Inverse Trigonometric Functions
Define sin⁻¹, cos⁻¹, and tan⁻¹ as the restricted inverses of the trig functions with domains that pass the horizontal line test. Evaluate inverse trig expressions exactly and apply to solving right triangle problems.
A lesson of Algebra 2, in Introduction to Trigonometry, about 10 minutes.
By the end of this lesson, you will be able to evaluate $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ (and their reciprocals), understand domain restrictions needed for invertibility, and apply inverse trig to solve equations and geometry problems.
The Big Idea
Sine, cosine, and tangent are not one-to-one on their natural domains (they repeat every period). To create inverse functions, we restrict each to an interval where it is one-to-one. The resulting inverse functions give the angle corresponding to a given trig value — always within a specific range.
Restricted Domains and Ranges
| Function | Restricted domain | Range of inverse |
|---|---|---|
| $\sin^{-1}$ (arcsin) | $[-\pi/2, \pi/2]$ | $[-1, 1]$ |
| $\cos^{-1}$ (arccos) | $[0, \pi]$ | $[-1, 1]$ |
| $\tan^{-1}$ (arctan) | $(-\pi/2, \pi/2)$ | $(-\infty, \infty)$ |
The output of $\sin^{-1}$ is always in $[-\pi/2, \pi/2]$ — the first and fourth quadrants (never QII or QIII).
The output of $\cos^{-1}$ is always in $[0, \pi]$ — QI and QII.
Notation
$\sin^{-1}(x) = \arcsin(x)$: the angle whose sine is $x$.
Warning: $\sin^{-1}(x) \neq [\sin(x)]^{-1} = \csc(x)$.
Example 1 — Evaluating Inverse Trig
$\sin^{-1}(1/2)$: "what angle in $[-\pi/2, \pi/2]$ has sine $1/2$?" → $\pi/6$
$\cos^{-1}(-\sqrt{2}/2)$: angle in $[0,\pi]$ with cosine $-\sqrt{2}/2$ → $3\pi/4$ (QII, where cos is negative)
$\tan^{-1}(-1)$: angle in $(-\pi/2, \pi/2)$ with tangent $-1$ → $-\pi/4$
Example 2 — Composing Trig with Inverse
$\sin(\cos^{-1}(3/5))$
Let $\theta = \cos^{-1}(3/5)$, so $\cos\theta = 3/5$ and $\theta \in [0, \pi]$.
$\sin^2\theta = 1 - 9/25 = 16/25 \implies \sin\theta = 4/5$ (positive because $\theta \in [0, \pi]$ → Q1 or Q2, where sin > 0)
$\sin(\cos^{-1}(3/5)) = 4/5$
Example 3 — More Complex Composition
$\tan(\sin^{-1}(-2/3))$
$\theta = \sin^{-1}(-2/3)$: $\theta \in [-\pi/2, 0]$ (QIV since sine is negative)
$\cos\theta = \sqrt{1 - 4/9} = \sqrt{5}/3$ (positive in QIV)
$\tan\theta = \sin/\cos = \frac{-2/3}{\sqrt{5}/3} = \frac{-2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}$
Solving Equations with Inverse Trig
General solution vs. principal value:
$\sin x = 1/2$ has infinitely many solutions: $x = \pi/6 + 2n\pi$ and $x = 5\pi/6 + 2n\pi$.