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Implicit and Inverse Differentiation · Tutorial 580 of 1495

Principal Values And Domains Of Inverse Trigonometric Functions

Learn the domains and principal-value ranges of all six inverse trigonometric functions, and distinguish them from reciprocal trigonometric functions.

Advanced 10 min read

What You'll Learn

  • State the domain and principal-value range of each inverse trigonometric function.
  • Explain why restricting a trigonometric function is necessary to define its inverse.
  • Distinguish arcsine notation from the reciprocal of sine.
  • Evaluate inverse trigonometric functions by selecting angles in the designated range.
  • Check inverse-function answers by applying the corresponding trigonometric function.

Why Principal Values Are Necessary

The trigonometric functions repeat their values. For example, many different angles have the same sine, cosine, or tangent. Consequently, a statement such as “the angle whose sine is \(1/2\)” does not specify a unique angle unless we also say which interval the angle must belong to.

An inverse function must assign exactly one output to each input in its domain. The standard inverse trigonometric functions achieve this by restricting each trigonometric function to a particular interval on which it is one-to-one. The output is called the principal value.

Definition: An inverse trigonometric function is the inverse of a trigonometric function restricted to a specified interval on which that function is one-to-one. Its domain is the range of the restricted trigonometric function, and its range is the interval of angles used for the restriction.

The angles in this tutorial are in radians. The selected intervals are standard conventions; in particular, we use \((0,\pi)\) for arccotangent. Other conventions for arccotangent exist, so stating the convention makes its values unambiguous.

The Six Domains and Principal-Value Ranges

The notation \(\arcsin x\), \(\arccos x\), and so on names inverse functions. Each function returns an angle in its stated range. The domain tells us which input values are possible.

Inverse function Domain Principal-value range
\(\arcsin x\) \([-1,1]\) \([-\pi/2,\pi/2]\)
\(\arccos x\) \([-1,1]\) \([0,\pi]\)
\(\arctan x\) \(\mathbb{R}\) \((-\pi/2,\pi/2)\)
\(\arccot x\) \(\mathbb{R}\) \((0,\pi)\)
\(\arcsec x\) \((-\infty,-1]\cup[1,\infty)\) \([0,\pi]\setminus\{\pi/2\}\)
\(\arccsc x\) \((-\infty,-1]\cup[1,\infty)\) \([-\pi/2,\pi/2]\setminus\{0\}\)

The domains for arcsecant and arccosecant exclude inputs strictly between \(-1\) and \(1\), because a real secant or cosecant value cannot lie strictly between those numbers. The endpoints are allowed: \(\sec 0=1\), \(\sec\pi=-1\), \(\csc(\pi/2)=1\), and \(\csc(-\pi/2)=-1\).

The inverse relationships describe both the input-output rule and the importance of the selected interval. For example, if \(y=\arcsin x\), then \(\sin y=x\), and \(y\) must lie in \([-\pi/2,\pi/2]\). Conversely, for every \(\theta\) in that interval, \(\arcsin(\sin\theta)=\theta\). The interval condition cannot be dropped: \(\arcsin(\sin\theta)\) need not equal \(\theta\) when \(\theta\) lies outside the principal range.

$$ \sin(\arcsin x)=x \quad (-1\le x\le 1), \qquad \arcsin(\sin\theta)=\theta \quad \left(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\right). $$

The same pattern applies to the other five functions: applying a trigonometric function after its inverse recovers an input in the inverse’s domain; applying the inverse after a trigonometric function recovers the angle when that angle is in the chosen principal-value range.

How the Restricted Intervals Make Inverses

The reason for restricting the angles is straightforward. A function has an inverse only when it does not assign the same output to two different inputs in its domain. The sine function, for instance, repeats values across the real line. On \([-\pi/2,\pi/2]\), however, sine is one-to-one and takes every value from \(-1\) to \(1\). Its restriction therefore has an inverse defined on \([-1,1]\).

Here is the same reasoning for each function. The trigonometric behavior on these intervals is standard: sine and tangent are strictly increasing on their chosen intervals, while cosine and cotangent are strictly decreasing. Secant and cosecant are one-to-one on the chosen ranges as well: their positive and negative branches have disjoint output ranges.

1
Choose one-to-one angle intervals.
Use \([-\pi/2,\pi/2]\) for sine, \([0,\pi]\) for cosine, \((-\pi/2,\pi/2)\) for tangent, and \((0,\pi)\) for cotangent. For secant and cosecant, exclude the angle where the corresponding function is undefined and use the ranges shown in the table.
2
Identify the outputs on those intervals.
Sine and cosine take values in \([-1,1]\); tangent and cotangent take every real value. Secant and cosecant take values in \((-\infty,-1]\cup[1,\infty)\).
3
Reverse input and output.
The selected angle interval becomes the range of the inverse, and the trigonometric output set becomes its domain.

This establishes the domains and ranges in the table. For instance, tangent on \((-\pi/2,\pi/2)\) is one-to-one and takes all real values, so its inverse has domain \(\mathbb R\) and range \((-\pi/2,\pi/2)\). The endpoints are excluded because tangent is undefined at \(\pm\pi/2\); it approaches arbitrarily large positive and negative values near those endpoints.

For secant, the range \([0,\pi]\setminus\{\pi/2\}\) consists of the two intervals \([0,\pi/2)\) and \((\pi/2,\pi]\). On the first, secant takes values in \([1,\infty)\); on the second, it takes values in \((-\infty,-1]\). These output sets do not overlap, so the whole restricted function remains one-to-one. This also explains both the range of arcsecant and its domain.

Worked Examples: Selecting Principal Values

Worked Example: Evaluating an Arcsine

Evaluate \(\arcsin(1/2)\).

