Evaluating Principal Values

MATPrincipal Values

Here you compute the single exact value of an inverse-trig expression (often a sum of several), where each term must be its principal value. The method is to ask, for $\cos^{-1}t$ say, "which angle in $[0,\pi]$ has cosine $t$ ?" and to handle negative arguments with the sign rules.

These are guaranteed easy marks provided you keep every term inside its own principal-value branch.

Negative-argument rules

\sin^{-1}(-x)=-\sin^{-1}x,\qquad \tan^{-1}(-x)=-\tan^{-1}x

Odd functions: the answer stays in

\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]

(arcsin) or

\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)

(arctan).

Negative argument for

\cos^{-1}

and

\cot^{-1}

\cos^{-1}(-x)=\pi-\cos^{-1}x,\qquad \cot^{-1}(-x)=\pi-\cot^{-1}x

NOT odd: a negative argument gives an obtuse angle in

[0,\pi]

; e.g.

\cot^{-1}(-1)=\dfrac{3\pi}{4}

Standard reference values

\cos^{-1}\!\left(-\dfrac{1}{2}\right)=\dfrac{2\pi}{3},\qquad \tan^{-1}(1)=\dfrac{\pi}{4},\qquad \sin^{-1}\!\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{6}

Each angle lies in that function's principal range; memorise the

30^{\circ},45^{\circ},60^{\circ}

table.

Reciprocal functions

\sec^{-1}(2)=\dfrac{\pi}{3},\qquad \operatorname{cosec}^{-1}(-2)=-\dfrac{\pi}{6},\qquad \sec^{-1}(-2)=\dfrac{2\pi}{3}

\sec^{-1}x\in[0,\pi]\setminus\left\{\dfrac{\pi}{2}\right\}

;

\operatorname{cosec}^{-1}x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\setminus\{0\}

For each term identify the function, then state its principal range, then pick the unique angle in that range with the given ratio.
$\cos^{-1}$ , $\cot^{-1}$ , $\sec^{-1}$ take values in $[0,\pi]$ , so a negative argument gives a SECOND-quadrant (obtuse) answer, never a negative one.
$\sin^{-1}$ , $\tan^{-1}$ , $\operatorname{cosec}^{-1}$ are odd, so a negative argument gives a negative (fourth-quadrant) answer.
Evaluate term by term, then add: e.g. $\tan^{-1}(1)+\cos^{-1}\!\left(-\dfrac{1}{2}\right)+\sin^{-1}\!\left(-\dfrac{1}{2}\right)=\dfrac{\pi}{4}+\dfrac{2\pi}{3}-\dfrac{\pi}{6}=\dfrac{3\pi}{4}$ .
For mixed expressions like $\sin\!\left(\dfrac{\pi}{3}-\sin^{-1}\!\left(-\dfrac{1}{2}\right)\right)$ , first reduce the inner inverse term to a number, then evaluate the outer trig function.
$\sec^{-1}$ and $\operatorname{cosec}^{-1}$ require $|x|\ge1$ ; there is no $\sec^{-1}(0.5)$ . Convert via $\sec^{-1}x=\cos^{-1}\dfrac{1}{x}$ if the cosine value is easier.
Keep answers as exact multiples of $\pi$ ; do not convert to degrees unless the question explicitly asks for degrees.

Where the marks go

Writing $\cos^{-1}\!\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{3}$ (treating arccos as odd) instead of the correct $\dfrac{2\pi}{3}$ .
Giving $\sec^{-1}(-2)$ as $-\dfrac{\pi}{3}$ or $\dfrac{\pi}{3}$ instead of $\dfrac{2\pi}{3}$ — its range is $[0,\pi]$ , not symmetric about $0$ .
Forgetting the domain restriction $|x|\ge1$ for $\sec^{-1}$ / $\operatorname{cosec}^{-1}$ and "evaluating" an out-of-domain argument.
Leaving a term as an angle outside the principal branch (e.g. choosing $\dfrac{7\pi}{6}$ for an arcsine), which is never a valid principal value.

How the board asks it

Numericalstandard reference values and odd/obtuse sign rules
Evaluate $\tan^{-1}(1)+\cos^{-1}\!\left(-\dfrac{1}{2}\right)+\sin^{-1}\!\left(-\dfrac{1}{2}\right)$ , giving your answer as an exact multiple of $\pi$ .
Numericalnegative arguments of the $[0,\pi]$ -range functions
Find the principal value of $\sec^{-1}(-2)+\cot^{-1}\!\left(-\dfrac{1}{\sqrt{3}}\right)$ .
Numericalreducing the inner inverse term first
Evaluate $\sin\!\left(\dfrac{\pi}{3}-\sin^{-1}\!\left(-\dfrac{1}{2}\right)\right)$ .
Give reasonsthe principal range $[0,\pi]$ for $\cos^{-1}$
A student writes $\cos^{-1}\!\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{3}$ . State, with reason, why this is incorrect and give the correct principal value.
Multiple choicethe odd nature of $\operatorname{cosec}^{-1}$ for negative arguments
The principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ is: (a) $\dfrac{\pi}{4}$ (b) $-\dfrac{\pi}{4}$ (c) $\dfrac{3\pi}{4}$ (d) not defined.

Practise this topic

PreviousPrincipal Value Branch, Domains and Ranges NextProperties and Identities of Inverse Trig Functions

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.