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In trigonometry it is conventional when defining $\cos\theta$, $\sin\theta$, etc., to parametrize the circle by arc length $\theta$. Some trigonometric identities don't depend at all on which parametrization of the circle is used, e.g. $\cos^2\theta+\sin^2\theta=1$. For others that choice of parametrization is crucial, e.g. $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$. To me it's amazing how MUCH follows from the simple identities taught to 15-year-olds, things like this and this and this and this, and if you want something yet more sophisticated, this item bearing a famous name. By current stereotype the term "trigonometric identity" connotes something rather humble, but look at that last item.

In a previous question I asked about the role of addition and subtraction. But now I will pose a different question:

Does anything similarly elaborate (the variety of different sorts of identities, some simple; others more involved; some that could bear famous names) happen with any other parametrizations of the circle?

Perhaps the most notable among other parametrizations --- almost surely one of the most well known --- is this: $$ \begin{align} x & = \frac{1-t^2}{1+t^2} \\[8pt] y & = \frac{2t}{1+t^2} \end{align} $$ (The parameter space should be the one-point compactification $\mathbb{R}\cup\{\infty\}$.)

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Since the second one identifies rational points with rational $t$, it is more useful for number-theoretic purposes, and with some work gets us Euclid's characterization of Pythagorean triples. And in effect it gives the Weierstrass substitution for integrals. (I was dropping names.) –  André Nicolas May 24 '12 at 21:18
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Probably not. The conventional parameterization is rather special as it arises from the exponential map $\mathfrak{o}(1) \to \text{O}(1)$. –  Qiaochu Yuan May 25 '12 at 0:13
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I've done it! I've shown that the answer is affirmative with close relatives of "the most notable" one. –  Michael Hardy Dec 23 '12 at 1:08
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up vote 6 down vote accepted

Incidentally, I spent some time a few years ago trying to find connections between parametrisation and natural trigonometry of closed plain curves. My hypothesis is that the answer to your question is generally negative.

Since the statement of your question is rather loose, I shall confine my argument to a certain understanding of trigonometry which I believe to be of relevance.

So I suppose trigonometry is really about finding relations between arcs of a closed curve in general and circumference in particular. It is probably a coincidence that the fundamental relation between sine and cosine gives the corresponding parametrisation.

Now I will try to support my generic statements in a moderately rigorous way by constructing a counter-example.

Basing on the above understanding of trigonometry we shall adopt the following definitions of sine an cosine on terms of the arc length:

$$\phi=\int_0^{\sin \phi}\frac{dt}{\sqrt{1-t^2}} \qquad \phi'=\int_{\cos \phi'}^1\frac{dt}{\sqrt{1-t^2}}$$

Now take the second integral, say, and substitute:

$$t^2=1-z^2$$

$$\phi=\int_{\cos\phi}^{1}\frac{dt}{\sqrt{1-t^2}}=-\int_{\sqrt{1-\cos^2\phi}}^0\frac{z\,dz}{z\sqrt{1-z^2}}=\int_{0}^{\sqrt{1-\cos^2\phi}}\frac{dz}{\sqrt{1-z^2}}$$

Comparing the limits we deduce the fundamental identity as a direct implication of the definitions.

Now consider the following generalisation. Define lemniscate sine and cosine (see Wolfram article for details) as follows:

$$\phi=\int_0^{\operatorname{sinlemn}\phi}\frac{dt}{\sqrt{1-t^4}}\qquad\phi=\int_{\operatorname{coslemn}\phi}^{1}\frac{dt}{\sqrt{1-t^4}}$$

Now in a similar fashion let:

$$t^2=\frac{1-z^2}{1+z^2}$$

$$dt=-\sqrt{\frac{1+z^2}{1-z^2}}\frac{2z^3\,dz}{(1+z^2)^2}$$ $$\phi=-\int_{\sqrt{\frac{1-\operatorname{coslemn}^2\phi}{1+\operatorname{coslemn}^2\phi}}}^0\frac{1}{z}\frac{(1+z^2)}{2z^2}\sqrt{\frac{1+z^2}{1-z^2}}\frac{2z^3\,dz}{(1+z^2)^2}=\int_0^{\sqrt{\frac{1-\operatorname{coslemn}^2\phi}{1+\operatorname{coslemn}^2\phi}}}\frac{dz}{\sqrt{\frac{1-z^2}{1+z^2}}(1+z^2)} =\int_0^{\sqrt{\frac{1-\operatorname{coslemn}^2\phi}{1+\operatorname{coslemn}^2\phi}}}\frac{dz}{\sqrt{1-z^4}}=\int_0^{\sqrt{\frac{1-\operatorname{coslemn}^2\phi}{1+\operatorname{coslemn}^2\phi}}}\frac{dt}{\sqrt{1-t^4}}$$

