Inverse trigonometric functions

Sine and cosine already name the height of a point on the unit circle. Given a height between $-1$ and $1$, which radian angle produced it? Many angles share that height, so the unrestricted sine does not name one of them.

Freeze one branch, then reflect

This picture keeps the sine on $[-\pi /2,\pi /2]$. That interval is extra. Drag the height $y$. The green point sits on the frozen branch. Reflect it across the diagonal $y=x$: the blue point's height is the unique angle on that interval.

For every real $y$ with $-1\le y\le 1$, the equation $y=\sin x$ with $x$ restricted to $[-\pi /2,\pi /2]$ has exactly one solution. Call that solution

$$x=\arcsin y.$$

Swap the letters formally: for every real $x$ with $-1\le x\le 1$,

$$y=\arcsin x$$

means $\sin y=x$ and $y\in [-\pi /2,\pi /2]$. Reflecting the restricted graph across the diagonal produces this graph.

The same freeze, one new cut each time

Each partner is that reflection after one further cut:

Cosine is frozen on $[0,\pi ]$, where it decreases. The reflected range stays the closed interval $[0,\pi ]$.

Tangent is frozen on the open interval $(-\pi /2,\pi /2)$. The input of arctan is then every real number, and the height never reaches $\pm \pi /2$. Drag the real height $t$. The green point sits on the central branch. The blue point is the reflection, so its height is $\arctan t$, strictly between the dashed lines.

Cotangent is frozen on $(0,\pi )$. The same reflection gives $\mathrm{arccot}$, whose height stays strictly between $0$ and $\pi$.

Drop the freeze and the periods return

The principal value is the one angle on the frozen branch. Drop that cut and the full sine meets the same height again after a full turn, and also at the supplementary angle. Those other solutions are not new function names. For $-1\le y\le 1$ and every integer $k$,

$$\sin x=y\iff x=\arcsin y+2k\pi \text{or}x=\pi -\arcsin y+2k\pi .$$

For cosine, $x=\pm \arccos y+2k\pi$. For tangent and cotangent, a single family suffices:

$$\tan x=y\iff x=\arctan y+k\pi ,\cot x=y\iff x=\mathrm{arccot}y+k\pi .$$

Drag the height and the integer $k$. The green point is the principal $\arcsin y$. The blue point is that angle plus $2k\pi$. The orange point is the supplementary angle plus $2k\pi$.

Arcsin, arccos, arctan, and arccot are the unique angles on those frozen branches. A solution of the unrestricted equation starts from that angle and adds the period. How these branches rewrite into one another, and what their slopes are, is the question the inverse trigonometric identities answer.