Arccosine Function

Introduction

1.The entry arcsine and arccosine defines the arccosine as the angle whose cosine is a given value. Here the arccosine is a real function of a real variable.

2.The cosine is periodic, so it takes every value of $[-1, 1]$ infinitely many times and has no inverse on the whole of $\mathbb{R}.$ Restricted to the interval $[0, \pi],$ it is continuous and strictly decreasing, and it maps that interval onto $[-1, 1].$ The restricted cosine is a bijection, and its inverse function is the arccosine:

$$\arccos : [-1, 1] \to [0, \pi]$$

3.The function $f(x) = \arccos(x)$ assigns to each $x \in [-1, 1]$ the unique angle in $[0, \pi],$ measured in radians, whose cosine equals $x.$ Its graph is the reflection of the restricted cosine branch in the line $y = x.$

4.The two compositions agree with the identity function on different sets. Applying the cosine after the arccosine returns $x$ for every $x \in [-1, 1],$ whereas applying the arccosine after the cosine returns $\theta$ only when $\theta$ lies in the restricted interval:

$$ \begin{align} \cos(\arccos(x)) &= x \quad \forall x \in [-1, 1] \\[6pt] \arccos(\cos(\theta)) &= \theta \quad \iff \quad \theta \in [0, \pi] \end{align} $$

5.

Outside that interval the arccosine returns the angle in $[0, \pi]$ having the same cosine as $\theta.$ For $\theta = -\frac{\pi}{3}$ the cosine equals $\frac{1}{2},$ so $\arccos(\cos(-\pi/3)) = \frac{\pi}{3}.$

Properties

6.As the inverse of the restricted cosine, the arccosine has the following properties.

8.The arccosine is bounded, and its minimum and maximum are attained at the endpoints of the domain. The relation $\arccos(-x) = \pi - \arccos(x)$ makes the graph symmetric about the point $\left(0, \frac{\pi}{2}\right).$ For every $x \in [-1, 1],$ the arccosine and the arcsine satisfy the identity:

$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$

9.Since the sine is non-negative on $[0, \pi],$ the Pythagorean identity gives algebraic expressions for the sine and tangent of the angle $\arccos(x):$

$$ \begin{align} \sin(\arccos(x)) &= \sqrt{1 - x^2} \\[6pt] \tan(\arccos(x)) &= \frac{\sqrt{1 - x^2}}{x} \end{align} $$

10.The first identity holds for every $x \in [-1, 1],$ whereas the second holds for $x \in [-1, 1]$ with $x \neq 0,$ since $\tan(\arccos(0)) = \tan\left(\frac{\pi}{2}\right)$ is undefined.

11.The arccosine also has a representation in terms of the complex logarithm. For $x \in [-1, 1],$ set $w = \arccos(x)$ and $u = e^{iw}.$ Euler's formula turns $\cos(w) = x$ into the quadratic equation $u^2 - 2xu + 1 = 0.$ Since $\sin(w) \geq 0,$ the corresponding root is $u = x + i\sqrt{1 - x^2} = e^{i\arccos(x)}.$ Taking the principal value of the complex logarithm gives:

$$\arccos(x) = -i\mathrm{Log}\left(x + i\sqrt{1 - x^2}\right)$$

12.Here $\mathrm{Log}$ is the principal value of the complex logarithm, defined by $\mathrm{Log}(z) = \ln|z| + i\mathrm{Arg}(z),$ where $\mathrm{Arg}(z) \in (-\pi, \pi].$ For $w \in [0, \pi],$ the principal argument of $e^{iw}$ is $w,$ including at $w = \pi.$ Thus the formula also holds for $x = -1,$ where $\mathrm{Log}(-1) = i\pi.$

Limits, derivatives, and integrals of the arccosine function

13.The arccosine is continuous on $[-1, 1]$ because it is the inverse of the continuous, strictly monotone restriction of the cosine. To determine its behavior near the origin, set $y = \frac{\pi}{2} - \arccos(x).$ Then $y \to 0$ as $x \to 0,$ and $x = \sin(y).$ The quotient below is therefore equal to $-y/\sin(y),$ so the standard limit $\sin(y)/y \to 1$ gives:

$$\lim_{x \to 0} \frac{\arccos(x) - \frac{\pi}{2}}{x} = -1$$

14.Thus the line $y = \frac{\pi}{2} - x$ is the first-order approximation to the graph of the arccosine near the origin.

