Domain: [-1, 1], Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$, increasing function, symmetric about the origin.

Domain: [-1, 1], Range: $[0, \pi]$, decreasing function, no symmetry.

Domain: $(-\infty, \infty)$, Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$, increasing function, symmetric about the origin, horizontal asymptotes at $y = \pm \frac{\pi}{2}$.

The graph is bounded between $y = -\frac{\pi}{2}$ and $y = \frac{\pi}{2}$, ensuring it's a function (passes the vertical line test).

The graph is bounded between $y = 0$ and $y = \pi$, ensuring it's a function (passes the vertical line test).

The graph is bounded between $y = -\frac{\pi}{2}$ and $y = \frac{\pi}{2}$, ensuring it's a function (passes the vertical line test).

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.