We seek an angle \(y\) in the principal range \([-\pi/2,\pi/2]\) such that \(\sin y=1/2\). The special angle \(\pi/6\) is in that range, and

$$ \sin\left(\frac{\pi}{6}\right)=\frac{1}{2}. $$

Therefore, \(\arcsin(1/2)=\pi/6\). To check, applying sine to the answer gives \(\sin(\pi/6)=1/2\), the original input. The range condition is satisfied because \(0\le\pi/6\le\pi/2\).

Worked Example: Evaluating an Arccosine

Evaluate \(\arccos(-\sqrt{3}/2)\).

We need an angle \(y\in[0,\pi]\) whose cosine is \(-\sqrt{3}/2\). The reference angle with cosine \(\sqrt{3}/2\) is \(\pi/6\). Cosine is negative in the second quadrant, so the angle in the required range is \(y=\pi-\pi/6=5\pi/6\).

$$ \cos\left(\frac{5\pi}{6}\right) = \cos\left(\pi-\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}. $$

Thus \(\arccos(-\sqrt{3}/2)=5\pi/6\). The check recovers the input, and \(5\pi/6\in[0,\pi]\), as required.

Worked Example: Evaluating Arcsecant and Arccosecant

Evaluate \(\arcsec(-2)\) and \(\arccsc(-2)\).

For \(\arcsec(-2)\), seek \(y\in[0,\pi]\setminus\{\pi/2\}\) with \(\sec y=-2\). This is equivalent to \(\cos y=-1/2\). The angle in the required range is \(y=2\pi/3\), since

$$ \sec\left(\frac{2\pi}{3}\right) = \frac{1}{\cos(2\pi/3)} = \frac{1}{-1/2} = -2. $$

Therefore, \(\arcsec(-2)=2\pi/3\), and the value is in the principal range.

For \(\arccsc(-2)\), seek \(y\in[-\pi/2,\pi/2]\setminus\{0\}\) with \(\csc y=-2\). This is equivalent to \(\sin y=-1/2\). The angle in the required range is \(y=-\pi/6\). Checking directly,

$$ \csc\left(-\frac{\pi}{6}\right) = \frac{1}{\sin(-\pi/6)} = \frac{1}{-1/2} = -2. $$

Hence \(\arccsc(-2)=-\pi/6\). The angle lies in the specified range and is not zero.

Inverse Functions Are Not Reciprocals

The notation can be misleading. The inverse sine is written \(\arcsin x\) (also sometimes \(\sin^{-1}x\)), but it is not the reciprocal of \(\sin x\). The reciprocal of \(\sin x\) is \(\csc x=1/\sin x\). In the same way, \(\arctan x\) is an inverse function, while \(\tan^{-1}x\), when used in inverse-function notation, does not mean \(1/\tan x\); the reciprocal of tangent is cotangent.

Meaning Example notation What it does
Inverse sine \(\arcsin x\) Returns the principal angle whose sine is \(x\)
Reciprocal sine \(\csc x\) Returns \(1/\sin x\), when \(\sin x\ne 0\)
Inverse tangent \(\arctan x\) Returns the principal angle whose tangent is \(x\)
Reciprocal tangent \(\cot x\) Returns \(1/\tan x\), when \(\tan x\ne 0\)

Writing \(\arcsin\) or \(\arctan\) avoids the ambiguity of a superscript \(-1\). When a text does use \(\sin^{-1}\) or \(\tan^{-1}\) for an inverse function, context matters; the reciprocal functions have their own standard names, cosecant and cotangent.

Common Mistake: Ignoring the Principal Range

A familiar angle may have the correct trigonometric value but still be the wrong inverse-function output if it is outside the specified range. For example, \(\sin(5\pi/6)=1/2\), but it would be incorrect to conclude that \(\arcsin(1/2)=5\pi/6\). The arcsine range is \([-\pi/2,\pi/2]\), and \(5\pi/6\) is outside it.

$$ \arcsin\left(\sin\frac{5\pi}{6}\right) = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \ne \frac{5\pi}{6}. $$

The correction is to choose the angle in the principal range that has the given trigonometric value. Here that angle is \(\pi/6\). Before accepting any inverse-trigonometric answer, check both conditions: its trigonometric value must match the input, and the angle must lie in the inverse function’s principal-value range.

Why These Conventions Matter

The domains and ranges are not merely bookkeeping. They determine whether an expression is defined and which angle it represents. For example, \(\arcsin(2)\) has no real value because \(2\notin[-1,1]\). By contrast, \(\arctan(2)\) is defined because the domain of arctangent is all real numbers, and its output must lie between \(-\pi/2\) and \(\pi/2\).

These conventions also make inverse-function identities precise. The statement \(\arctan(\tan\theta)=\theta\) holds for \(-\pi/2<\theta<\pi/2\), but not for every real \(\theta\). Outside that range, the composition returns the angle in the principal range with the same tangent. As inverse trigonometric functions are used in differentiation, these domain and range restrictions remain essential for interpreting the functions correctly.

Key takeaway: An inverse trigonometric function returns a principal angle, not an arbitrary angle with the same trigonometric value. Check the input against the function’s domain and the output against its principal-value range.

Check Your Understanding

For each question, identify the relevant domain or principal-value restriction before evaluating.

  1. State the domain and range of \(\arctan x\).
  2. Evaluate \(\arccos(1/2)\), and verify the result.
  3. Why is \(\arcsin(3/2)\) not defined as a real number?
  4. Evaluate \(\arccot(0)\) using the convention \((0,\pi)\), and check the answer.
  5. Explain the difference between \(\arcsin x\) and \(\csc x\).

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