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.

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 $\cos(\theta) = \frac{1}{\sqrt{5}}$.