how to evaluate inverse trig functions

before you start, remember the domain and range of the inverse trig functions, shown below, and in the trig functions page.

$\sin (x)$
domain: $\mathbb{R}$
range: $[-1,1]$

${\sin }^{-1}(x)$
domain: $[-1,1]$
range: $[-\frac{\pi }{2},\frac{\pi }{2}]$

$\cos (x)$
domain: $\mathbb{R}$
range: $[-1,1]$

${\cos }^{-1}(x)$
domain: $[-1,1]$
range: $[0,\pi ]$

$\tan (x)$
domain: $\mathbb{R}$
range: $\mathbb{R}$

${\tan }^{-1}(x)$
domain: $\mathbb{R}$
range: $[-\frac{\pi }{2},\frac{\pi }{2}]$

we evaluate inverse trig functions by looking at the common angles table

$\theta$	$0$	$\frac{\pi }{6}$	$\frac{\pi }{4}$	$\frac{\pi }{3}$	$\frac{\pi }{2}$
$\sin (\theta )$	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
$\cos (\theta )$	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
$\tan (\theta )$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	undefined

worked example: evaluate ${\sin }^{-1}(\frac{1}{2})$
answer will be in the range: $[-\frac{\pi }{2},\frac{\pi }{2}]$
$\sin (\theta )=\frac{1}{2}$
$\theta =\frac{\pi }{6}$

worked example: evaluate ${\cos }^{-1}(-\frac{\sqrt{3}}{2})$
answer will be in the range: $[0,\pi ]$
$\cos (\theta )=-\frac{\sqrt{3}}{2}$
firstly we need an angle that gets cosine of $\frac{\sqrt{3}}{2}$
we know that $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$
now we need to reflect that angle across the y axis, because we want $-\frac{\sqrt{3}}{2}$
$\frac{\pi }{6}$ reflected across the y axis results in an angle in the second quadrant, found with $\pi -\frac{\pi }{6}=\frac{5\pi }{6}$
$\cos (\frac{5\pi }{6})=-\frac{\sqrt{3}}{2}$

worked example: evaluate ${\cos }^{-1}(-\frac{\sqrt{3}}{2})$
we’re going to do this worked example again, this time with a quicker way
$\cos (\theta )=-\frac{\sqrt{3}}{2}$
firstly we need an angle that gets cosine of $\frac{\sqrt{3}}{2}$
we know that $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$
the output of inverse cosine is always non-negative
so $\theta \ge 0$
$\theta =\frac{\pi }{6}$

worked example: evaluate ${\tan }^{-1}(-\frac{1}{\sqrt{3}})$
answer will be in the range: $[-\frac{\pi }{2},\frac{\pi }{2}]$
$\tan (\theta )=-\frac{1}{\sqrt{3}}$

$\tan (\frac{\pi }{6})=\frac{1}{\sqrt{3}}$
the output of inverse tan always has the same sign as the input
$\theta =-\frac{\pi }{6}$