Trig functions are periodic, so domain restrictions ensure unique inverse values.

$[-\frac{\pi}{2}, \frac{\pi}{2}]$

$[0, \pi]$

$(-\frac{\pi}{2}, \frac{\pi}{2})$

It visually represents angles and their corresponding sine, cosine, and tangent values, making it easy to find the angle for a given ratio.

Inverse trig functions 'undo' what trig functions do. If $\sin(x) = y$, then $\arcsin(y) = x$.

Find the angle within the range of arcsine ($[-\frac{\pi}{2}, \frac{\pi}{2}]$) whose sine is $\frac{1}{2}$. Answer: $\frac{\pi}{6}$.

Find the angle within the range of arccosine ($[0, \pi]$) whose cosine is 0. Answer: $\frac{\pi}{2}$.

Find the angle within the range of arctangent $(-\frac{\pi}{2}, \frac{\pi}{2})$ whose tangent is 1. Answer: $\frac{\pi}{4}$.

Let $\theta = \arccos(x)$. Then $\cos(\theta) = x$. Draw a right triangle, find the missing side using the Pythagorean theorem, and then find $\sin(\theta)$.

Let $\theta = \arcsin(\frac{3}{5})$. Then $\sin(\theta) = \frac{3}{5}$. Draw a right triangle, find the adjacent side using the Pythagorean theorem (which is 4), and then find $\tan(\theta) = \frac{3}{4}$.

Let $\theta = \arctan(2)$. Then $\tan(\theta) = 2$. Draw a right triangle, find the hypotenuse using the Pythagorean theorem (which is $\sqrt{5}$), and then find $\cos(\theta) = \frac{1}{\sqrt{5}}$.

The inverse function of sine; finds the angle whose sine is a given value.

The inverse function of cosine; finds the angle whose cosine is a given value.

The inverse function of tangent; finds the angle whose tangent is a given value.

The angle whose sine is x.

The angle whose cosine is x.

The angle whose tangent is x.