The trigonometric functions are not one-one on their natural domains — $\sin 30^\circ = \sin 150^\circ = \tfrac12$ — so they are not invertible as they stand. To define an inverse we first restrict the domain to an interval on which the function is one-one and still covers the entire range. That restricted interval is called the principal value branch.
The standard principal branches
On these branches each inverse function returns the single principal value:
| Function | Domain | Principal value branch (range) |
|---|---|---|
| $\sin^{-1}x$ | $[-1,1]$ | $\left[-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right]$ |
| $\cos^{-1}x$ | $[-1,1]$ | $[0,\ \pi]$ |
| $\tan^{-1}x$ | $\mathbb{R}$ | $\left(-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right)$ |
| $\csc^{-1}x$ | $|x|\ge 1$ | $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$ |
| $\sec^{-1}x$ | $|x|\ge 1$ | $[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ |
| $\cot^{-1}x$ | $\mathbb{R}$ | $(0,\ \pi)$ |
Reading the symbol correctly
$\sin^{-1}x$ is the angle whose sine is $x$ — it is not $\dfrac{1}{\sin x}$. So $\sin^{-1}x = \theta$ means $\sin\theta = x$ and $\theta$ lies in the principal branch $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. The two conditions together pin down a unique answer.
A common trap: $\sin^{-1}\!\left(\sin\tfrac{2\pi}{3}\right)\ne \tfrac{2\pi}{3}$, because $\tfrac{2\pi}{3}$ is outside $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. You must bring the angle back into the branch (here the answer is $\tfrac{\pi}{3}$).