Inverse Trigonometric Functions

By the end of this lesson, you will be able to evaluate all six inverse trig functions, graph arcsin/arccos/arctan, compose trig functions with their inverses, and apply to triangle-solving.

Review: Restricted Domains

Function	Restricted domain	Range (of inverse)	Output range
$\sin$	$[-\pi/2, \pi/2]$	$[-1, 1]$	$[-\pi/2, \pi/2]$
$\cos$	$[0, \pi]$	$[-1, 1]$	$[0, \pi]$
$\tan$	$(-\pi/2, \pi/2)$	$\mathbb{R}$	$(-\pi/2, \pi/2)$

Graphs of Inverse Functions

$y = \sin^{-1}x$: increasing from $(-1, -\pi/2)$ to $(1, \pi/2)$; passes through $(0, 0)$.

$y = \cos^{-1}x$: decreasing from $(-1, \pi)$ to $(1, 0)$; passes through $(0, \pi/2)$.

$y = \tan^{-1}x$: increasing from $-\pi/2$ to $\pi/2$; horizontal asymptotes at $\pm\pi/2$.

Each is the reflection of the restricted trig function over $y = x$.

Cancellation Identities

$$\sin^{-1}(\sin\theta) = \theta \quad \text{for } \theta \in [-\pi/2, \pi/2]$$

$$\cos^{-1}(\cos\theta) = \theta \quad \text{for } \theta \in [0, \pi]$$

$$\sin(\sin^{-1}x) = x \quad \text{for } x \in [-1, 1]$$

Caution: $\sin^{-1}(\sin(5\pi/4)) \neq 5\pi/4$ because $5\pi/4 \notin [-\pi/2, \pi/2]$.

$5\pi/4$ in QIII → reference angle $\pi/4$ → $\sin(5\pi/4) = -\sqrt{2}/2$ → $\sin^{-1}(-\sqrt{2}/2) = -\pi/4$.

Composing Trig with Inverse Trig

$\tan(\sin^{-1}(3/5))$:

Let $\theta = \sin^{-1}(3/5)$: $\sin\theta = 3/5$, $\theta \in [0, \pi/2]$ (QI since $3/5 > 0$).

$\cos\theta = 4/5$ (Pythagorean); $\tan\theta = 3/4$.

Example 1 — Evaluating Inverse Trig

$\cos^{-1}(0) = \pi/2$ (the angle in $[0,\pi]$ with cosine 0)

$\sin^{-1}(-\sqrt{3}/2) = -\pi/3$ (angle in $[-\pi/2,\pi/2]$ with sine $-\sqrt{3}/2$; it's in QIV at $-\pi/3$)

$\tan^{-1}(\sqrt{3}) = \pi/3$

Example 2 — Complex Composition

$\sin(\cos^{-1}(5/13))$: