# Do "Parabolic Trigonometric Functions" exist?

The parametric equation

\begin{align*} x(t) &= \cos t\\ y(t) &= \sin t \end{align*}

traces the unit circle centered at the origin ($x^2+y^2=1$). Similarly,

\begin{align*} x(t) &= \cosh t\\ y(t) &= \sinh t \end{align*}

draws the right part of a regular hyperbola ($x^2-y^2=1$). The hyperbolic trigonometric functions are very similar to the standard trigonometric function.

Do similar functions exist that trace parabolas (because it is another conic section) when set up as parametric equations like the above functions? If so, are they also similar to the standard and hyperbolic trigonometric functions?

• I think you probably have to consider ellipses, not just the circle. There are a couple of ways of looking at the process of moving from circle to ellipse to parabola to hyperbola. There is a very geometric way of taking a right circular cone extended in both directions with axis vertical. A horizontal plane will cut the cone in a circle. Rotate the plane until it is vertical, and the section will change as indicated. Algebraically this is changing the eccentricity of the conic. The parabola is a limiting/special case, so you might expect to see something "in between". Apr 21 '12 at 21:28
• Apr 22 '12 at 0:03
• @J.M. I agree with the rest of your edits, but I don't think it was necessary to replace the British spelling of "centred" with the American one.
– user856
Jul 7 '12 at 4:14
• @Rahul, sorry. Old habits... Jul 7 '12 at 4:17
• Sine and cosine turning into sinh and cosh: desmos.com/calculator/50zuz23ic4 Feb 18 at 18:58

Indeed there are, but they're not usually called that. What ordinary trigonometric and hyperbolic functions have in common is that they are solutions to the differential equation $$f''(t) = af(t)$$ When $a$ is negative, the solutions are ordinary sines and cosines, scaled horizontally by a factor that depends on $a$. If you take a solution $f$ and draw the parametric plot $(x,y)=(f'(t), f(t))$, the result is an ellipse whose eccentricity depends on $a$. For $a=-1$ the ordinary sine and cosine are solutions, and you get a circle.

On the other hand, when $a$ is positive the solutions are either hyperbolic sines or hyperbolic cosines, again with a horizontal scaling factor that depends on $a$. A plot of $(x,y)=(f'(t), f(t))$ is one arm of a hyperbola with a central angle that depends on $a$. For $a=1$ the hyperbolic sine and cosine are solutions, and the hyperbola is right-angled.

Intuitively, then, since a parabola is the limiting case between an ellipse and a hyperbola, we should expect to get a "parabolic function" by setting $a=0$. Unfortunately the differential equation then becomes $$f''(t)=0$$ whose solutions are first-degree polynomials, and it is hard to make those create a parabola. However, there's a way out (many thanks to Qiaochu Yuan for pointing this out!): Instead of $f''(t)=af(t)$ we can take the basic differential equation to be $$f'''(t)=af'(t)$$ In the $a\ne 0$ case all this changes is to allow us to add a constant term to solutions, which just moves the conic about in the plane. But for $a=0$, the solutions are now all the polynomials of degree $\le 2$. And when we take any quadratic polynomial $f$ and plot $(x,y)=(f'(t), f(t))$, what we get is indeed a parabola centered around the $y$-axis!

If we take $f$ to be a first-degree polynomial, the parametric plot is just a straight (vertical) line, another limiting case of conic sections.

In all of the above cases, plotting $(f_1(t),f_2(t))$ for two unrelated solutions (for the same $a$) generally produces a conic of the same general kind, but perhaps moved and rotated. And the dependency on $a$ of the eccentricity/angle disappears; that was mediated through the derivative in the $x$ position.

So a "parabolic function" is simply another (redundant) term for a quadratic polynomial. It is not quite clear which should be counted as the parabolic sine and cosine, though. Cases could be made for either $\operatorname{sinp}(t) = t$ and $\operatorname{cosp}(t) = 1+\frac12 t^2$ or the other way around -- but worrying too much about that is just silly.

• @Qiaochu: Thanks for the salvaging suggestion! Apr 21 '12 at 21:52
• No problem. The issue here is that if you restrict yourself to things centered at the origin, then an ellipse or a hyperbola naturally degenerates into a pair of lines rather than a parabola, which is why you were getting linear rather than quadratic functions. The "center" of a parabola is at infinity so you need to let the center vary as well. (Note that a pair of lines is also a conic section!) Apr 21 '12 at 22:07
• @QiaochuYuan: Makes sense. That also explains why in the $(f'(t),f(t))$ case it is enough to allow vertical translations. Apr 21 '12 at 22:11
• Somewhat interestingly, $\operatorname{cosp}(x+y)-\operatorname{cosp}(x-y)= 2\operatorname{sinp}x\operatorname{sinp}y$. Apr 21 '12 at 22:39
• @sdcvvc: Nice observation. Unfortunately, the other trigonometric sum/product identities cannot generalize to any nontrivial polynomials, since the degrees wouldn't match up. Apr 21 '12 at 23:06

The problem of generalized forms of trigonmetry has been touched in the past by several authors, E. Ferrari (Rome university) proposed different forms and proved the link with elliptic functions.

