Inverse trigonometric functions

Sine fails the horizontal line test spectacularly: every output between −1 and 1 is hit infinitely often. A function must return one output per input, so "the inverse of sine" cannot mean "all angles with that sine." The fix is the same one used for $y = x^2$: restrict the original function to a piece where it is one-to-one, and invert only that piece.

The chosen pieces are standardized, and they are chosen so that every possible output value is still reachable exactly once:

1. arcsin inverts sine on $[-\pi/2, \pi/2]$: the right half of the unit circle. Range of arcsin: $[-\pi/2, \pi/2]$.
2. arccos inverts cosine on $[0, \pi]$: the top half of the circle. Range of arccos: $[0, \pi]$.
3. arctan inverts tangent on $(-\pi/2, \pi/2)$: one full branch. Range of arctan: $(-\pi/2, \pi/2)$, endpoints excluded because tangent never reaches them.

These ranges are not stylistic preferences. They are the definition. An answer outside the range is not an alternative answer; it is a wrong answer.

Evaluating an inverse trig expression is a two-step discipline: recall the reference angle, then place it inside the range.

$\arcsin(1/2)$: which angle in $[-\pi/2, \pi/2]$ has sine 1/2? Only $\pi/6$. The angle $5\pi/6$ also has sine 1/2, but it lives outside the contract, so it is not the value of arcsin. Negative inputs pull arcsin below zero: $\arcsin(-1/2) = -\pi/6$, not $11\pi/6$ (coterminal, but outside the range) and not $7\pi/6$.

$\arccos(-1/2)$: which angle in $[0, \pi]$ has cosine $-1/2$? That is $2\pi/3$, up in the second quadrant, because arccos handles negative inputs by climbing past $\pi/2$, never by going negative. $\arccos$ never returns a negative angle; $\arcsin$ and $\arctan$ never return an angle beyond $\pm\pi/2$.

Domains matter too: arcsin and arccos only accept inputs in $[-1, 1]$, because no angle has a sine or cosine of 2. $\arcsin(2)$ is not a hard question; it is an undefined expression. Arctan, by contrast, accepts every real number, since slopes come in all sizes.

The superscript in $\sin^{-1}x$ is function-inverse notation, the same $-1$ as in $f^{-1}$. It is not an exponent. The reciprocal of $\sin x$ is a different object with its own name:

$$\sin^{-1}x = \arcsin x \qquad \text{but} \qquad (\sin x)^{-1} = \frac{1}{\sin x} = \csc x$$

The two disagree everywhere they are both defined. Try $x = 1/2$: $\arcsin(1/2) = \pi/6 \approx 0.524$, while $1/\sin(1/2) \approx 1/0.479 \approx 2.09$. Not close. The parentheses in $(\sin x)^{-1}$ are what turn the superscript back into an ordinary exponent. When in doubt, write arcsin, arccos, arctan and the ambiguity disappears.

Composing a function with its inverse returns the input, but only when the input honors the contract. $\sin(\arcsin x) = x$ for every x in $[-1, 1]$: arcsin produces an angle, sine measures it, and you get your number back.

The reverse trip is the trap. $\arcsin(\sin\theta) = \theta$ only when $\theta$ is already in $[-\pi/2, \pi/2]$. Feed it $\theta = 5\pi/6$ and the machine returns the in-range angle with the same sine: $\arcsin(\sin(5\pi/6)) = \arcsin(1/2) = \pi/6$. The composition does not remember which angle you started with; it only knows the sine value, and it always answers from inside the range.