Does the function $x \mapsto \sqrt{1-x^2}$ have a name?

Define a function

$$[-1,1] \rightarrow \mathbb{R}$$

$$x \mapsto x^\perp$$

as follows:

$$x^\perp = \sqrt{1-x^2}.$$

I find this function to be pedagogically useful, to help students grasp certain facts about the circular functions and relationships between circles and right-angled triangles.

Question. Does $x^\perp$ have an accepted name and/or notation?

I might as well tell you guys how I use it.

I begin by drawing a right-angled triangle with hypotenuse $1$. One side is labelled $x$. I ask the student to write find the length of the other side, and to begin by labelling the other side $y$. So they solve $x^2+y^2=1^2$ for $y$, and obtain $y = \sqrt{1-x^2}$. I tell them: lets call $\sqrt{1-x^2}$ the "Pythagorean complement" of $x$, and denote it $x^\perp$. I get them to compute a few examples until they've internalized the meaning of the formula. Then we graph the whole thing an it ends up looking like a half-circle. Hmmm, why is that? I get them to compute the length of the vector $(x,x^\perp),$ and sure enough it's $1$. That makes sense; we obtained $x^\perp$ by solving the equation $|(x,y)| = 1$ subject to the constraint $y \geq 0$, and then defining $y= x^\perp$. So it stands to reason that $(x,x^\perp)$ is always a point on the upper half of the unit circle.

Then we move to trigonometry. I complete the half-circle to a unit circle, and draw a vertical line at $x=\frac{1}{2}$. What are the $y$-values of the points of intersection? The student puzzles out that these are $\pm x^\perp$. I get them to write those two points as $(x,x^\perp)$ and $(x,-x^\perp)$. To see why this makes sense, we go back to the equation $x^2+y^2 = 1^2$ and solve it properly, with no assumptions on $y$ beyond being a real number and satisfying this equation. We get $$x^2+y^2 = 1^2 \iff y^2=1-x^2 \iff y \in \pm \sqrt{1-x^2} \iff y \in \pm x^\perp.$$

Seems to make sense.

I get the student to compute a few examples:

$$0^\perp = 1, \qquad \left(\frac{1}{2}\right)^\perp = \frac{\sqrt{3}}{2}, \qquad \left(\frac{1}{\sqrt{2}}\right)^\perp = \frac{1}{\sqrt{2}}, \qquad \mathrm{etc.} $$

I explain that this pairs off each point in $[0,1]$ with another point, such that the ordered pair $(x,x^\perp)$ is always on the unit circle in the first quadrant. We discuss whether or not $x \mapsto x^\perp$ is an involution. On the domain $[0,1]$, it is. But on $[-1,1]$, all we can say is that $(x^\perp)^\perp = |x|.$

Finally, we get to solving some trigonometric equations. In order to avoid complex numbers, I define $\mathrm{wrap} \,\theta = (\cos \theta, \sin \theta).$ The student computes $\|\mathrm{wrap}\,\theta\|$ and checks that this is always $1$. So $\mathrm{wrap}$ lands us on the unit circle; it "wraps" the real line around the circle. Finally we try to solve some trigonometric equations. We proceed like so:

TFAE

$\cos \theta = \frac{1}{2}$

$\mathrm{wrap}\, \theta = \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) \vee \mathrm{wrap} \,\theta = \left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

$\left(\theta \in \frac{\pi}{3}+2\pi \mathbb{Z}\right) \vee \left(\theta \in \frac{5\pi}{3}+2\pi \mathbb{Z}\right)$

$\theta \in \left(\frac{\pi}{3}+2\pi \mathbb{Z}\right) \cup \left(\frac{5\pi}{3}+2\pi \mathbb{Z}\right)$

I think students appreciate this kind of reasoning, because it helps them understand why there's two families of solutions to each of these problems.

For more advanced students, you can show them things like:

$$(\cos\theta)^\perp = |\sin \theta|, \qquad (\sin\theta)^\perp = |\cos \theta|$$

$$\cos (\mathrm{arcsin}(x)) = \sin (\mathrm{arccos}(x)) = x^\perp$$

$$\frac{d}{dx} \mathrm{arcsin}(x) = \frac{1}{x^\perp}, \qquad \frac{d}{dx} \mathrm{arccos}(x) = -\frac{1}{x^\perp}, \quad \frac{d}{dx} \mathrm{arcsec} = \frac{1}{|x|x^\perp}$$

Etc.

You can also give them some challenge problems related to the geometry of the circle. For instance, ask them to conjecture a value for $\int_{x=-1}^1 x^\perp$.

Addendum. I just learned that this function shows up when sums of inverse trigonometric functions are involved.

In particular:

$$\sin^{-1}(a) \pm \sin^{-1}(b) = \sin^{-1}(ab^\perp \pm a^\perp b)$$

$$\cos^{-1}(a) \pm cos^{-1}(b) = \cos^{-1}(ab \mp a^\perp b^\perp)$$

Solution 1:

I would call it a semicircle.

But I need thirty characters.