Arcsine Function

Introduction

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

2.The sine 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 $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],$ it is continuous and strictly increasing, and it maps that interval onto $[-1, 1].$ The restricted sine is a bijection, and its inverse function is the arcsine:

$$\arcsin : [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

3.The function $f(x) = \arcsin(x)$ assigns to each $x \in [-1, 1]$ the unique angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],$ measured in radians, whose sine equals $x.$ Its graph is the reflection of the restricted sine branch in the line $y = x.$

4.The two compositions have different domains on which they return the argument. Applying the sine after the arcsine returns every $x \in [-1, 1],$ whereas applying the arcsine after the sine returns $\theta$ only when $\theta$ lies in the restricted interval:

$$ \begin{align} \sin(\arcsin(x)) &= x \quad \forall x \in [-1, 1] \\[6pt] \arcsin(\sin(\theta)) &= \theta \quad \iff \quad \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{align} $$

5.

Outside that interval the arcsine returns the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ having the same sine as $\theta.$ For $\theta = \frac{3\pi}{4}$ the sine equals $\frac{\sqrt{2}}{2},$ so $\arcsin(\sin(3\pi/4)) = \frac{\pi}{4}.$

Properties

6.As the inverse of the restricted sine, the arcsine has the following properties.

8.The arcsine is bounded, and its minimum and maximum are attained at the endpoints of the domain rather than at interior stationary points. For every $x \in [-1, 1],$ the arcsine and the arccosine satisfy the identity:

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

9.Since the cosine is non-negative on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],$ the Pythagorean identity expresses the trigonometric functions of the angle $\arcsin(x)$ in algebraic form:

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

10.The first identity holds for every $x \in [-1, 1],$ whereas the second holds for $x \in (-1, 1),$ where the denominator does not vanish.

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

$$\arcsin(x) = -i\ln\left(ix + \sqrt{1 - x^2}\right)$$

12.Here $\ln$ denotes the principal branch.

Limits, derivatives, and integrals of the arcsine function

13.Near the origin both the arcsine and sine curves have the line $y = x$ as their first-order approximation. For the arcsine, set $y = \arcsin(x).$ Then $x = \sin(y)$ and $y \to 0$ as $x \to 0,$ so the standard limit $\sin(y)/y \to 1$ gives:

$$\lim_{x \to 0} \frac{\arcsin(x)}{x} = 1$$

14.The function is continuous on $[-1, 1]$ and 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 $y = \arcsin(x).$ Then $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\sin(y) = x.$ Differentiating this identity with respect to $x$ gives $\cos(y)y' = 1.$ Since the cosine is positive on this interval, $\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2},$ and the derivative is:

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

15.The derivative is an algebraic function, although the arcsine is not. As $x$ approaches either endpoint from within the domain, the derivative tends to $+\infty:$

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

16.The graph therefore reaches the points $\left(1, \frac{\pi}{2}\right)$ and $\left(-1, -\frac{\pi}{2}\right)$ with a vertical tangent. The sine has horizontal tangents at $x = \pm\frac{\pi}{2},$ and the reflection in the line $y = x$ turns them into vertical ones.

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

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

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

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

19.Over a symmetric interval the definite integral vanishes, since the arcsine is odd. Over the positive half of the domain the primitive gives:

$$\int_0^1 \arcsin(x) \ dx = \frac{\pi}{2} - 1$$

Monotonicity and convexity

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

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

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

22.The denominator is positive on $(-1, 1),$ so the second derivative has the sign of $x.$ The graph is concave on $(-1, 0)$ and convex on $(0, 1).$ The origin is an inflection point because the second derivative vanishes at $x = 0$ and changes sign there. Its tangent is the line $y = x.$

Maclaurin series

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

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

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

25.A power series can be integrated term by term inside its interval of convergence. Integrating from $0$ to $x$ and using $\arcsin(0) = 0$ gives a power series of radius $1:$

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

26.Only odd powers appear because the arcsine is odd. The coefficients are all positive, in contrast with the alternating coefficients of the sine series. Retaining the first term gives the approximation $\arcsin(x) \approx x$ for small $x.$ Dividing the series by $x$ and taking $x \to 0$ yields $\dfrac{\arcsin(x)}{x} \to 1.$

27.At $x = 1$ the series still converges, and its sum is the value of the function at that point:

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

28.The general term behaves like $\dfrac{1}{2\sqrt{\pi}n^{3/2}}$ as $n$ grows, so the series converges slowly at $x = 1$ and is inefficient for computing $\pi.$ When $|x|$ is well below $1,$ the powers of $x$ decay quickly, and a few terms give an accurate value of $\arcsin(x).$