Inverse Trigonometric Functions — Formulas & Graphs

Inverse trigonometric functions return an angle from a trig value. In practice, $\arcsin x$, $\arccos x$, and $\arctan x$ each return one standard angle, called the principal value, not every angle that works.

That restriction is essential. Sine, cosine, and tangent repeat values on their full graphs, so they only have inverses after we limit them to intervals where each output comes from exactly one angle.

What $\arcsin x$, $\arccos x$, and $\arctan x$ mean

These definitions show both the trig relationship and the allowed output range:

$$\arcsin x = y \quad \text{means} \quad \sin y = x \text{ and } -\frac{\pi}{2} \le y \le \frac{\pi}{2}$$ $$\arccos x = y \quad \text{means} \quad \cos y = x \text{ and } 0 \le y \le \pi$$ $$\arctan x = y \quad \text{means} \quad \tan y = x \text{ and } -\frac{\pi}{2} < y < \frac{\pi}{2}$$

Those interval conditions are not extra detail. They are what make the inverse single-valued.

Domains and ranges you actually need

For the three inverse trig functions students use most often:

$$\arcsin x: \quad -1 \le x \le 1, \quad -\frac{\pi}{2} \le y \le \frac{\pi}{2}$$ $$\arccos x: \quad -1 \le x \le 1, \quad 0 \le y \le \pi$$ $$\arctan x: \quad x \in \mathbb{R}, \quad -\frac{\pi}{2} < y < \frac{\pi}{2}$$

Read each line as input first, output second. For example, $\arcsin x$ only accepts $-1 \le x \le 1$ because sine never produces a value outside that interval.

How inverse trig graphs work

Inverse trig graphs are reflections across the line $y = x$, but only after the original trig function is restricted to a one-to-one interval.

For example, $y = \arcsin x$ is the reflection of the restricted sine graph

$$y = \sin x \quad \text{for} \quad -\frac{\pi}{2} \le x \le \frac{\pi}{2}$$

across the line $y = x$.

The same idea gives these matching pairs:

$$y = \arccos x \leftrightarrow y = \cos x \quad \text{for} \quad 0 \le x \le \pi$$ $$y = \arctan x \leftrightarrow y = \tan x \quad \text{for} \quad -\frac{\pi}{2} < x < \frac{\pi}{2}$$

Do not reflect the full repeating sine, cosine, or tangent graph. The full graph fails the horizontal line test, so it cannot have an inverse function.

One worked example with principal range

Evaluate

$$\arccos\left(-\frac{1}{2}\right)$$

We want the angle $y$ such that $\cos y = -\frac{1}{2}$. Many angles work, but $\arccos x$ must return the angle in the principal range

$$0 \le y \le \pi$$

Inside that interval, the correct angle is $y = \frac{2\pi}{3}$, so

$$\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$$

That is the main habit to build: do not ask for any angle that works. Ask for the angle in the correct range.

Common inverse trig mistakes

The most common mistake is confusing inverse trig with reciprocal trig. $\arcsin x$ is not the same as $\csc x$, and $\sin^{-1} x$ usually means inverse sine, not $1/\sin x$.

Another common mistake is ignoring the principal range. For instance, $\sin \left(\frac{5\pi}{6}\right) = \frac{1}{2}$, but

$$\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$

because $\frac{\pi}{6}$ is the angle in the allowed range for $\arcsin x$.

Students also sometimes forget the domain. Expressions like $\arcsin 2$ and $\arccos(-3)$ are not real-valued because sine and cosine do not produce outputs outside $[-1,1]$.

When inverse trigonometric functions are used

Inverse trig functions show up whenever you know a ratio and need the angle back. That happens in right-triangle geometry, navigation, slope and direction problems, vector components, and triangle-based modeling.

They also matter in calculus. You see them in derivatives, antiderivatives such as $\int \frac{1}{1+x^2}\,dx = \arctan x + C$, and substitutions involving trig expressions.

A 2-step way to think about them

When you evaluate an inverse trig expression, do these two checks:

1. Which trig function matches the value I was given?
2. What is the angle in that function's principal range?

If you keep those two checks together, the formulas and graphs become much easier to read.

Try your own version

Try evaluating $\arcsin\left(-\frac{\sqrt{2}}{2}\right)$ and $\arctan(1)$. If you choose the principal range first, both answers fall into place quickly.