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

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: the two always add up to \frac{\pi}{2}.

Definition of arccos x

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 is one-to-one there and hits every value in [-1, 1].

Graph of y = arccos x

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

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

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 with u = \arccos x and dv = dx, so du = -\frac{dx}{\sqrt{1-x^2}} and v = x:

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):

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).
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. Its derivative mirrors arcsine, and the third inverse trig function you need for integration is arctangent.