Dattoli, Migliorati and Ricci used the Ferrari's approach to study the parabolic trigonometric functions and the relevant link with Chebyshev polynomials. The relevant papers have appeared on arXiv:

You could write down some equivalent functions, but they wouldn't be very useful. For instance, for $y = x^2 -1$: $$x=\frac{-1\pm\sqrt{1+4\tan^2 (t)}}{2\tan (t)}$$ $$y=\frac{1\mp\sqrt{1+4\tan^2 (t)}}{2\tan^2 (t)}$$

Why these functions are less useful than the usual hyperbolic and trigonometric functions, I don't know.

$\DeclareMathOperator{\cosp}{cosp}$ $\DeclareMathOperator{\sinp}{sinp}$ $\DeclareMathOperator{\tanp}{tanp}$ $\DeclareMathOperator{\cotp}{cotp}$ $\DeclareMathOperator{\secp}{secp}$ $\DeclareMathOperator{\cscp}{cscp}$

This is similar to another post, so I'll repeat my answer:

Defining $$\cosp u = \cosh 2u\quad\text{ and }\quad\sinp u = \sqrt{2}\sinh u$$ gives the identities $$\cosp u - \sinp^2 u = 1,\quad \cotp u - \sinp u = \cscp u,\quad\text{ and }\quad\cscp u - \tanp u = \secp u \cscp u,$$ corresponding to the parabola $x - y^2 = 1$. These are analogous to the circular functions with defining equation $x^2 + y^2 =1$ or the hyperbolic functions, $x^2-y^2 = 1$.

Let's think about trig functions as being convenient ways of going from cartesian to polar coordinates. Converting a "unit" parabola to polar, you would get something like

$$r = \frac{\sin\theta}{\cos^2\theta} = \frac{\sin\theta}{1 - \sin^2\theta}$$

where one could go on ad nauseum with trig identities

• Welcome to MathSE! Your posts will be better received if you follow a few guidelines. Learning $\LaTeX$ will make your equations much more clear. Also, you should answer the question as explicitly as possible. Your answer does not seem to address the issue of functions other than trigonometric ones relating to parabolas. Could you do further clarification in your post? Dec 21 '14 at 23:29

One place where hyperbolic functions appear is, of course, hyperbolic geometry. There is a phenomenon that formulas of hyperbolic geometry are similar to the formulas of the spherical (elliptic) geometry, except that the hyperbolic formulas use sinh and cosh, while the spherical formulas use sin and cos, and some signs are changed (because of the opposite curvature).

In particular:

spherical geometry:

curvature is $$K=1$$

the circumference of a circle of radius $$r$$ is $$2\pi \sin(r)$$

Trigonometric identity: $$\cos^2(x)−K\sin^2(x)=1$$

spherical Pythagorean theorem: $$\sin(a)\sin(b) = \sin(c)$$

differential equations: $$\cos(0)=1, \sin(0)=0, \cos′(x)=−K⋅\sin(x),\sin′(x)=\cos(x)$$

hyperbolic geometry:

curvature is $$K=−1$$

the circumference of a circle of radius $$r$$ is $$2π\sinh(r)$$

Trigonometric identity: $$\cosh^2(x)−K\sinh^2(x)=1$$

hyperbolic Pythagorean theorem: $$\sinh(a)\sinh(b) = \sinh(c)$$

differential equations: $$\cosh(0)=1, \sinh(0)=0, \cosh′(x)=−K⋅\sinh(x),\sinh′(x)=\cosh(x)$$

parabolic geometry:

The parabolic geometry would be the Euclidean geometry (but we usually just say Euclidean geometry). In the Euclidean geometry we have $$K=0$$. So the parabolic sine and the parabolic cosine would be the functions which make the similar formulas true in Euclidean geometry.

The only choice is: sinp$$(x)=x$$, cosp $$(x)=1$$. Not very interesting, but this is what you should be using if you want to use the same formulas for spherical, parabolic, and hyperbolic geometry (see e.g. functions sin_auto and cos_auto here). The formulas above work, just as many others.

