arcsin

Consider set

$S = \{(x,y) | \sin(y)=x, -1

Since

$\sin(\underbrace{\frac{\pi}{2}}_{y}) = \underbrace{1}_{x}, \quad \sin(\underbrace{\frac{5\pi}{2}}_{y}) = \underbrace{1}_{x},$

it does not define a function.

However, redefine $S$ as

$S = \{(x,y) | \sin(y)=x, -1 < x < 1, \boxed{-\frac{\pi}{2} \le y \le \frac{\pi}{2}}\},\quad\quad\quad(1)$

we have

$\forall (x, y_1), (x, y_2) \in S, \sin(y_1)-\sin(y_2) = \left( \frac{d}{dy}\sin(y)\bigg|_{y=\xi}\right)\cdot(y_1-y_2)$

where $-\frac{\pi}{2} < \xi < \frac{\pi}{2}.$

That is,

$\sin(y_1)-\sin(y_2) = \cos(\xi)\cdot(y_1-y_2), \quad-\frac{\pi}{2} < \xi < \frac{\pi}{2}.\quad\quad\quad(2)$

(see “A Sprint to FTC“)

From (1),

$\sin(y_1)-\sin(y_2) = x - x =0$

and it simplifies (2) to

$0 = \cos(\xi)\cdot(y_1-y_2), \quad-\frac{\pi}{2} < \xi < \frac{\pi}{2}.\quad\quad\quad(3)$

Since for $-\frac{\pi}{2} < \xi < \frac{\pi}{2}, \cos(\xi) \ne 0,$ (3) gives $y_1 = y_2.$ It means

$\forall (x, y_1), (x, y_2) \in S \implies y_1=y_2.$

i.e.,

It is true that $\forall (x_1, y_1), (x_2, y_2) \in S, x_1=x_2\implies y_1=y_2.$

And so,

The set $S$ defines a function.

In fact, $S$ defines $\arcsin$, the inverse function of $\sin.$

Let us now examine $\arcsin$ qualitatively.

Differentiate $\sin(y) = x$ gives

$\cos(y)\frac{dy}{dx} = 1.$

Since

$\cos(y) > 0$ for $-\frac{\pi}{2} < y <\frac{\pi}{2}$,

we have

$\frac{dy}{dx} = \frac{1}{\cos(y)} =\frac{1}{\sqrt{1-(\sin(y))^2}}\overset{\sin(y)=x}{=}\frac{1}{\sqrt{1-x^2}}$.

That is,

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

It follows that

$\arcsin(x)$ is an increase function on $-1 < x < 1$.

Moreover,

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

i.e.,

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

And so,

for $0 \le x < 1, \frac{x}{\sqrt{1-x^2}} >0 \implies \arcsin(x)$ is concave on $0 \le x < 1$. It is convex on $-1< x < 0.$

Fig. 1 $\arcsin$ illustrated qualitatively

To compute $\arcsin(x)$ for any given $x$, see “A Cautionary Tale of Compute Inverse Trigonometric Functions“, “A Mathematical Allegory“.

Exercise-1 Define $\arccos$, the inverse function of $\cos.$