The six trigonometric functions are periodic, so on their natural domains they are not one-one: $\sin 30^\circ=\sin 150^\circ=\tfrac12$, and infinitely many angles share any given sine. A function can be inverted only when it is one-one and onto, so to define an inverse we first restrict the domain to an interval on which the function takes every value of its range exactly once. That chosen interval is the principal value branch, and the value the inverse returns from it is the principal value.
Why restrict, and why these particular intervals
The branch is chosen by two rules: it must make the function one-one, and it must still cover the entire range of the original function. For $\sin$ the interval $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ works because $\sin$ climbs steadily from $-1$ to $1$ there, hitting each value once. For $\cos$ we cannot use that interval (it is symmetric, so $\cos$ repeats), so we take $[0,\pi]$, where $\cos$ falls steadily from $1$ to $-1$. The branches are conventions, but they are the ones used universally and the ones examiners expect.
Reading the symbol correctly
$\sin^{-1}x$ is the angle whose sine is $x$ — it is not $\dfrac{1}{\sin x}$ (that would be $(\sin x)^{-1}=\csc x$). So $\sin^{-1}x=\theta$ packs two conditions: $\sin\theta=x$ and $\theta$ lies in the principal branch $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Both together pin down a single answer. The same double condition defines all six inverse functions.
The standard principal branches
Learn this table cold — almost every question in the chapter begins here. The domain of each inverse function is the range of the original trig function; its range (the branch) is the restricted domain we picked.
| Function | Domain | Principal value branch (range) |
|---|---|---|
| $\sin^{-1}x$ | $[-1,1]$ | $\left[-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right]$ |
| $\cos^{-1}x$ | $[-1,1]$ | $[0,\ \pi]$ |
| $\tan^{-1}x$ | $\mathbb{R}$ | $\left(-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right)$ |
| $\csc^{-1}x$ | $|x|\ge 1$ | $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$ |
| $\sec^{-1}x$ | $|x|\ge 1$ | $[0,\pi]\setminus\left\{\tfrac{\pi}{2}\right\}$ |
| $\cot^{-1}x$ | $\mathbb{R}$ | $(0,\ \pi)$ |
Two patterns make the table easy to recall. The functions that pair $-1$ and $+1$ symmetrically — $\sin^{-1}$, $\tan^{-1}$, $\csc^{-1}$ — have a branch centred on $0$ (negative angles allowed). The "co" partners $\cos^{-1}$, $\cot^{-1}$, $\sec^{-1}$ have a branch in $[0,\pi]$ (no negative angles; a negative input pushes the answer into the second quadrant). Note the two deletions: $0$ is removed from the $\csc^{-1}$ branch (because $\csc$ is undefined at $0$) and $\tfrac{\pi}{2}$ from the $\sec^{-1}$ branch (because $\sec$ is undefined there).
Evaluating a principal value — the method
To find, say, $\cos^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)$: (1) call the answer $\theta$ and write the defining equation $\cos\theta=-\tfrac{\sqrt3}{2}$; (2) restrict to the correct branch, here $[0,\pi]$; (3) recall the related acute angle ($\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$) and place it in the right quadrant of the branch (cosine is negative in the second quadrant, so $\theta=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}$). Always confirm the final angle actually lies in the branch.
The trap with $f^{-1}(f(\theta))$
$\sin^{-1}(\sin\theta)$ equals $\theta$ only when $\theta$ already lies in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. For example $\sin^{-1}\!\left(\sin\tfrac{2\pi}{3}\right)\ne\tfrac{2\pi}{3}$, because $\tfrac{2\pi}{3}$ is outside the branch. Reduce first: $\sin\tfrac{2\pi}{3}=\sin\!\left(\pi-\tfrac{2\pi}{3}\right)=\sin\tfrac{\pi}{3}$, and $\tfrac{\pi}{3}$ is in the branch, so the value is $\tfrac{\pi}{3}$. The same caution applies to $\cos^{-1}(\cos\theta)$ (reduce into $[0,\pi]$) and $\tan^{-1}(\tan\theta)$ (reduce into $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$).