Consider set

S = \{(x,y) | \sin(y)=x, -1<x<1\}.


\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.


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.


\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.


\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}}.


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


One thought on “arcsin

  1. Pingback: Newton’s Pi Simplified |

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