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Q.sin⁡−1x\sin^{-1}x is a function whose domain is __________.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2026Subjective· 1mImportance★★★★★
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Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

Take sin⁡x=12\sin x=\tfrac12. It has infinitely many solutions: x=π6,5π6,13π6,−7π6,…x=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},-\tfrac{7\pi}{6},\dots — every angle whose sine is 12\tfrac12. So if we want an inverse that returns a single angle for sin⁡−1(0.5)\sin^{-1}(0.5), we must first agree on one angle to report. A function is allowed only one output per input, and sin⁡x\sin x over all of R\mathbb{R} is many-to-one — it fails the horizontal line test and cannot be inverted as it stands.

The idea: restrict to one clean interval

For each trigonometric ratio we restrict the angle to a single standard interval on which the function is one-to-one while still covering its entire range exactly once. On that interval the inverse becomes well-defined and single-valued. That interval — the set of angles the inverse is allowed to return — is the principal value branch (also called the principal value domain).

The interval is chosen to be strictly monotonic, to hit every output once, and to sit as close to 00 as possible. For sine that is [−π2,π2]\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], where sin⁡\sin increases from −1-1 to 11.

The principal value branch of an inverse trig function is the interval of angles it returns — the restricted interval on which the original ratio is one-to-one and onto its range.

Inverse functionDomain (allowed inputs xx)Principal value branch (angles returned)
sin⁡−1x\sin^{-1} x[−1,1][-1,1][−π2,π2]\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]
cos⁡−1x\cos^{-1} x[−1,1][-1,1][0,π][0,\pi]
tan⁡−1x\tan^{-1} xR\mathbb{R}(−π2,π2)\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)
cot⁡−1x\cot^{-1} xR\mathbb{R}(0,π)(0,\pi)
sec⁡−1x\sec^{-1} x(−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty)[0,π]∖{π2}[0,\pi]\setminus\{\tfrac{\pi}{2}\}
csc⁡−1x\csc^{-1} x(−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty)[−π2,π2]∖{0}\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}

Why the intervals differ …

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