JEE Mathematics Inverse Trigonometric Functions Notes Set 03

Principal Values and Domains

\(\sin^{-1}x\): domain \([-1,1]\), range \(\left[-\frac\pi2,\frac\pi2\right]\). \(\cos^{-1}x\): domain \([-1,1]\), range \([0,\pi]\). \(\tan^{-1}x\): domain R, range \(\left(-\frac\pi2,\frac\pi2\right)\). \(\cot^{-1}x\): range \((0,\pi)\). \(\sec^{-1}x\) and \(\operatorname{cosec}^{-1}x\) are defined for \(|x|\ge1\), with ranges \([0,\pi]\setminus\left\{\frac\pi2\right\}\) and \(\left[-\frac\pi2,\frac\pi2\right]\setminus\{0\}\).

\(\sin(\sin^{-1}x)=x\) only for \(x\in[-1,1]\), while \(\sin^{-1}(\sin\theta)=\theta\) only for \(\theta\in\left[-\frac\pi2,\frac\pi2\right]\); outside this range first rewrite the angle with an equivalent sine. The same care applies to \(\cos^{-1}(\cos\theta)\) and \(\tan^{-1}(\tan\theta)\).

Standard Relations

\(\sin^{-1}(-x)=-\sin^{-1}x\), \(\cos^{-1}(-x)=\pi-\cos^{-1}x\), \(\tan^{-1}(-x)=-\tan^{-1}x\). \(\sin^{-1}x+\cos^{-1}x=\frac\pi2\), \(\tan^{-1}x+\cot^{-1}x=\frac\pi2\). For \(x>0\): \(\tan^{-1}x+\tan^{-1}\frac1x=\frac\pi2\).

\(\tan^{-1}x\pm\tan^{-1}y=\tan^{-1}\frac{x\pm y}{1\mp xy}\) (with \(+\pi\) or \(-\pi\) when the principal range is exceeded). Double angle forms: \(2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}=\cos^{-1}\frac{1-x^2}{1+x^2}=\tan^{-1}\frac{2x}{1-x^2}\) for \(|x|<1\), with suitable ranges of x for each form.

Telescoping and Special Results

\(\tan^{-1}\frac{b-a}{1+ab}=\tan^{-1}b-\tan^{-1}a\), which turns sums such as \(\sum\tan^{-1}\frac{1}{1+r+r^2}\) into differences. If \(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi\) then \(x+y+z=xyz\). If \(\cos^{-1}x+\cos^{-1}y+\cos^{-1}z=\pi\) then \(x^2+y^2+z^2+2xyz=1\). Machin: \(4\tan^{-1}\frac15-\tan^{-1}\frac1{239}=\frac\pi4\).

Worked Examples

Example 1. \(\sin^{-1}(\sin10)\): \(10\) is outside the principal range, and \(\sin10=\sin(3\pi-10)\), so the value is \(3\pi-10\).

Example 2. Solve \(\tan^{-1}2x+\tan^{-1}3x=\frac\pi4\): \(\frac{5x}{1-6x^2}=1\) gives \(x=\frac16\) or \(-1\); the root \(-1\) is rejected because both angles would be negative.

Example 3. \(\tan\left(\cos^{-1}\frac1{5\sqrt2}-\sin^{-1}\frac4{\sqrt{17}}\right)=\frac{7-4}{1+28}=\frac3{29}\).

Quick Tips

Convert every inverse function to \(\tan^{-1}\) or to a right triangle, then use the addition formula. Always verify the answer against the principal range and discard extra roots.