A Cautionary Tale of Compute Inverse Trigonometric Functions

October 27, 2021January 5, 2023 / Michael Xue / Leave a comment

From $\arcsin(x)$ ‘s definition:

$\{(x, y) | \sin(y) = x, -\frac{\pi}{2} \le y \le \frac{\pi}{2}\},$

we have

$\arcsin(0) = 0, \arcsin(\frac{1}{2})= \frac{\pi}{6},\arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}, \arcsin(1) = \frac{\pi}{2}$

and,

$\arcsin(-\frac{1}{2})= -\frac{\pi}{6},\arcsin(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}, \arcsin(-1) = -\frac{\pi}{2}.$

To obtain other values of $\arcsin(x)$ , we may simply solve

$\sin(y) - x = 0$

for $y$ where $-1 \le x \le 1.$

For example (see Fig. 1), solving $\sin(y) = \frac{1}{\sqrt{2}}$ for $y$ gives $y \approx \frac{\pi}{4}.$ It is in agreement with the fact that $\arcsin(\frac{1}{\sqrt{2}})= \frac{\pi}{4}.$

Fig. 1

In Fig. 2, we compute $\arcsin(x)$ from repeatedly solving $\sin(y) = x$ for $y$ where

$x=-1+i \cdot\frac{1-(-1)}{n}, i=1,2,...n.$

Fig. 2

Since $\sin(y)$ is a periodic function, $\sin(y) = x$ has infinitely many solutions. It is possible that the solution obtained by Newton’s method lies outside of $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , the range of $\arcsin(x)$ by its definition. Such solution cannot be considered the value of $\arcsin(x).$

Fig. 3

Exercise-1 Compute $\arccos(x)$ by solving $\cos(y) = x$ for $y.$

Exercise-2 Explain Fig. 3.

A Mathematical Allegory

October 20, 2021January 5, 2023 / Michael Xue / Leave a comment

We have defined function $\arcsin$ as a set:

$\{(x, y) | \sin(y) =x, \frac{-\pi}{2} \le y \le \frac{\pi}{2}\}.$

By definition,

$\arcsin(-1) = \frac{-\pi}{2}, \arcsin(0)=0, \arcsin(1) = \frac{\pi}{2}$

and

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

It means that $\arcsin(x)$ is the unique solution of

$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\quad\quad\quad(\star)$

where $y(-1)=-\frac{\pi}{2}, y(0)=0$ and $y(1)=\frac{\pi}{2}.$

To compute $\arcsin(x)$ , we solve $(\star)$ for $y(x)$ as follows:

Integrate $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ from $-1$ to $x$ gives

$\displaystyle\int\limits_{-1}^{x}\frac{dy}{dx} \,dx=\displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\; d\xi\overset{\textbf{FTC}}{\implies}y(x) - y(-1) = \displaystyle\int\limits_{-1}^{x} \frac{1}{\sqrt{1-\xi^2}}\, d\xi.$

Therefore,

$y(x) = \displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt(1-\xi^2}\,d\xi + y(-1) \overset{y(-1)=\frac{-\pi}{2}}{=} \displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\,d\xi - \frac{\pi}{2}.$

That is,

$\arcsin(x) = \displaystyle\int\limits_{-1}^{x} \frac{1}{\sqrt{1-\xi^2}}\;d\xi - \frac{\pi}{2}.$

To obtain $\arcsin(x), -1 < x < 1$ , we numerically evaluate $\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\,d\xi$ , using function ‘quad_qags’.

Fig. 1

The result is visually validated in Fig. 2.

Fig. 2

Note: ‘romberg’, another function that computes the numerical integration by Romberg’s method will not work since it evaluates $\frac{1}{\sqrt{1-x^2}}$ at $x=-1.$

Fig. 3

An alternate approach is to solve $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(0)=0$ as an initial-value problem of ODE using ‘rk’ , the function that implements the classic Runge-Kutta algorithm.