Euler's Eyes
2026-10-21
Arcsine, arccosine and arctangent
Why is it necessary to restrict?
An inverse needs to retrieve a single input from the output. Sine, cosine and tangent repeat values. To define inverses, we choose intervals in which each function is one-to-one and reaches its entire desired image.
| Restricted function | Chosen domain | Image |
|---|---|---|
| Sine | $[-\pi/2,\pi/2]$ | $[-1,1]$ |
| Cosine | $[0,\pi]$ | $[-1,1]$ |
| Tangent | $(-\pi/2,\pi/2)$ | $\mathbb R$ |
In these intervals, sine and tangent are strictly increasing, and cosine is strictly decreasing. Each output appears only once; therefore, the inverses are well defined.
Definitions and domains
$$y=\arcsin t\iff\sin y=t,\quad-\frac\pi2\le y\le\frac\pi2,$$$$y=\arccos t\iff\cos y=t,\quad0\le y\le\pi,$$
$$y=\arctan t\iff\tan y=t,\quad-\frac\pi2
Compose in the correct order
For $t\in[-1,1]$, $\sin(\arcsin t)=t$. But $\arcsin(\sin x)=x$ is only guaranteed when $x$ is in the main range. For example:
$$\arcsin\left(\sin\frac{5\pi}{6}\right) =\arcsin\frac12=\frac\pi6.$$The value $5\pi/6$ has the same sine, but is outside the main range of the arcsine. The inverse function gives a leading answer; The trigonometric equation can have infinite answers.
An identity between inverses
If $y=\arcsin t$, then $y\in[-\pi/2,\pi/2]$ and $\pi/2-y\in[0,\pi]$. Like $\cos(\pi/2-y)=\sin y=t$, the definition of arc cosine gives:
$$\arcsin t+\arccos t=\frac\pi2,\qquad -1\le t\le1.$$The notation $\arcsin t$ represents an inverse of a restricted function, not the reciprocal $1/\sin t$. The reciprocal of the sine is the cosecant.
In everyday life
- Ramps and streets. From the slope in percentage, the angle is taken: 20% corresponds to $\operatorname{arc\,tg} 0{,}2 \approx 11{,}3^\circ$.
- Ladder. With the foot 1/4 length apart, the angle with the floor is $\operatorname{arc\,cos}\frac{1}{4} \approx 75{,}5^\circ$.
- Sensors. Cell phones and drones calculate tilt angles from sensor measurements using arc tangents.
Exercises with resolution
1. Calculate $\arccos(-\sqrt2/2)$ and $\arctan(-\sqrt3)$.
Check resolution 1
Arc cosine returns $3\pi/4$; the tangent arc returns $-\pi/3$, respecting the respective main images.
2. What is the real domain of $f(x)=\arcsin(2x-1)$?
Check resolution 2
$-1\le2x-1\le1$ is required. Adding 1 and dividing by 2, $0\le x\le1$.
In the interactive book
The same subjects, with all the demonstrations and the complete list of exercises, are in the interactive Trigonometry book:
This class is a shorter introduction to the topic; In the book, the theory is accompanied by solved and proposed exercises from the chapter.