The equation $\operatorname{sen} x = \frac{1}{3}$ has infinitely many solutions, but none of them is a notable arc. How to write the solution? We need a name for “the arc whose sine is $\frac{1}{3}$”. The problem is that there are infinitely many arcs like this; so that the name designates a number, we choose an interval where the sine is not repeated. This choice defines the inverse circular functions.

The story

The notation $\operatorname{arc\,sen} x$ — “the arc whose sine is $x$” — was used by Daniel Bernoulli in 1729 and by the Austrian mathematician Joseph Scherffer in 1772, and gained traction with Lagrange and other authors from the end of the 18th century. The notation $\operatorname{sen}^{-1} x$, common in calculators, was proposed by the English astronomer John Herschel in 1813, by analogy with the inverse of a function; it has the defect of being confused with $\frac{1}{\operatorname{sen} x}$, which is why in this book we use $\operatorname{arc\,sen}$.

  1. 1729
    Daniel Bernoulli writes “A S.” for the arc whose sine is given.
  2. 1772
    Joseph Scherffer uses arc. sin.
  3. end of the century. XVIII
    Lagrange and others spread arc sin.
  4. 1813
    John Herschel proposes $\sin^{-1} x$.
  5. today
    Calculators use $\sin^{-1}$; Brazilian books, $\operatorname{arc\,sen}$.

Sources: John Herschel (MacTutor) · F. Cajori, A History of Mathematical Notations, vol. II (1929), §§ 532–535 · Earliest Uses of Function Symbols (Jeff Miller).

In everyday life

  • Steep streets. A street with 20% slope rises 20 m every 100 m horizontally; the angle is $\operatorname{arc\,tg} 0{,}2 \approx 11{,}3^\circ$.
  • Ladder. The safety rule requires moving the foot of the ladder 1/4 away from the wall; the angle with the floor is $\operatorname{arc\,cos}\frac{1}{4} \approx 75{,}5^\circ$.
  • Cell phone as level. The accelerometer measures how much gravity falls on each axis of the device, and the level application calculates the inclination with the tangent arc.
From safety rule (length leg away 1/4) to angle: $\operatorname{arc\,cos}\frac14 \approx 75{,}5^\circ$. Use the button.

Theory

A function only has an inverse if it is bijective. The sine, from $\mathbb{R}$ to $[-1, 1]$, is surjective but not injective (each value is reached infinitely many times). By restricting the domain to an interval in which it is increasing or decreasing, we obtain a bijection — and then we can invert it.

Arcsine function

The function $f\colon \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1, 1]$, $f(x) = \operatorname{sen} x$, is bijective: it is increasing in this interval (1st and 4th quadrants) and takes all values ​​from $-1$ to $1$.

DefiniçãoArc sine

The inverse of this restriction is the function arc sine, $\operatorname{arc\,sen}\colon [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$:

$$y = \operatorname{arc\,sen} x \iff \operatorname{sen} y = x \;\;\text{e}\;\; -\frac{\pi}{2} \le y \le \frac{\pi}{2}.$$

As with every function and its inverse, the graphs of $\operatorname{sen}$ (restricted) and $\operatorname{arc\,sen}$ are symmetric with respect to the straight line $y = x$.

$\operatorname{sen} x$ into $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (blue), $\operatorname{arc\,sen} x$ into $[-1, 1]$ (red) and angle bisector $y = x$ (dashed).

Arc cosine function

The cosine is decreasing in $[0, \pi]$ (1st and 2nd quadrants) and there takes on all values ​​from $1$ to $-1$; therefore $f\colon [0, \pi] \to [-1, 1]$, $f(x) = \cos x$, is bijective.

DefiniçãoArc cosine

$\operatorname{arc\,cos}\colon [-1, 1] \to [0, \pi]$:

$$y = \operatorname{arc\,cos} x \iff \cos y = x \;\;\text{e}\;\; 0 \le y \le \pi.$$
$\operatorname{arc\,cos} x$: descending, from $\pi$ (in $x = -1$) to $0$ (in $x = 1$), passing through $\frac{\pi}{2}$ in $x = 0$.

PropriedadeTwo relationships

$$\operatorname{arc\,sen} x + \operatorname{arc\,cos} x = \frac{\pi}{2}$$$$\operatorname{arc\,cos}(-x) = \pi - \operatorname{arc\,cos} x \qquad (-1 \le x \le 1)$$
View the demos

I) Let $y = \operatorname{arc\,sen} x$: $\operatorname{sen} y = x$ and $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Then $\cos\left(\frac{\pi}{2} - y\right) = \operatorname{sen} y = x$ and $0 \le \frac{\pi}{2} - y \le \pi$; by definition, $\frac{\pi}{2} - y = \operatorname{arc\,cos} x$.

II) Let $y = \operatorname{arc\,cos} x$: $\cos(\pi - y) = -\cos y = -x$ and $0 \le \pi - y \le \pi$; logo $\pi - y = \operatorname{arc\,cos}(-x)$.

Arc tangent function

The tangent is increasing at $\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[$ and assumes all real values ​​there; therefore $f\colon \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[ \to \mathbb{R}$, $f(x) = \operatorname{tg} x$, is bijective.

DefiniçãoArc tangent

$\operatorname{arc\,tg}\colon \mathbb{R} \to \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[$:

$$y = \operatorname{arc\,tg} x \iff \operatorname{tg} y = x \;\;\text{e}\;\; -\frac{\pi}{2} < y < \frac{\pi}{2}.$$
$\operatorname{arc\,tg} x$ is defined throughout $\mathbb{R}$ and approaches the straight lines $y = \pm\frac{\pi}{2}$ (dashed) without ever touching them.

Atençãoarc sin (sin x) is not always x

$\operatorname{sen}(\operatorname{arc\,sen} x) = x$ to all $x \in [-1, 1]$. But $\operatorname{arc\,sen}(\operatorname{sen} x) = x$ only when $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$: for example, $\operatorname{arc\,sen}(\operatorname{sen}\pi) = \operatorname{arc\,sen} 0 = 0 \ne \pi$.

ExemploSolving sin x = 1/3

With arcsine we can write the solution: $x = \operatorname{arc\,sen}\frac{1}{3} + 2k\pi$ or $x = \pi - \operatorname{arc\,sen}\frac{1}{3} + 2k\pi$.

The exercises in this section are numbers 246 to 270 from chapter list.