JEE Mathematics Inverse Trigonometric Functions Notes Set 02

Important Notes and Formulas

Domains and Principal Value Ranges

The six inverse trigonometric functions are defined by restricting the trigonometric functions to intervals on which they are one-one. The principal values are:
\( \sin^{-1}x \): domain \( [-1,1] \), range \( [-\tfrac\pi2,\tfrac\pi2] \).
\( \cos^{-1}x \): domain \( [-1,1] \), range \( [0,\pi] \).
\( \tan^{-1}x \): domain R, range \( (-\tfrac\pi2,\tfrac\pi2) \).
\( \cot^{-1}x \): domain R, range \( (0,\pi) \).
\( \sec^{-1}x \): domain \( |x|\ge1 \), range \( [0,\pi]-\{\tfrac\pi2\} \).
\( \operatorname{cosec}^{-1}x \): domain \( |x|\ge1 \), range \( [-\tfrac\pi2,\tfrac\pi2]-\{0\} \).
For a composite expression, the domain is the intersection of the conditions: for example \( \sin^{-1}(f(x)) \) needs \( |f(x)|\le1 \), and \( \sec^{-1}(f(x)) \) needs \( |f(x)|\ge1 \).

Example 1. Find the domain of \( \sin^{-1}(2x^2+1) \).
Solution:
We need \( -1\le2x^2+1\le1 \), so \( x^2\le0 \). The domain is \( \{0\} \).

Basic Properties

Complementary pairs: \( \sin^{-1}x+\cos^{-1}x=\tfrac\pi2 \) for \( x\in[-1,1] \), \( \tan^{-1}x+\cot^{-1}x=\tfrac\pi2 \) for all x, and \( \sec^{-1}x+\operatorname{cosec}^{-1}x=\tfrac\pi2 \) for \( |x|\ge1 \).
Negative arguments: \( \sin^{-1}(-x)=-\sin^{-1}x \), \( \tan^{-1}(-x)=-\tan^{-1}x \), \( \operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}x \), while \( \cos^{-1}(-x)=\pi-\cos^{-1}x \), \( \cot^{-1}(-x)=\pi-\cot^{-1}x \), \( \sec^{-1}(-x)=\pi-\sec^{-1}x \).
Reciprocals: \( \operatorname{cosec}^{-1}x=\sin^{-1}\tfrac1x \), \( \sec^{-1}x=\cos^{-1}\tfrac1x \), and \( \cot^{-1}x=\tan^{-1}\tfrac1x \) for \( x>0 \) (it equals \( \pi+\tan^{-1}\tfrac1x \) for \( x<0 \)).
Also \( \tan^{-1}x+\tan^{-1}\tfrac1x=\tfrac\pi2 \) for \( x>0 \) and \( -\tfrac\pi2 \) for \( x<0 \).

Trigonometric of Inverse and Inverse of Trigonometric

\( \sin(\sin^{-1}x)=x \) for \( |x|\le1 \), but \( \sin^{-1}(\sin x)=x \) only when \( x\in[-\tfrac\pi2,\tfrac\pi2] \). For other x, shift into the principal interval:
\( \sin^{-1}(\sin x) \) equals \( \pi-x \) on \( [\tfrac\pi2,\tfrac{3\pi}2] \) and \( x-2\pi \) on \( [\tfrac{3\pi}2,\tfrac{5\pi}2] \) (a zig-zag graph with period \( 2\pi \)).
\( \cos^{-1}(\cos x) \) equals x on \( [0,\pi] \) and \( 2\pi-x \) on \( [\pi,2\pi] \).
\( \tan^{-1}(\tan x) \) equals x on \( (-\tfrac\pi2,\tfrac\pi2) \) and \( x-\pi \) on \( (\tfrac\pi2,\tfrac{3\pi}2) \).
To evaluate expressions such as \( \tan(\cos^{-1}\tfrac45) \) or \( \sin(2\tan^{-1}\tfrac12) \), draw a right triangle for the inverse function and read off the other ratios.

