JEE Mathematics Inverse Circular Functions Notes Set 01

Principal Values and Domains

Each inverse trigonometric function is defined on a restricted branch so that it becomes a function.
\( \sin^{-1}x:[-1,1]\to\left[-\frac\pi2,\frac\pi2\right] \); \( \cos^{-1}x:[-1,1]\to[0,\pi] \); \( \tan^{-1}x:\mathbb R\to\left(-\frac\pi2,\frac\pi2\right) \); \( \cot^{-1}x:\mathbb R\to(0,\pi) \); \( \sec^{-1}x:|x|\ge1\to[0,\pi]\setminus\{\frac\pi2\} \); \( \operatorname{cosec}^{-1}x:|x|\ge1\to\left[-\frac\pi2,\frac\pi2\right]\setminus\{0\} \).
Always check that the argument lies in the domain before using a formula. For example \( \cos^{-1}(\ln y) \) needs \( -1\le\ln y\le1 \).

Basic Identities

\( \sin^{-1}x+\cos^{-1}x=\frac\pi2 \) for \( x\in[-1,1] \); \( \tan^{-1}x+\cot^{-1}x=\frac\pi2 \) for all real x; \( \sec^{-1}x+\operatorname{cosec}^{-1}x=\frac\pi2 \) for \( |x|\ge1 \).
\( \sin^{-1}(-x)=-\sin^{-1}x \), \( \tan^{-1}(-x)=-\tan^{-1}x \), \( \cos^{-1}(-x)=\pi-\cos^{-1}x \), \( \cot^{-1}(-x)=\pi-\cot^{-1}x \).
\( \sin(\sin^{-1}x)=x \) only for \( x\in[-1,1] \), but \( \sin^{-1}(\sin x)=x \) only for \( x\in[-\frac\pi2,\frac\pi2] \). Outside that interval first reduce x to an equivalent angle in the principal range. The graphs of \( \sin^{-1}(\sin x) \) and \( \cos^{-1}(\cos x) \) are zigzag (triangular) waves with period 2π.

Addition Formulas

\( \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \) if \( xy<1 \); add \( \pi \) if \( xy>1 \) and \( x,y>0 \); subtract \( \pi \) if \( xy>1 \) and \( x,y<0 \).
\( 2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}=\cos^{-1}\frac{1-x^2}{1+x^2}=\tan^{-1}\frac{2x}{1-x^2} \), each valid on its own range of x (for example \( |x|\le1 \) for the first two).
A sum such as \( \sum\tan^{-1}\frac{1}{1+n(n+1)} \) telescopes using \( \tan^{-1}\frac{a-b}{1+ab}=\tan^{-1}a-\tan^{-1}b \).

Solving Equations and Inequalities

Take a trigonometric function of both sides (or substitute \( x=\cos\theta \) or \( x=\tan\theta \)), solve, then verify every root in the original equation because the operation can introduce extra solutions.
For inequalities involving integer-part functions of inverse trigonometric functions, use the bounded range of the inverse function to cut the cases down to a few integers.

Worked Examples

Example 1. Find the value of \( \sin^{-1}\left(-\sin\frac{50\pi}{9}\right) \).
Answer:
\( -\sin\frac{50\pi}{9}=\sin\left(6\pi-\frac{50\pi}{9}\right)=\sin\frac{4\pi}{9} \).
Since \( \frac{4\pi}{9}\in\left[-\frac\pi2,\frac\pi2\right] \), the value is \( \frac{4\pi}{9} \).

Example 2. Find \( \tan\left(\tan^{-1}\frac12+\tan^{-1}\frac13\right) \).
Answer:
\( \frac{\frac12+\frac13}{1-\frac16}=1 \), so the value is 1.

Example 3. Sum the series \( \cot^{-1}\left(n^2+\frac34\right) \) for n = 1, 2, 3, ... to infinity.
Answer:
\( \cot^{-1}\left(n^2+\frac34\right)=\tan^{-1}\left(n+\frac12\right)-\tan^{-1}\left(n-\frac12\right) \), so the series telescopes.
The sum is \( \frac\pi2-\tan^{-1}\frac12=\)\( \tan^{-1}2 \).