\frac{x-y}{1+xy}\right)$, if $xy > -1$
• $\sin^{-1} x \pm \sin^{-1} y = \sin^{-1}\left(x\sqrt{1-y^2} \pm y\sqrt{1-x^2}\right)$
  • Applicable directly when $x, y \ge 0$ and $x^2 + y^2 \le 1$. If $x^2 + y^2 > 1$, subtract from $\pi$.
• $\cos^{-1} x \pm \cos^{-1} y = \cos^{-1}\left(xy \mp \sqrt{1-x^2}\sqrt{1-y^2}\right)$
  • Applicable when $x, y \ge 0$.

6. Multiple Angle Formulas (Domain Restricted)

• $2\tan^{-1} x$ Conversions:
  • $= \sin^{-1}\left(\frac{2x}{1+x^2}\right)$, valid for $|x| \le 1$
  • $= \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, valid for $x \ge 0$
  • $= \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, valid for $|x| < 1$
• $2\sin^{-1} x$ and $3\sin^{-1} x$:
  • $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1} x$, for $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$
  • $\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1} x$, for $\frac{1}{\sqrt{2}} \le x \le 1$
  • $3\sin^{-1} x = \sin^{-1}(3x - 4x^3)$, for $x \in [-\frac{1}{2}, \frac{1}{2}]$
• $3\cos^{-1} x$:
  • $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)$, for $x \in [\frac{1}{2}, 1]$

Standard Derivations & Step-by-Step Problem Solving

Simplifying Complex Inverse Expressions

When simplifying expressions, use standard trigonometric substitutions:

• For $\sqrt{a^2 - x^2}$, substitute $x = a\sin\theta$ or $a\cos\theta$.
• For $\sqrt{a^2 + x^2}$, substitute $x = a\tan\theta$ or $a\cot\theta$.
• For $\sqrt{x^2 - a^2}$, substitute $x = a\sec\theta$ or $a\text{cosec}\theta$.
• For $\sqrt{\frac{a-x}{a+x}}$, substitute $x = a\cos 2\theta$.

Example 1: Simplify $\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right)$ for $-\frac{3\pi}{2} < x < \frac{\pi}{2}$.

1. Use half-angle identities: $\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$ and $1 - \sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2$.
2. Substitute and factor: $\tan^{-1}\left( \frac{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})}{(\cos\frac{x}{2} - \sin\frac{x}{2})^2} \right)$.
3. Divide numerator and denominator by $\cos\frac{x}{2}$: $\tan^{-1}\left( \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} \right)$.
4. Recognize tangent addition formula: $\tan^{-1}\left(\tan(\frac{\pi}{4} + \frac{x}{2})\right) = \frac{\pi}{4} + \frac{x}{2}$.

Example 2: Evaluate $\sin^{-1}(\sin \frac{3\pi}{5})$.

1. Check bounds: $\frac{3\pi}{5} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$.
2. Use $\sin(\pi - \theta) = \sin\theta$: $\sin(\frac{3\pi}{5}) = \sin(\pi - \frac{2\pi}{5}) = \sin(\frac{2\pi}{5})$.
3. Since $\frac{2\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\sin^{-1}(\sin \frac{2\pi}{5}) = \frac{2\pi}{5}$.

Infinite Series Summation (Method of Differences)

[JEE TIP] For a series like $S = \sum_{n=1}^{\infty} \tan^{-1}\left( \frac{k}{1 + A_n} \right)$, factor $A_n$ into the product of two terms $(x_n \cdot x_{n-1})$ such that their difference $x_n - x_{n-1}$ exactly matches the numerator $k$. Then apply: $\tan^{-1}\left( \frac{x_n - x_{n-1}}{1 + x_n x_{n-1}} \right) = \tan^{-1} x_n - \tan^{-1} x_{n-1}$. By summing, interior terms cancel out (telescoping series).

Conditions & Limitations

• Non-algebraic Nature: $\sin^{-1}(x+y) \neq \sin^{-1} x + \sin^{-1} y$. Inverse trigonometric functions are transcendental, not linear.
• Variable Bounds: Never manipulate an identity without checking the domain of the variable $x$. For instance, writing $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ outside $x \in [-1/2, 1/2]$ is mathematically invalid and requires piecemeal branch adjustments (e.g., adding/subtracting $\pi$).

