Imagine a robot arm. You know exactly how far along and how far up the gripper needs to reach — 3 units across, 3 units up — and now the motor at the shoulder needs to know one thing: what angle to turn to. You know the ratio; you need the angle back. That backwards question is exactly what an inverse trig function answers.

Forward, the trig functions turn an angle into a ratio: \sin 30^\circ = \tfrac12. Inverse, they turn a ratio back into an angle: \arcsin \tfrac12 = 30^\circ. The three inverses have two spellings each — pick whichever you like, they mean the same thing:

A ramp, a staircase, a satellite dish, the recoil angle of a billiard ball — anywhere you measure a ratio of sides and want the angle, you are pressing the \sin^{-1} button. This page is about the one subtle catch that makes those buttons behave, and how to read exact values off them.

The problem: sine hits the same value over and over

Here is the snag. To undo a function you need each output to come from exactly one input — the function must pass the horizontal-line test (no horizontal line may cross it twice). But \sin fails this spectacularly: \sin 30^\circ = \tfrac12, and so does \sin 150^\circ, and \sin 390^\circ, and infinitely many more. The wave is many-to-one. If you ask "which angle has sine \tfrac12?", there is no single honest answer.

The fix is a deal we strike with the function: we restrict its domain to just one clean stretch where it climbs (or falls) once through every value, passing the horizontal-line test. On that stretch it is one-to-one, so it can be inverted. The chosen stretch is called the set of principal values, and the output of the inverse always lives there:

Notice \arccos's range is different from the other two — it lives entirely at or above zero, in [0,\pi], because that is where cosine falls once from 1 to -1. Angles here are in radians; the exact values below come straight from the special-angle table.

An inverse is a mirror image in the line y = x

There is a beautiful way to see the inverse. Take the restricted trig graph and reflect it in the diagonal line y = x — swap every (a, b) for (b, a) — and you get the inverse graph. Inputs and outputs trade places, which is precisely what "undo" means.

Choose an inverse below. The faint dashed curve is the trig function restricted to its principal stretch; the thin diagonal is the mirror y = x; and the bold curve is the inverse — the dashed curve flipped across the mirror. Watch how the inverse's whole height stays trapped inside its principal-value range: \arcsin and \arctan between \pm\tfrac{\pi}{2}, \arccos between 0 and \pi.

Worked example 1 — \arccos\!\left(-\tfrac12\right)

We want the angle whose cosine is -\tfrac12, and the answer must land in [0, \pi].

• Recall the special value \cos 60^\circ = \cos\tfrac{\pi}{3} = \tfrac12. We need cosine to be negative, so we are looking in the second quadrant, where cosine is below zero.
• The angle in [0,\pi] with cosine -\tfrac12 is the supplement of \tfrac{\pi}{3}: \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}.

Check it lands home: \tfrac{2\pi}{3} is indeed inside [0,\pi]. Tempting wrong answer: -\tfrac{2\pi}{3} also has cosine -\tfrac12, but it is outside the range, so it is not the principal value.

Worked example 2 — a composition: \cos(\arctan 2)

Now the classic move: the inside is an inverse (an angle), and the outside is a different trig function of that angle. You almost never need a calculator — a right triangle does it exactly.

Let \theta = \arctan 2. Then \tan\theta = 2 = \dfrac{2}{1} = \dfrac{\text{opposite}}{\text{adjacent}}, so build a right triangle whose side opposite \theta is 2 and whose side adjacent is 1. Pythagoras fills in the hypotenuse:

Now just read cosine straight off the triangle:

The identical trick handles \cos\!\left(\arcsin\tfrac35\right): set \sin\theta = \tfrac35 = \tfrac{\text{opp}}{\text{hyp}}, so opposite = 3, hypotenuse = 5, and the adjacent side is \sqrt{5^2 - 3^2} = 4 — the famous 3\text{–}4\text{–}5 triangle. Hence \cos\!\left(\arcsin\tfrac35\right) = \tfrac45.

Undo and redo: when does it cancel?

Inverse and function should cancel — and one direction always does. The other direction has a catch that trips up almost everyone.

• \sin(\arcsin x) = x — always (for every valid x in [-1,1]). You feed in a ratio, get the principal angle, take its sine, and land back on the ratio. Clean.
• \arcsin(\sin x) = x — only when x is already in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]. If your starting angle lives outside that range, arcsin cannot give it back — it is only allowed to return an answer inside the range.

With the angle recovered, you're ready to hunt down every solution of a trig equation — the inverse gives the first one, and symmetry (from Properties of sin(x)) supplies the rest. And every reshaped inverse graph is just a mirror image of the transforms you met in Transforming Trig Graphs.