Inverse Trig Functions: Domain, Range, Principal Values

Quick answer: Each inverse trigonometric function returns an angle from a fixed principal branch: $\sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\cos^{-1}x \in [0, \pi]$ , both for $x \in [-1, 1]$ ; $\tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for all real $x$ . Because of these branches, $\sin^{-1}(\sin\theta) = \theta$ only when θ already lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

Why the ranges are restricted

$\sin x$ is not one-to-one on $\mathbb{R}$ : $\sin\frac{\pi}{6} = \sin\frac{5\pi}{6} = \frac{1}{2}$ . To define an inverse, we restrict sine to an interval where it is one-to-one and takes every value in $[-1, 1]$ once. That interval, $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , becomes the range of $\sin^{-1}x$ . The same idea gives the other principal branches.

Domain and range table

Function	Domain	Range (principal branch)
$\sin^{-1}x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\cos^{-1}x$	$[-1, 1]$	$[0, \pi]$
$\tan^{-1}x$	$\mathbb{R}$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\cot^{-1}x$	$\mathbb{R}$	$(0, \pi)$
$\sec^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$[0, \pi] - \left\{\frac{\pi}{2}\right\}$
$\csc^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$

A memory aid: sin, tan and cosec have ranges symmetric about 0; cos, cot and sec have ranges starting at 0.

Principal values: how to evaluate

To find $\sin^{-1}x$ , ask: "Which angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has sine equal to $x$ ?"

Expression	Value	Reason
$\sin^{-1}\left(-\frac{1}{2}\right)$	$-\frac{\pi}{6}$	$-\frac{\pi}{6}$ is in the sin⁻¹ range
$\cos^{-1}\left(-\frac{1}{2}\right)$	$\frac{2\pi}{3}$	Must lie in $[0, \pi]$
$\tan^{-1}(-1)$	$-\frac{\pi}{4}$	Must lie in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\cot^{-1}(-\sqrt{3})$	$\frac{5\pi}{6}$	Must lie in $(0, \pi)$
$\sec^{-1}(-2)$	$\frac{2\pi}{3}$	$\cos\frac{2\pi}{3} = -\frac{1}{2}$

Negative arguments

Odd-type	Supplement-type
$\sin^{-1}(-x) = -\sin^{-1}x$	$\cos^{-1}(-x) = \pi - \cos^{-1}x$
$\tan^{-1}(-x) = -\tan^{-1}x$	$\cot^{-1}(-x) = \pi - \cot^{-1}x$
$\csc^{-1}(-x) = -\csc^{-1}x$	$\sec^{-1}(-x) = \pi - \sec^{-1}x$

sin⁻¹(sin θ) versus sin(sin⁻¹ x)

$\sin(\sin^{-1}x) = x$ for every $x \in [-1, 1]$ .
$\sin^{-1}(\sin\theta) = \theta$ only for $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Otherwise, find the angle in the principal range with the same sine.

Example: $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$ . Since $\frac{2\pi}{3}$ is outside the range, use $\sin\frac{2\pi}{3} = \sin\left(\pi - \frac{2\pi}{3}\right) = \sin\frac{\pi}{3}$ . Answer: $\frac{\pi}{3}$ .

Example: $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ . Since $\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}$ , and $\frac{5\pi}{6} \in [0, \pi]$ , the answer is $\frac{5\pi}{6}$ .

Example: $\tan^{-1}(\tan 2)$ . Here 2 radians is about 114.6°, outside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Since tan has period π, $\tan 2 = \tan(2 - \pi)$ , and $2 - \pi \approx -1.14$ lies in the range. Answer: $2 - \pi$ .

Finding the domain of a composite expression

Example: Find the domain of $f(x) = \sin^{-1}(2x - 3)$ .

Step 1: The argument must lie in $[-1, 1]$ : $-1 \le 2x - 3 \le 1$ .

Step 2: $2 \le 2x \le 4$ , so $1 \le x \le 2$ . Domain: $[1, 2]$ .

Example: Find the domain of $\cos^{-1}\left(\frac{x}{2}\right) + \sqrt{x}$ .

The first term needs $-2 \le x \le 2$ ; the square root needs $x \ge 0$ . Domain: $[0, 2]$ .

Summary table

Idea	Key point
sin⁻¹, tan⁻¹, cosec⁻¹	Ranges centred on 0
cos⁻¹, cot⁻¹, sec⁻¹	Ranges starting at 0
Domain of sin⁻¹ and cos⁻¹	$[-1, 1]$
$\sin^{-1}(\sin\theta)$	Equals θ only if θ is in the principal range
Negative arguments	Odd-type change sign; cos⁻¹, cot⁻¹, sec⁻¹ use $\pi - (\cdot)$

Exam tips

Before writing any answer, check that it lies in the principal range of that function.
For $f^{-1}(f(\theta))$ questions, shift θ by π or 2π (or reflect it) until it lands in the range.
Domain questions usually reduce to solving $-1 \le g(x) \le 1$ .
Use the negative-argument rules to avoid sign errors, especially for cos⁻¹.

For more practice, see inverse trigonometric functions solved PYQs for NIMCET and the NIMCET Trigonometry chapter.

Common mistakes

Writing $\sin^{-1}x = \frac{1}{\sin x}$ . The inverse function is not the reciprocal; the reciprocal is $\csc x$ .
Answering $\cos^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{3}$ . Negative angles are not in the range of cos⁻¹.
Assuming $\sin^{-1}(\sin\theta) = \theta$ for every θ.
Forgetting that $\sin^{-1}x$ is undefined for $\lvert x\rvert > 1$ .

Frequently Asked Questions

What is the range of sin⁻¹x?

The principal range of sin⁻¹x is [−π/2, π/2], and its domain is [−1, 1].

What is the range of cos⁻¹x?

The principal range of cos⁻¹x is [0, π], and its domain is [−1, 1].

What is the value of sin⁻¹(sin 2π/3)?

It is π/3, because 2π/3 is outside [−π/2, π/2] and sin(2π/3) = sin(π/3).

What is cos⁻¹(−x) in terms of cos⁻¹x?

cos⁻¹(−x) = π − cos⁻¹x.

Why are inverse trigonometric functions restricted to principal values?

Trigonometric functions are periodic and not one-to-one. Restricting them to an interval where they are one-to-one makes a single-valued inverse possible.