JEE Mathematics Inverse Trigonometric Functions Notes Set 01

Why principal values are needed

Trigonometric functions repeat their values, so none of them is one-one on all of R. Restricting each function to a suitable interval makes it a bijection, and the inverse of this restricted function is the inverse trigonometric function. The restricted interval is called the principal branch and the outputs are called principal values.

Domains and ranges

Composition with the original function

\( \sin(\sin^{-1}x)=x \) for \( x\in[-1,1] \), and similarly \( \cos(\cos^{-1}x)=x \), \( \tan(\tan^{-1}x)=x \) for all real x.
The other order needs care: \( \sin^{-1}(\sin\theta)=\theta \) only for \( \theta\in[-\frac\pi2,\frac\pi2] \); \( \cos^{-1}(\cos\theta)=\theta \) only for \( \theta\in[0,\pi] \); \( \tan^{-1}(\tan\theta)=\theta \) only for \( \theta\in(-\frac\pi2,\frac\pi2) \).
Outside these intervals, shift the angle by a suitable multiple of \( \pi \) or reflect it. For example \( \sin^{-1}(\sin10)=3\pi-10 \) and \( \cos^{-1}\left(\cos\frac{7\pi}6\right)=\frac{5\pi}6 \).

Negative arguments

\( \sin^{-1}(-x)=-\sin^{-1}x \), \( \tan^{-1}(-x)=-\tan^{-1}x \), \( \operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}x \)
\( \cos^{-1}(-x)=\pi-\cos^{-1}x \), \( \cot^{-1}(-x)=\pi-\cot^{-1}x \), \( \sec^{-1}(-x)=\pi-\sec^{-1}x \)

Reciprocal and complementary relations

\( \sin^{-1}x=\operatorname{cosec}^{-1}\frac1x \), \( \cos^{-1}x=\sec^{-1}\frac1x \) for \( |x|\le1,\ x\ne0 \)
\( \tan^{-1}x=\cot^{-1}\frac1x \) for \( x>0 \), and \( \tan^{-1}x=\cot^{-1}\frac1x-\pi \) for \( x<0 \)
\( \sin^{-1}x+\cos^{-1}x=\frac\pi2 \), \( \tan^{-1}x+\cot^{-1}x=\frac\pi2 \), \( \sec^{-1}x+\operatorname{cosec}^{-1}x=\frac\pi2 \) in their domains

Sum and difference of tan inverse terms

\( \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \) if \( xy<1 \)
\( \tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\frac{x+y}{1-xy} \) if \( x,y>0 \) and \( xy>1 \); it is \( -\pi+\tan^{-1}\frac{x+y}{1-xy} \) if \( x,y<0 \) and \( xy>1 \)
\( \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \) if \( xy>-1 \)
For three terms: if \( xy+yz+zx=1 \) then \( \tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\frac\pi2 \); if \( x+y+z=xyz \) then the sum is \( \pi \) (for positive x, y, z)

Sum and difference of sin inverse and cos inverse terms

\( \sin^{-1}x+\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right) \) when \( x,y\ge0 \) and \( x^2+y^2\le1 \); if \( x,y>0 \) and \( x^2+y^2>1 \) the sum is \( \pi-\sin^{-1}(\cdots) \)
\( \cos^{-1}x+\cos^{-1}y=\cos^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right) \) when \( x+y\ge0 \)
\( \cos^{-1}x-\cos^{-1}y=\cos^{-1}\left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right) \) when \( x\le y \)

Multiple-angle conversions

\( 2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2} \) for \( |x|\le1 \)
\( 2\tan^{-1}x=\cos^{-1}\frac{1-x^2}{1+x^2} \) for \( x\ge0 \)
\( 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2} \) for \( |x|<1 \)
\( 3\sin^{-1}x=\sin^{-1}(3x-4x^3) \) for \( |x|\le\frac12 \), and \( 3\cos^{-1}x=\cos^{-1}(4x^3-3x) \) for \( \frac12\le x\le1 \)
\( 2\sin^{-1}x=\sin^{-1}\left(2x\sqrt{1-x^2}\right) \) for \( |x|\le\frac1{\sqrt2} \)

Solving equations: a checklist

1. Write down the domain of every inverse function in the equation.
2. Reduce the equation to one inverse function on each side (or use the complementary identities).
3. Take the same trigonometric function on both sides and solve the algebraic equation.
4. Substitute every root back, because taking a trigonometric function can bring in extra roots.