Principal Values

Learn what principal values are for inverse trigonometric functions, why they are needed, and the standard principal value ranges for sin⁻¹x, cos⁻¹x, tan⁻¹x and others.

1. What Are Principal Values?

Inverse trigonometric functions such as sin⁻¹x, cos⁻¹x, and tan⁻¹x return angles. But since trigonometric functions are periodic, their inverses can have infinitely many possible answers.

To make each inverse trig function return one unique angle, we select a fixed interval. The value returned in that interval is called the principal value.

Thus, the principal value is simply the standard angle we choose to represent the infinite possible angles.

2. Why Principal Values Are Needed

The function sin x repeats every \(2\pi\), so if:

\( \sin y = \dfrac{1}{2} \)

Possible values are:

  • \(y = \dfrac{\pi}{6}\)
  • \(y = \dfrac{5\pi}{6}\)
  • \(y + 2n\pi\) for any integer n

To make sin⁻¹ produce a unique answer, we choose one specific interval where the function behaves predictably (monotonic and one-to-one). This chosen interval gives principal values.

3. Principal Value Ranges for Inverse Trigonometric Functions

The following intervals are the standard principal value ranges used in mathematics:

FunctionPrincipal Value Range
\(\sin^{-1} x\)\(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\)
\(\cos^{-1} x\)\([0, \pi]\)
\(\tan^{-1} x\)\(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)
\(\cot^{-1} x\)\((0, \pi)\)
\(\sec^{-1} x\)\([0, \pi], \; y \neq \dfrac{\pi}{2}\)
\(\csc^{-1} x\)\([-\dfrac{\pi}{2}, \dfrac{\pi}{2}], \; y \neq 0\)

4. How Principal Values Work

Whenever you calculate inverse trig values, the answer must lie inside the principal value interval of that function—even if many angles satisfy the corresponding trigonometric equation.

4.1. Example

Example: Evaluate \( \sin^{-1}(\dfrac{\sqrt{3}}{2}) \).

The angles whose sine equals \(\dfrac{\sqrt{3}}{2}\) are:

  • \(60^\circ\) or \(\dfrac{\pi}{3}\)
  • \(120^\circ\) or \(\dfrac{2\pi}{3}\)

But the principal value of sin⁻¹ lies in:

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

Hence, the principal value is:

\( \sin^{-1}(\dfrac{\sqrt{3}}{2}) = \dfrac{\pi}{3} \)

5. More Examples of Principal Values

1. Find \(\cos^{-1}(-1)\)

The only angle in \([0, \pi]\) whose cosine is –1 is:

\(\pi\)


2. Find \(\tan^{-1}(\sqrt{3})\)

tan is \(\sqrt{3}\) at \(60^\circ\), but principal values for tan⁻¹ lie in:

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

So the principal value is:

\( \dfrac{\pi}{3} \)