Not very formal, but you could also say that the curvature is infinitesimal ($$K=\epsilon$$) -- then you get cosp$$(x)=1+\epsilon x^2/2$$, correct and usable Pythagorean theorem, and the graph (cosp($$x$$), sinp($$x$$)) is actually a parabola.

Note in the case of the unit circle and the regular hyperbola x and y appear to the same degree. What qualifies as a "regular parabola" and what symmetries do we find in the x and y paremeterizations as functions of t?

We might get such a symmetric paremeterization generalizing hmakholm left over Monica's method.

Instead of the condition stated, we can use $$x\frac{d^2y}{dt^2}=y\frac{d^2x}{dt^2}$$.

Without loss of generality, all parabolas in the plane are the parabola $$y=Ax^2+Bx+C$$ coupled with a rotation through some angle around the z axis. We can place a further restriction if we swap all x's with y's and all y's with x's we want to produce the same expression for the parabola implying a symmetry about the line $$y=x$$.

This yields $$0=A(x-y)^2-(x+y)\sqrt{2}+2C$$

Using the Quadratic formula one can obtain $$y$$ as a function of $$x$$. Afterwards, one can apply the above modified condition yielding a differential equation for $$x$$ in terms of $$t$$.

There is a way to define parabolic trigonometric functions, the parabolic sine and parabolic cosine. To do this we understand the generalized trigonometric functions as the projections on the Cartesian axis of some curve parametrized by arc-length.

First we will make some definitions, let $$P:=\{(x,y,z)\in \mathbb{R}^3:z=1\}$$ and $$\operatorname{rot}_{\alpha }\in \mathcal{L}(\mathbb{R}^3,\mathbb{R}^3)$$ the rotation around the $$Y$$-axis defined by the matrix

$$[\operatorname{rot}_\alpha ]=\begin{bmatrix} \cos \alpha &0&-\sin \alpha \\0&1&0\\\sin \alpha &0&\cos \alpha \end{bmatrix}\tag1$$

in standard coordinates. Also we will set $$C:=\{(x,y,z)\in \mathbb{R}^3: x^2+y^2=z^2\}$$. Finally we will set the parametrization

$$g_{\alpha }:\mathbb{R}^2\to \mathbb{R}^3,\, (x,y)\mapsto \operatorname{rot}_\alpha(x,y,1) \tag2$$

Then the conics, in local coordinates, are given by $$C_{\alpha }:=g^{-1}_\alpha (C\cap \operatorname{rot}_\alpha (P))$$. If we use the Lorentz metric $$\mathfrak{m}=(dx)^2+(dy)^2-(dz)^2$$ then in local coordinates it take the form

$$g_\alpha ^* \mathfrak{m}=(dy)^2+(\cos ^2\alpha -\sin^2 \alpha )(dx)^2\tag3$$

Then $$C_0$$ is the circle, that have the arc-length parametrization $$t\mapsto (\cos t,\sin t)$$, and $$C_{\pi/2}$$ is the half of a hyperbola parametrized by arc-lenght by $$t\mapsto (\cosh t,\sinh t)$$.

Now, we are interested in the parabola $$C_{\pi/4}$$, what have the arc-length parametrization given by the map $$t\mapsto (t^2/2,t)$$, thus we can define the parabolic sine and cosine by

$$\operatorname{cosp}(t):=\frac{t^2}{2},\quad \operatorname{sinp}(t):=t\tag4$$

Similarly, for any $$\alpha$$, we can define trigonometric functions associated to $$C_\alpha$$. Note that these trigonometric functions $$(x,y)$$ solve the differential equation

$$(\dot y)^2+(\cos ^2\alpha -\sin ^2\alpha )(\dot x)^2=1\tag5$$

(In this model the parabola in local coordinates is given by the equation $$2x-y^2=0$$.)

Yes, indeed they exist as the parabola "shallow" parametric expressions of lowest order sine/cos power series when originating circular and hyperbolic functions are lost in linearization.

A short truncation of sine/cosine series only up to second order expansions in parametric form result in a parabolas.

Circular trig

$$x = \cos t \approx 1- t^2/2!+...,\quad y = t - t^3/3!+.. \approx t$$ Eliminate $$t$$ and the parabola whose symmetry axis parallel to x-axis opening to left t: $$\frac{y^2}{2}= (1-x)$$

Hyperbolic trig

$$x = \cosh t \approx 1+ t^2/2!+...,\quad y = t + t^3/3!+.. \approx t$$ Eliminate $$t$$ and the parabola whose symmetry axis parallel to x-axis opening to right:

$$\frac{y^2}{2}= (x-1)$$

So their "shallow" parabolic parametrization is

$$(y=t,\; x= 1\pm t^2/2\;).$$