15.The function is differentiable on the open interval $(-1, 1).$ Its derivative follows from the rule for the derivative of an inverse function. For $x \in (-1, 1),$ set $y = \arccos(x).$ Then $y \in (0, \pi)$ and $\cos(y) = x.$ Differentiating this identity with respect to $x$ gives $-\sin(y)y' = 1.$ Since the sine is positive on this interval, $\sin(y) = \sqrt{1 - \cos^2(y)} = \sqrt{1 - x^2},$ and the derivative is:

$$\frac{d}{dx}\arccos(x) = -\frac{1}{\sqrt{1 - x^2}}$$

16.The derivative is an algebraic function, although the arccosine is not. It is the negative of the derivative of the arcsine, as the identity $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$ requires. As $x$ approaches either endpoint from within the domain, the derivative tends to $-\infty:$

$$\lim_{x \to 1^-} -\frac{1}{\sqrt{1 - x^2}} = -\infty \qquad \lim_{x \to -1^+} -\frac{1}{\sqrt{1 - x^2}} = -\infty$$

17.The graph therefore has one-sided vertical tangents at the points $(1, 0)$ and $(-1, \pi).$ The cosine has horizontal tangents at $x = 0$ and $x = \pi,$ and reflection in the line $y = x$ turns them into vertical tangents.

18.The indefinite integral is computed by parts, differentiating the arccosine and integrating the constant factor $1:$

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

19.Since the numerator equals $-\dfrac{1}{2}$ times the derivative of $1 - x^2,$ the substitution $u = 1 - x^2$ reduces the remaining integral to the integral of a power. The antiderivative is:

$$\int \arccos(x) \ dx = x\arccos(x) - \sqrt{1 - x^2} + c$$

20.The arccosine is not odd, but its point symmetry determines the definite integral over the whole domain. The values at $x$ and $-x$ have sum $\pi$ and average $\dfrac{\pi}{2},$ so the area is equal to that of the rectangle with base $2$ and height $\dfrac{\pi}{2}.$ On the positive half of the domain, evaluating the antiderivative gives the value $1:$

$$\int_{-1}^{1} \arccos(x) \ dx = \pi \qquad \int_0^1 \arccos(x) \ dx = 1$$

Monotonicity and convexity

21.The derivative $-\dfrac{1}{\sqrt{1 - x^2}}$ is negative at every point of $(-1, 1),$ so the arccosine is strictly decreasing on its whole domain and has no stationary points. Its maximum is $\pi$ and its minimum is $0.$ Both are attained at the endpoints, where the function is continuous but not differentiable.

22.Convexity and concavity are determined by the second derivative:

$$\frac{d^2}{dx^2}\arccos(x) = -\frac{x}{\left(1 - x^2\right)^{3/2}}$$

23.The denominator is positive on $(-1, 1),$ so the second derivative has the sign of $-x.$ The graph is convex on $(-1, 0)$ and concave on $(0, 1).$ The point $\left(0, \frac{\pi}{2}\right)$ is an inflection point because the second derivative vanishes at $x = 0$ and changes sign there. The tangent to the graph at this point is the line $y = \frac{\pi}{2} - x.$

Maclaurin series

24.The Maclaurin series of the arccosine is obtained from its derivative. For $|t| < 1,$ the derivative has the binomial expansion:

$$-\frac{1}{\sqrt{1 - t^2}} = -\sum_{n=0}^{\infty} \frac{1}{4^n}\binom{2n}{n}t^{2n}$$

25.Here $\binom{2n}{n}$ is the central binomial coefficient of order $n.$

26.A power series can be integrated term by term inside its interval of convergence. Integrating from $0$ to $x$ and using $\arccos(0) = \dfrac{\pi}{2}$ gives the following power series, whose radius of convergence is $1:$

$$\arccos(x) = \frac{\pi}{2} - \sum_{n=0}^{\infty} \frac{1}{4^n}\binom{2n}{n}\frac{x^{2n+1}}{2n+1} = \frac{\pi}{2} - x - \frac{x^3}{6} - \frac{3x^5}{40} - \frac{5x^7}{112} - \cdots$$

27.Apart from the constant term only odd powers appear, because $\arccos(x) - \dfrac{\pi}{2}$ is odd. Every coefficient after the constant is negative, in agreement with the strict decrease of the function. Retaining the first two terms gives the approximation $\arccos(x) \approx \dfrac{\pi}{2} - x$ for small $x.$

28.At $x = 1$ the general term is asymptotic to $\dfrac{1}{2\sqrt{\pi}n^{3/2}},$ so the series converges absolutely. Abel's theorem allows us to take the limit as $x \to 1^-$ in the power series identity. Since $\arccos(1) = 0,$ the positive series subtracted from the constant term has sum $\dfrac{\pi}{2}:$

$$\sum_{n=0}^{\infty} \frac{1}{4^n}\binom{2n}{n}\frac{1}{2n+1} = \frac{\pi}{2}$$

29.The asymptotic estimate also shows that the series converges slowly at $x = 1.$ When $|x|$ is well below $1,$ the powers of $x$ decay quickly, and a few terms give an accurate value of $\arccos(x).$