Previous Year JEE Topics

• Roots of equations involving ITFs: Equating functions with different domains requires taking intersections of domains. (e.g., finding $x$ satisfying $\sin^{-1} x + \cos^{-1} (1-x) = \dots$).
• Telescoping Series of $\tan^{-1}$: Almost guaranteed to appear in JEE Advanced Paper 1 or 2 every alternate year.
• Calculus of ITFs: Limits and derivatives involving composite functions like $\frac{d}{dx}(\sin^{-1}(\sin x))$ at $x = \pi/2$ (It is non-differentiable here due to the sharp corner in the sawtooth graph).
• Integration of ITFs: Often requires integration by parts where the inverse function is set as the first function ($u$) according to the ILATE rule.

Memory Aids & JEE Traps

• JEE TIPWhen solving $\cos^{-1}(-x) = y$, students often write $y = -\cos^{-1}(x)$. Remember the 'C' functions with range $[0, \pi]$ (i.e., $\cos^{-1}, \cot^{-1}, \sec^{-1}$) pull out negatives as $\pi - f^{-1}(x)$.
• JEE TIPWhen $x>0, y>0$ and $x y = 1$, $\tan^{-1} x + \tan^{-1} y$ is exactly $\frac{\pi}{2}$, not undefined, even though the standard addition formula yields $\frac{x+y}{0}$.

Top 10 JEE MCQ Traps (Misconception $\rightarrow$ Correct Understanding)

1. Misconception $\rightarrow$ $\sin^{-1}(\sin x) = x$ for any real number $x$. Correct Understanding $\rightarrow$ $\sin^{-1}(\sin x) = x$ ONLY if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Outside this, you must fold the value back into the domain using $\pi$ or $2\pi$ shifts.
2. Misconception $\rightarrow$ $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ under all conditions. Correct Understanding $\rightarrow$ This only holds if $xy < 1$. If $xy > 1$ and $x,y > 0$, you must add $\pi$. If $xy > 1$ and $x,y < 0$, you must subtract $\pi$.
3. Misconception $\rightarrow$ $\tan^{-1}(1/x) = \cot^{-1} x$ for all $x \neq 0$. Correct Understanding $\rightarrow$ This is only true for $x > 0$. If $x < 0$, $\tan^{-1}(1/x) = -\pi + \cot^{-1} x$.
4. Misconception $\rightarrow$ Domain of an inverse trig function is the same as the original trig function. Correct Understanding $\rightarrow$ The domain of an inverse trigonometric function is strictly the range of the originally restricted trigonometric function.
5. Misconception $\rightarrow$ The range of $\sec^{-1} x$ is $[0, \pi]$. Correct Understanding $\rightarrow$ The range of $\sec^{-1} x$ is $[0, \pi] - \{\frac{\pi}{2}\}$ because $\sec(\frac{\pi}{2})$ is not defined.
6. Misconception $\rightarrow$ Expanding $\sin^{-1}(2x\sqrt{1-x^2})$ automatically to $2\sin^{-1} x$. Correct Understanding $\rightarrow$ It equals $2\sin^{-1} x$ only when $x \in [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$. For $x \in [\frac{1}{\sqrt{2}}, 1]$, it equals $2\cos^{-1} x$.
7. Misconception $\rightarrow$ $\sin^{-1} x^2$ is the same as $(\sin^{-1} x)^2$. Correct Understanding $\rightarrow$ The former applies the square to the argument $x$. The latter squares the angle outcome. They are entirely different functions.
8. Misconception $\rightarrow$ If $y = \text{cosec}^{-1} x$, the interval includes $0$. Correct Understanding $\rightarrow$ The principal value branch strictly removes $0$ because $\text{cosec}(0)$ is undefined. The range is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.
9. Misconception $\rightarrow$ The limit $\lim_{x\to 0} \frac{\sin^{-1}x}{x} = 1$ applies for $x \to \infty$. Correct Understanding $\rightarrow$ This limit is standard for $x \to 0$. For $x \to \infty$, $\sin^{-1} x$ is fundamentally undefined because its domain is restricted to $[-1, 1]$.
10. Misconception $\rightarrow$ $\cos^{-1}(\cos \frac{7\pi}{6}) = \frac{7\pi}{6}$. Correct Understanding $\rightarrow$ Since $\frac{7\pi}{6}$ is outside $[0, \pi]$, we evaluate $\cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2}$. Then $\cos^{-1}(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6}$. The correct output is $\frac{5\pi}{6}$.