Inverse trig functions and basic equations | Trigonometry

Going backwards

$\sin\frac{\pi}{6} = \frac{1}{2}$ . The reverse question is: which angle has sine $\frac{1}{2}$ ? There are infinitely many ( $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ , $\frac{13\pi}{6}$ , ...), so to define an inverse function we have to pick one. Each inverse trig function picks its answer from a fixed range.

DefinitionInverse trig functions

Function	Input	Output (range)
$\arcsin x$ , also $\sin^{-1}x$	$-1 \le x \le 1$	$-\frac{\pi}{2} \le y \le \frac{\pi}{2}$
$\arccos x$ , also $\cos^{-1}x$	$-1 \le x \le 1$	$0 \le y \le \pi$
$\arctan x$ , also $\tan^{-1}x$	any real $$x$$	$-\frac{\pi}{2} < y < \frac{\pi}{2}$

Solving $\sin x = a$

The inverse function gives one solution. To find all of them, use the unit circle: the horizontal line $$y = a$$ crosses it at two points (when $$-1 < a < 1$$ ).

Solving $\cos x = a$

The vertical line $$x = a$$ crosses the unit circle at two points that mirror each other across the $$x$$ -axis.

Solving $\tan x = a$

Tangent has period $\pi$ and takes every real value exactly once per period, so one family covers everything.

Practice

Quick check

What is the range of $\arccos x$ ?

Quick check

How many solutions does $\tan x = 5$ have on $[0, 2\pi)$ ?