Inverse Trigonometric Functions — Study Notes

Inverse Trigonometric Functions — Study Notes

The six trigonometric functions are periodic, so they keep handing back the same outputs over and over. That repetition is exactly what stops them from being one-to-one: a single value such as \(\sin\theta=\frac{1}{2}\) comes from infinitely many angles. To build a genuine inverse we first chop the domain down to one stretch where the function climbs (or falls) without repeating, and only then reverse it. The chosen stretch is the principal domain, and the outputs of the inverse on that stretch are the principal values.

Principal-value branches (the table to memorise)

Learning this table cold is the single most useful thing you can do for the chapter — almost every evaluation question leans on it.

How to read the table

• The input column tells you when an expression even exists. For example \(\sin^{-1}(2)\) is undefined, because \(2\) is outside \([-1,1]\).
• The range column tells you which angle to pick when several share the same trig value. The principal value is the one inside that range; and when a positive and a negative candidate are equal in size, the convention keeps the positive one — that is why \(\cos^{-1}\) and \(\sec^{-1}\) live in \([0,\pi]\).
• \(\sin^{-1}, \tan^{-1}, \operatorname{cosec}^{-1}\) and \(\cot^{-1}\) behave oddly (their graphs are origin-symmetric where defined), while \(\cos^{-1}\) and \(\sec^{-1}\) are neither even nor odd.

Cancellation rules — and their fine print

"The inverse undoes the function" is only half the story. The cancellation \(f^{-1}(f(\theta))=\theta\) works only when \(\theta\) already sits in the principal range; otherwise you must shift \(\theta\) by a multiple of the period (or reflect it) until it does.

• \(\sin^{-1}(\sin\theta)=\theta\) only if \(\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).
• \(\cos^{-1}(\cos\theta)=\theta\) only if \(\theta\in[0,\pi]\).
• \(\tan^{-1}(\tan\theta)=\theta\) only if \(\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
• The reverse direction is friendlier: \(\sin(\sin^{-1}x)=x\), \(\cos(\cos^{-1}x)=x\), and so on hold for every allowed \(x\), because the inner inverse already lands in the right place.

Identities worth carrying in your head

• Complementary pairs: \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\), \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\), \(\sec^{-1}x+\operatorname{cosec}^{-1}x=\frac{\pi}{2}\). These three settle a surprising number of MCQs in a single line.
• Reciprocal links: \(\sec^{-1}x=\cos^{-1}\frac{1}{x}\), \(\operatorname{cosec}^{-1}x=\sin^{-1}\frac{1}{x}\), and \(\cot^{-1}x=\tan^{-1}\frac{1}{x}\) for \(x\gt 0\). They let you swap an awkward inverse for a familiar one.
• Negative arguments: \(\sin^{-1}(-x)=-\sin^{-1}x\) and \(\tan^{-1}(-x)=-\tan^{-1}x\), whereas \(\cos^{-1}(-x)=\pi-\cos^{-1}x\) and \(\cot^{-1}(-x)=\pi-\cot^{-1}x\).
• Double-angle shortcuts: \(2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^{2}}=\cos^{-1}\frac{1-x^{2}}{1+x^{2}}=\tan^{-1}\frac{2x}{1-x^{2}}\), each on its own restricted set of \(x\).

Common exam traps

• Forgetting the range. Writing \(\sin^{-1}(\sin 5)=5\) is wrong, because \(5\) radians is outside \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\); reduce it to \(5-2\pi\) first.
• Reading \(\sin^{-1}\) as \(\frac{1}{\sin}\). The superscript \(-1\) means "inverse function", not "reciprocal". The reciprocal of \(\sin\) is \(\operatorname{cosec}\).
• Using an addition formula out of range. \(\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\) needs \(ab\lt 1\); when \(ab\gt 1\) you must add or subtract \(\pi\).
• Domain slips with composites. For something like \(\sin^{-1}\sqrt{x-1}\), both the inner expression and the arcsine input must be legal — check them separately and intersect.
• Sign of the principal value. Always land the candidate angle inside the listed range; for \(\cos^{-1}\) and \(\sec^{-1}\) the answer is never negative.