Hence, we deduce

$$\operatorname{coslemn}^2\phi=\frac{1-\operatorname{coslemn}^2\phi}{1+\operatorname{coslemn}^2\phi}$$

which is an analogue to the fundamental identity for lemniscate functions. Whittaker and Watson offer the reader to prove this using the connection with Jacobi elliptic functions (see Wolfram page). Using the same connection it is possible to prove, for example, the addition theorem

$$\operatorname{sinlemn}(x+y)=\frac{\operatorname{sinlemn}x \operatorname{coslemn} y +\operatorname{coslemn}x \operatorname{sinlemn} y}{1-\operatorname{sinlemn}x \operatorname{sinlemn}y \operatorname{coslemn} x \, \operatorname{coslemn} y}$$

Some other identities which are not exactly relevant to the present discussion can be deduced from definitions.

Lemniscate is a well-known $\infty$-shaped algebraic curve of the fourth order

lemniscate

given in Cartesian coordinates by the equation:

$$(x^2+y^2)^2=a^2(x^2-y^2)$$

and in polar coordinates by

$$\rho=\sqrt{\cos2\theta}$$

Setting $$y=xt$$ it is also possible to obtain a rational parametrisation.

Now if we plot the identity for the lemniscate functions, we get something that is nowhere near that shape:

lemniscate identity

Still, lemniscate functions do give a representation for the lemniscate in the following polar form: $$\rho=\operatorname{sinlemn} \theta$$ in the same fashion as the circle is given by $$\rho=\sin \theta$$

Thus, the point is that natural definition of trigonometric functions for a certain curve and parametrisation of the same curve are two different tasks whose results are not necessarily linked to each other.

Some interesting implications follow.

1) Obviously, the definition of a trigonometric system can be generalised to other closed curves with the apparent definition: $$\phi=\int_0^{\mathscr{S}(\phi)}dl\qquad \phi=\int_{\mathscr{C}(\phi)}^1 dl$$

where it is easy to whow that, if

$$\omega=\int_0^1 dl$$

then

$$\mathscr{S}\left(\phi-\frac{\omega}{2}\right)=\mathscr{C}\left(\phi\right)$$

2) Generalized hyperbolic functions can be defined via area under a curve. In the lemniscate case it is easy to show ($x\to ix$) that the conventional relation with the "trigonometric" functions holds and the responding "hyperbola" ($y^2\to-y^2$) has 4 branches.

3) Looking at the tricks used to transform integrals (which are largely reverse engineering) we may also say that trigonometric functions have the property of leaving a differential form $p(x) \, dx$ invariant under the change of coordinates. Perhapse this could be generalised to the case $\sum p_i \, dx_i\ldots$

Hope I was not taken too much off the tangent, but then one can do sophisticated things in trigonometry

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Quite interesting---tomorrow I'm going to print this out and sufficiently digest it to figure out what to think. I've taken a liberty with your $\TeX$ code and changed \text{sinlemn} to \operatorname{sinlemn}. One effect is the spacing between "$\operatorname{sinlemn}$" and what comes before and after it. Thus: $\operatorname{sinlemn}\alpha\operatorname{coslemn}\beta$. You see space between $\alpha$ and $\operatorname{coslemn}$ and between that and $\beta$, without any need for things like "\,". –  Michael Hardy May 25 '12 at 4:29
    
Thanks for the note. I have a very superficial knowledge of Tex hoping to improve it here. –  Valentin May 25 '12 at 20:03
    
I finally got around to printing this. I don't have a printer at home; hence some delay...... –  Michael Hardy May 28 '12 at 2:09
    
no worries, i've been enjoying my first week on SE :) –  Valentin May 28 '12 at 22:49
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If the lemniscatic functions being described by Valentin caught your fancy, you'll enjoy A.I. Markushevich's The Remarkable Sine Functions; in that book, he discusses that and a number of other analogs of the usual sine. –  J. M. Jul 7 '12 at 19:36
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