A Cautionary Tale of Compute Inverse Trigonometric Functions

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}


\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.

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