Principal Value Branch, Domains and Ranges

MATPrincipal Values

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 $f^{-1}(f(\theta))$ for angles $\theta$ 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}

and

\cos^{-1}

\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}

is odd,

\cos^{-1}

is not.

\tan^{-1}

and

\cot^{-1}

\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

x

\sec^{-1}

and

\operatorname{cosec}^{-1}

\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

(-\infty,-1]\cup[1,\infty)

; the excluded angle is where the ratio is undefined.

Cancellation only inside the branch

\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]

\theta

is outside the branch, first replace

\sin\theta

(or

\cos\theta

) by the equal value of an angle inside it.

The principal-value branch is the agreed output interval; e.g. $\sin^{-1}x$ never returns an angle outside $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ .
$\cos^{-1}(\cos\theta)=\theta$ only if $\theta\in[0,\pi]$ ; for $\theta\in[\pi,2\pi]$ use $\cos^{-1}(\cos\theta)=2\pi-\theta$ , or reduce other $\theta$ using periodicity and evenness first.
For $\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)$ : since $\dfrac{2\pi}{3}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ , write $\sin\dfrac{2\pi}{3}=\sin\!\left(\pi-\dfrac{2\pi}{3}\right)=\sin\dfrac{\pi}{3}$ , giving $\dfrac{\pi}{3}$ .
For $\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)$ : $\dfrac{7\pi}{6}\notin[0,\pi]$ , so use $\cos\dfrac{7\pi}{6}=\cos\dfrac{5\pi}{6}$ , giving $\dfrac{5\pi}{6}$ .
For $\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)$ : $\dfrac{3\pi}{4}\notin\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ , so subtract $\pi$ : $\tan\dfrac{3\pi}{4}=\tan\!\left(-\dfrac{\pi}{4}\right)$ , giving $-\dfrac{\pi}{4}$ .
The inner cancellation $f(f^{-1}(x))=x$ always holds on the domain (no quadrant issue); only the OUTER cancellation $f^{-1}(f(\theta))$ 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 $\sin^{-1}(\sin\theta)=\theta$ for every $\theta$ — true only inside $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ .
Using the wrong reduction angle: e.g. writing $\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)=\dfrac{3\pi}{4}$ instead of $-\dfrac{\pi}{4}$ .
Confusing the open vs closed endpoints — $\tan^{-1}$ and $\cot^{-1}$ ranges exclude their endpoints, while $\sin^{-1}$ / $\cos^{-1}$ include theirs.
Forgetting that $\sec^{-1}/\operatorname{cosec}^{-1}$ have domain $|x|\ge1$ only, then claiming a value for an argument in $(-1,1)$ .

How the board asks it

Numericalouter cancellation $f^{-1}(f(\theta))$ for $\theta$ outside the principal branch
Find the principal value of $\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)$ .
Numericalcombining branch reductions across two inverse functions
Evaluate $\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right) + \cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)$ .
Define / stateprincipal-value branch (range) and domain of the inverse-trig functions
State the principal-value branch (range) of $\cot^{-1}x$ and of $\sec^{-1}x$ , and write the domain of each.
Give reasonscancellation $\cos^{-1}(\cos\theta)=\theta$ valid only on $[0,\pi]$
Explain why $\cos^{-1}(\cos\theta)=\theta$ fails for $\theta=\dfrac{7\pi}{6}$ , and hence write the correct value of $\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)$ .
Multiple choicedomain restriction $|x|\ge 1$ for $\sec^{-1}$ and $\csc^{-1}$
The value of $\sec^{-1}\!\left(\dfrac{1}{2}\right)$ is: (a) $0$ (b) $\dfrac{\pi}{3}$ (c) does not exist (d) $\dfrac{\pi}{2}$ .
Assertion–Reasonrange of $\sin^{-1}$ forces the reduced angle
Assertion (A): $\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)=\dfrac{\pi}{3}$ . Reason (R): the range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ . 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.