Arccos — Cheat sheet — Mr. Integral

Definition

$y = \arccos x \iff \cos y = x$ , with $D = [-1, 1]$ and $R = [0, \pi]$ .

Arccos turns a number into an angle, and that angle is never negative.

Which quadrant

Input	Output lands in
positive	first quadrant, $0$ to $\dfrac{\pi}{2}$
negative	second quadrant, $\dfrac{\pi}{2}$ to $\pi$

$\arccos(-x) = \pi - \arccos x$ . Arccos reflects; it does not flip sign the way arcsin does.

Exact values

$x$	$-1$	$-\dfrac{\sqrt{3}}{2}$	$-\dfrac{\sqrt{2}}{2}$	$-\dfrac{1}{2}$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\arccos x$	$180^\circ$	$150^\circ$	$135^\circ$	$120^\circ$	$90^\circ$	$60^\circ$	$45^\circ$	$30^\circ$	$0^\circ$

As $x$ grows, $\arccos x$ falls. Arcsin climbs.

Identities

Identity	Condition
$\cos(\arccos x) = x$	any $x$ in $[-1, 1]$
$\arcsin x + \arccos x = \dfrac{\pi}{2}$	any $x$ in $[-1, 1]$
$\arccos x + \arccos y = \dfrac{\pi}{2}$	$x^2 + y^2 = 1$ , both positive

When there is no special angle

Name it $\alpha$ , so $\cos\alpha$ is the given ratio. That is adjacent over hypotenuse, so draw the triangle, let Pythagoras fill in the third side, and read off whatever is asked. The angle itself is never needed.

Domain of an arccos expression

Whatever sits inside must satisfy $-1 \le \text{inside} \le 1$ . Solve that compound inequality.