Example 2. Find \( \sin^{-1}(\sin7) \) and \( \sin^{-1}(\sin(-5)) \).
Solution:
7 lies in \( [\tfrac{3\pi}2,\tfrac{5\pi}2] \), so \( \sin^{-1}(\sin7)=7-2\pi \). For \( -5 \): \( \sin(-5)=\sin(2\pi-5) \) and \( 2\pi-5\approx1.28\in[-\tfrac\pi2,\tfrac\pi2] \), so \( \sin^{-1}(\sin(-5))=2\pi-5 \).

Conversion Between Inverse Functions

For \( x\in(0,1) \): \( \sin^{-1}x=\cos^{-1}\sqrt{1-x^2}=\tan^{-1}\dfrac{x}{\sqrt{1-x^2}} \) and \( \tan^{-1}x=\sin^{-1}\dfrac{x}{\sqrt{1+x^2}}=\cos^{-1}\dfrac{1}{\sqrt{1+x^2}} \). For negative x the quantity must be adjusted using the negative-argument rules.
Double angle forms: \( 2\tan^{-1}x=\sin^{-1}\dfrac{2x}{1+x^2} \) for \( |x|\le1 \), \( =\cos^{-1}\dfrac{1-x^2}{1+x^2} \) for \( x\ge0 \), \( =\tan^{-1}\dfrac{2x}{1-x^2} \) for \( |x|<1 \).

Example 3. Evaluate \( \tan\left(\cos^{-1}\tfrac45+\tan^{-1}\tfrac23\right) \).
Solution:
\( \tan(\cos^{-1}\tfrac45)=\tfrac34 \). Then \( \tan(A+B)=\dfrac{\frac34+\frac23}{1-\frac34\cdot\frac23}=\dfrac{17/12}{1/2}=\dfrac{17}{6} \).

Example 4. Find \( \sin(2\tan^{-1}\tfrac12) \).
Solution:
With \( x=\tfrac12 \): \( \sin(2\tan^{-1}x)=\dfrac{2x}{1+x^2}=\dfrac{1}{1+\frac14}=\dfrac45 \).

Addition and Subtraction Rules

\( \tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy} \) if \( xy<1 \); it equals \( \pi+\tan^{-1}\dfrac{x+y}{1-xy} \) if \( xy>1,\ x,y>0 \); and \( -\pi+\tan^{-1}\dfrac{x+y}{1-xy} \) if \( xy>1,\ x,y<0 \). Similarly \( \tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy} \) if \( xy>-1 \).
\( \sin^{-1}x\pm\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}\pm y\sqrt{1-x^2}\right) \) when the result stays in the principal range, and \( \cos^{-1}x\pm\cos^{-1}y=\cos^{-1}\left(xy\mp\sqrt{(1-x^2)(1-y^2)}\right) \) under the corresponding conditions.

Example 5. Show that \( \tan^{-1}\tfrac12+\tan^{-1}\tfrac13=\tfrac\pi4 \).
Solution:
Here \( xy=\tfrac16<1 \), so the sum is \( \tan^{-1}\dfrac{\frac12+\frac13}{1-\frac16}=\tan^{-1}\dfrac{5/6}{5/6}=\tan^{-1}1=\tfrac\pi4 \).

Equations and Inequalities

Solve equations by taking a trigonometric function of both sides (and then verify the roots in the original equation, because the step can add false roots), or by using range arguments: for example if \( \sin^{-1}a+\cos^{-1}b=\tfrac\pi2 \) then \( a=b \), and a sum of terms equal to the sum of their maximum values forces each term to be at its maximum. For inequalities in terms of \( [\cdot] \) or \( \{\cdot\} \) of inverse functions, use the monotonicity of the inverse function (\( \sin^{-1},\tan^{-1} \) increasing; \( \cos^{-1},\cot^{-1} \) decreasing).

Example 6. Solve \( \sin^{-1}x+\sin^{-1}2x=\tfrac\pi2 \).
Solution:
Then \( \sin^{-1}2x=\tfrac\pi2-\sin^{-1}x=\cos^{-1}x \), so taking the sine of both sides, \( 2x=\sin(\cos^{-1}x)=\sqrt{1-x^2} \). Squaring: \( 4x^2=1-x^2 \), so \( x=\tfrac1{\sqrt5} \) (the positive root, as \( 2x\ge0 \)), which satisfies \( |2x|\le1 \).