# Arccosine arccos(x): Graph, Domain, Range, Derivative and Integral

> The arccosine function arccos(x), the inverse of cosine: domain [−1, 1], range [0, π], graph, derivative −1/√(1−x²), integral and examples.

Source: https://practicecalculus.com/functions/arccosine
Updated: 2026-11-03

$$ f(x) = \arccos x = \cos^{-1} x $$

## At a glance

- **Domain:** $[-1, 1]$
- **Range:** $[0, \pi]$
- **Derivative:** $-\frac{1}{\sqrt{1 - x^2}}$
- **Integral:** $x\arccos x - \sqrt{1 - x^2} + C$
- **Meaning:** The angle in $[0, \pi]$ whose cosine is $x$
- **Identity:** $\arccos x = \frac{\pi}{2} - \arcsin x$

## Key takeaways

- $\arccos x$ is the angle in $[0, \pi]$ whose cosine is $x$, defined for $-1 \le x \le 1$.
- Its derivative is $-\frac{1}{\sqrt{1-x^2}}$, the negative of the arcsine derivative.
- Arcsine and arccosine always add to $\frac{\pi}{2}$.

The **arccosine function** undoes cosine: $\arccos x$ is the angle in $[0, \pi]$ whose cosine is $x$. It is defined only for $-1 \le x \le 1$, and it is closely tied to [arcsine](/functions/arcsine): the two always add up to $\frac{\pi}{2}$.

## Definition of arccos x

$$y = \arccos x \quad\Longleftrightarrow\quad \cos y = x \text{ and } 0 \le y \le \pi$$

For example $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$, because $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\frac{2\pi}{3}$ lies in $[0, \pi]$. The interval $[0, \pi]$ is used because [cosine](/functions/cosine-function) is one-to-one there and hits every value in $[-1, 1]$.

## Graph of y = arccos x

```graph
title: 'Graph of y = arccos x on [−1, 1]'
caption: 'Arccosine falls from $\pi$ at $x = -1$ to $0$ at $x = 1$, passing through $\left(0, \frac{\pi}{2}\right)$.'
x: [-1.8, 1.8]
y: [-0.4, 3.6]
functions:
  - expr: arccos(x)
    label: y = arccos x
    domain: [-1, 1]
points:
  - [-1, 3.1416, '(−1, π)']
  - [0, 1.5708, '(0, π/2)']
  - [1, 0, '(1, 0)']
```

Unlike arcsine, arccosine is decreasing, and it is neither even nor odd. Its graph is symmetric about the point $\left(0, \frac{\pi}{2}\right)$, which reflects the identity $\arccos(-x) = \pi - \arccos x$.

## Key values of arccosine

| $x$ | $-1$ | $-\frac{\sqrt{3}}{2}$ | $-\frac{1}{2}$ | $0$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | $1$ |
| --- | --- | --- | --- | --- | --- | --- | --- | --- |
| $\arccos x$ | $\pi$ | $\frac{5\pi}{6}$ | $\frac{2\pi}{3}$ | $\frac{\pi}{2}$ | $\frac{\pi}{3}$ | $\frac{\pi}{4}$ | $\frac{\pi}{6}$ | $0$ |

## Properties of arccosine

- **Domain $[-1, 1]$, range $[0, \pi]$.**
- **Decreasing** on its whole domain.
- **Cancellation.** $\cos(\arccos x) = x$ for $x$ in $[-1, 1]$; $\arccos(\cos x) = x$ only for $x$ in $[0, \pi]$.
- **Complementary identity.** $\arcsin x + \arccos x = \frac{\pi}{2}$ for every $x$ in $[-1, 1]$.
- **Reflection.** $\arccos(-x) = \pi - \arccos x$.

## Derivative of arccos x

$$\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1$$

**Why.** Let $y = \arccos x$, so $\cos y = x$. Differentiating implicitly, $-\sin y\cdot y' = 1$, so $y' = -\frac{1}{\sin y}$. On $[0, \pi]$ we have $\sin y \ge 0$, so $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$.

A quicker route: differentiate $\arccos x = \frac{\pi}{2} - \arcsin x$. The constant disappears and the sign flips.

## Integral of arccos x

Use [integration by parts](/blog/integration-by-parts) with $u = \arccos x$ and $dv = dx$, so $du = -\frac{dx}{\sqrt{1-x^2}}$ and $v = x$:

$$\int\arccos x\,dx = x\arccos x + \int\frac{x}{\sqrt{1 - x^2}}\,dx = x\arccos x - \sqrt{1 - x^2} + C$$

Check: the derivative of $-\sqrt{1-x^2}$ is $\frac{x}{\sqrt{1-x^2}}$, which cancels the $-\frac{x}{\sqrt{1-x^2}}$ from the product rule on $x\arccos x$, leaving $\arccos x$.

In practice you rarely need an "arccos" antiderivative for $-\frac{1}{\sqrt{1-x^2}}$, since $-\arcsin x + C$ works just as well. The two answers differ by the constant $\frac{\pi}{2}$.

## Worked examples

**Example 1: chain rule.** Differentiate $f(x) = \arccos(x^2)$:

$$f'(x) = -\frac{2x}{\sqrt{1 - x^4}}$$

**Example 2: evaluate.** $\arccos\left(\cos\frac{5\pi}{3}\right)$. Since $\cos\frac{5\pi}{3} = \frac{1}{2}$, the answer is $\arccos\frac{1}{2} = \frac{\pi}{3}$, not $\frac{5\pi}{3}$.

**Example 3: composition.** Simplify $\sin(\arccos x)$. With $\theta = \arccos x$ in $[0, \pi]$, $\sin\theta \ge 0$, so $\sin(\arccos x) = \sqrt{1 - x^2}$.

**Example 4: a tangent line.** At $x = 0$, $\arccos 0 = \frac{\pi}{2}$ and the slope is $-\frac{1}{\sqrt{1}} = -1$. The tangent line is $y = \frac{\pi}{2} - x$.

## Common mistakes

1. **Using the arcsine range.** $\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$, which is in $[0, \pi]$, not $-\frac{\pi}{4}$.
2. **Dropping the minus sign in the derivative.** The arcsine derivative is positive; the arccosine derivative is negative.
3. **Treating $\cos^{-1}x$ as $\sec x$.** The reciprocal $\frac{1}{\cos x}$ is the secant (see [reciprocal trig functions](/functions/reciprocal-trig-functions)).
4. **Assuming $\arccos(\cos x) = x$ always.** It only holds for $x$ in $[0, \pi]$.

## How arccosine connects to other topics

Arccosine gives the angle between two vectors, $\theta = \arccos\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}$, which is why it shows up with the [dot product](/blog/vectors-dot-and-cross-product). Its derivative mirrors [arcsine](/functions/arcsine), and the third inverse trig function you need for integration is [arctangent](/functions/arctangent).

## Frequently asked questions

### What is the derivative of arccos x?

The derivative of $\arccos x$ is $-\frac{1}{\sqrt{1-x^2}}$ for $-1 < x < 1$. With the chain rule, $\frac{d}{dx}\arccos(u) = -\frac{u'}{\sqrt{1-u^2}}$.

### What is the range of arccos x?

The range is $[0, \pi]$. Cosine is one-to-one on that interval and takes every value from $-1$ to $1$ exactly once there.

### Why is the range of arccos different from the range of arcsin?

Sine is one-to-one on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ but cosine is not (it is even). Cosine is one-to-one on $[0, \pi]$, so arccosine uses that interval instead.

### What is the integral of arccos x?

Integration by parts gives $\int\arccos x\,dx = x\arccos x - \sqrt{1-x^2} + C$.