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Inverse Trigonometric Functions

6 articles26 formulas32 ways the board asks it
MATPrincipal Values

Principal Value Branch, Domains and Ranges

Inverse-trig functions are made one-to-one by restricting each parent function to a chosen interval; the inverse's range is that interval, called the principal-value branch. This subtopic asks you to state domains and principal ranges and to use them when simplifying f−1(f(θ))f^{-1}(f(\theta)) for angles θ\theta outside the branch.

InteractiveThis topic has a hand-built visualisation (inverse-trig-branch-graphs.html). It is not wired into the app yet.

It is examined because almost every later identity silently depends on these branches being respected.

sin⁡−1\sin^{-1} and cos⁡−1\cos^{-1}
sin⁡−1x: [−1,1]→[−π2,π2],cos⁡−1x: [−1,1]→[0,π]\sin^{-1}x:\ [-1,1]\to\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right],\qquad \cos^{-1}x:\ [-1,1]\to[0,\pi]
Domain →\to principal range; sin⁡−1\sin^{-1} is odd, cos⁡−1\cos^{-1} is not.
tan⁡−1\tan^{-1} and cot⁡−1\cot^{-1}
tan⁡−1x: R→(−π2,π2),cot⁡−1x: R→(0,π)\tan^{-1}x:\ \mathbb{R}\to\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right),\qquad \cot^{-1}x:\ \mathbb{R}\to(0,\pi)
Endpoints excluded (open intervals); both defined for all real xx.
sec⁡−1\sec^{-1} and cosec⁡−1\operatorname{cosec}^{-1}
sec⁡−1x: ∣x∣≥1→[0,π]∖{π2},cosec⁡−1x: ∣x∣≥1→[−π2,π2]∖{0}\sec^{-1}x:\ |x|\ge1\to[0,\pi]\setminus\left\{\dfrac{\pi}{2}\right\},\qquad \operatorname{cosec}^{-1}x:\ |x|\ge1\to\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\setminus\{0\}
Domain is (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty); the excluded angle is where the ratio is undefined.
Cancellation only inside the branch
sin⁡−1(sin⁡θ)=θ  ⟺  θ∈[−π2,π2],cos⁡−1(cos⁡θ)=θ  ⟺  θ∈[0,π]\sin^{-1}(\sin\theta)=\theta\iff\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right],\quad \cos^{-1}(\cos\theta)=\theta\iff\theta\in[0,\pi]
If θ\theta is outside the branch, first replace sin⁡θ\sin\theta (or cos⁡θ\cos\theta) by the equal value of an angle inside it.
  • The principal-value branch is the agreed output interval; e.g. sin⁡−1x\sin^{-1}x never returns an angle outside [−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].
  • cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos\theta)=\theta only if θ∈[0,π]\theta\in[0,\pi]; for θ∈[π,2π]\theta\in[\pi,2\pi] use cos⁡−1(cos⁡θ)=2π−θ\cos^{-1}(\cos\theta)=2\pi-\theta, or reduce other θ\theta using periodicity and evenness first.
  • For sin⁡−1 ⁣(sin⁡2π3)\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right): since 2π3∉[−π2,π2]\dfrac{2\pi}{3}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], write sin⁡2π3=sin⁡ ⁣(π−2π3)=sin⁡π3\sin\dfrac{2\pi}{3}=\sin\!\left(\pi-\dfrac{2\pi}{3}\right)=\sin\dfrac{\pi}{3}, giving π3\dfrac{\pi}{3}.
  • For cos⁡−1 ⁣(cos⁡7π6)\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right): 7π6∉[0,π]\dfrac{7\pi}{6}\notin[0,\pi], so use cos⁡7π6=cos⁡5π6\cos\dfrac{7\pi}{6}=\cos\dfrac{5\pi}{6}, giving 5π6\dfrac{5\pi}{6}.
  • For tan⁡−1 ⁣(tan⁡3π4)\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right): 3π4∉(−π2,π2)\dfrac{3\pi}{4}\notin\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right), so subtract π\pi: tan⁡3π4=tan⁡ ⁣(−π4)\tan\dfrac{3\pi}{4}=\tan\!\left(-\dfrac{\pi}{4}\right), giving −π4-\dfrac{\pi}{4}.
  • The inner cancellation f(f−1(x))=xf(f^{-1}(x))=x always holds on the domain (no quadrant issue); only the OUTER cancellation f−1(f(θ))f^{-1}(f(\theta)) needs the branch check.
  • Keep a quick reference of which quadrants each branch covers: arcsin/arccosec/arctan reach into Q4 (negative angles), while arccos/arccot/arcsec stay in Q1 to Q2 (non-negative angles).
Where the marks go
  • Asserting sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta)=\theta for every θ\theta — true only inside [−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].
  • Using the wrong reduction angle: e.g. writing tan⁡−1 ⁣(tan⁡3π4)=3π4\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)=\dfrac{3\pi}{4} instead of −π4-\dfrac{\pi}{4}.
  • Confusing the open vs closed endpoints — tan⁡−1\tan^{-1} and cot⁡−1\cot^{-1} ranges exclude their endpoints, while sin⁡−1\sin^{-1}/cos⁡−1\cos^{-1} include theirs.
  • Forgetting that sec⁡−1/cosec⁡−1\sec^{-1}/\operatorname{cosec}^{-1} have domain ∣x∣≥1|x|\ge1 only, then claiming a value for an argument in (−1,1)(-1,1).
How the board asks it
  • Numericalouter cancellation f−1(f(θ))f^{-1}(f(\theta)) for θ\theta outside the principal branch
    Find the principal value of sin⁡−1 ⁣(sin⁡2π3)\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right).
  • Numericalcombining branch reductions across two inverse functions
    Evaluate tan⁡−1 ⁣(tan⁡3π4)+cos⁡−1 ⁣(cos⁡7π6)\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right) + \cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right).
  • Define / stateprincipal-value branch (range) and domain of the inverse-trig functions
    State the principal-value branch (range) of cot⁡−1x\cot^{-1}x and of sec⁡−1x\sec^{-1}x, and write the domain of each.
  • Give reasonscancellation cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos\theta)=\theta valid only on [0,π][0,\pi]
    Explain why cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos\theta)=\theta fails for θ=7π6\theta=\dfrac{7\pi}{6}, and hence write the correct value of cos⁡−1 ⁣(cos⁡7π6)\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right).
  • Multiple choicedomain restriction ∣x∣≥1|x|\ge 1 for sec⁡−1\sec^{-1} and csc⁡−1\csc^{-1}
    The value of sec⁡−1 ⁣(12)\sec^{-1}\!\left(\dfrac{1}{2}\right) is: (a) 00 (b) π3\dfrac{\pi}{3} (c) does not exist (d) π2\dfrac{\pi}{2}.
  • Assertion–Reasonrange of sin⁡−1\sin^{-1} forces the reduced angle
    Assertion (A): sin⁡−1 ⁣(sin⁡2π3)=π3\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)=\dfrac{\pi}{3}. Reason (R): the range of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]. Choose the correct option.

Practise this topic

NextEvaluating Principal Values

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.