Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When proving identities like $$\cosh(2x)=\cosh^2(x)+\sinh^2(x)$$ $$\cosh^2(x)=\sinh^2(x)+1$$ algebraically, I am beset by the feeling that there should be a geometrical interpretation that makes them immediately obvious. Most of the analogous identities involving $\sin,$ $\cos$ have such interpretations.

Is it possible to use hyperbolic geometry to prove identities in hyperbolic trigonometry geometrically? Any examples would be greatly appreciated (if this is in fact possible).


Addendum: If there is no analogy in hyperbolic trigonometry, could the complex plane and the relationship between hyperbolic and non-hyperbolic functions be used instead (i.e. $\cos(x)= \cosh(ix), \sin(x)=-i\sinh(ix))$?

The second identity is equivalent to saying that for all $x\in\mathbb{R}$ there is a triangle with hypotenuse $\cosh(x)$ and sides $\sinh(x),1$. Is there a geometric interpretation of $x$?

share|improve this question
2  
I strictly dislike sending links to Wikipedia when it is about math and/or programming, but I remember that en.wikipedia.org/wiki/Hyperbolic_function has some nice illustrations on that topic. –  Evgeny Oct 14 '13 at 18:07
1  
@Evgeny I agree that it is excellent, but it's all either algebra or Euclidean geometry (I could be an idiot in thinking that hyperbolic geometry has anything to do with hyperbolic functions, Euclidean hyperbolae could be the only geometrical interpretation). –  Alyosha Oct 14 '13 at 18:14
3  
Illustrating (circular) trig identities with diagrams is aided by nature of similar triangles in Euclidean geometry. Drop a few perpendiculars here, add a parallel or two there, maybe inscribe an angle in a circle, and you can create lots of similar triangles to make your case. Non-Euclidean (in particular, hyperbolic) geometry lacks this utility. There are no similar-but-not-congruent triangles, so diagrams like ones you link just don't work. (The altitude to the hypotenuse of a right triangle does not create similar sub-triangles.) [continued] –  Blue Oct 14 '13 at 18:21
1  
(Part 2) When you note that the Law of Sines in hyperbolic geometry is $$\frac{\sin A}{\sinh a} = \frac{\sin B}{\sinh b} = \frac{\sin C}{\sinh c}$$ you realize that the notion of proportionality of (raw) side-lengths is decidedly unhelpful; even the impact of addition and subtraction of lengths is unclear. Every geometric formula wraps lengths in some hyperbolic trig function ... which makes relations really hard (impossible?) to represent in intuitive visuals. In fact, this difficulty leads me to muse often about the question "What are trig classes like in a hyperbolic universe?" –  Blue Oct 14 '13 at 18:34
3  
(Part 3) A final thought on this. Consider the Pythagorean Theorems in Euclidean and Hyperbolic geometries:$$a^2 + b^2 = c^2 \qquad \text{vs} \qquad \cosh a \; \cosh b = \cosh c$$ (I'm biased of course, but...) It seems that the notion of squaring numbers is fairly natural, so it's not too hard to accept that ancient geometers "noticed" the sum-of-squares formula. But what culture goes around cosh-ing numbers with such ease that the product-of-cosines is comparably "noticeable"? How does fluency in transcendental exponentials precede discovery of a (the?) fundamental geometric relation? –  Blue Oct 14 '13 at 18:55

2 Answers 2

The answer to your question is yes - by now you've probably already seen this and moved on, but I post this anyway for the benefit of other readers. The hyperbolic functions, such as $\sinh$ and $\cosh$, are one of those topics that is almost always skipped right over in undergraduate math, then if it happens to pop up after that everyone acts like you must have seen it a million times. As a result there seem to be a lot of questions like this on the web, often with overly complicated answers. Anyway, here's my attempt at aiding that situation.

The hyperbolic functions are so called because they arise from the hyperbola $x^2-y^2=1$, analogously to how the ordinary trigonometric functions relate to the circle $x^2+y^2=1$ (but not quite in the way you might think). Recall how if $\theta$ is an angle off the positive $x$-axis in $\mathbb{R}^2$, going counterclockwise, then the point on the unit circle intercepted by $\theta$ will be $(\cos\theta, \sin\theta)$. We want to do something similar, replacing the circle with the hyperbola as mentioned, but for the analogy to work we need to describe the unit circle scenario a little differently.

First of all, get a good picture in your head (or draw one, or look one up) of the hyperbola $x^2-y^2=1$. It has two components: one on either side of the $y$-axis, and if we consider the places where an angle $\theta$ as above would intercept a point on it, we would be limited to $\theta\in(-\frac{\pi}{4},\frac{\pi}{4})\cup(\frac{3\pi}{4},\frac{5\pi}{4})$ plus multiples of $2\pi$. But as you can see from the exponential definition of these functions, $\cosh$ and $\sinh$ have domain $\mathbb{R}$.

So, instead of thinking about the angle $\theta$, think about the area $\alpha$ bounded by its initial ray (the $x$-axis), its terminal ray, and the circle. Putting $\theta$ in radians, this area is $\alpha=\theta/2$ (for instance, the area of the whole circle is $\pi$, which is traversed by the full rotation $2\pi$). Looking now at the point intercepted by $\theta$ on the circle in terms of $\alpha$, we have $(\cos 2\alpha,\sin 2\alpha)$.

This is the correct setup for moving to the hyperbolic setting. Suppose $\alpha$ is now the area bounded by the $x$-axis, some other ray $\rho$ coming out of the origin, and the hyperbola $x^2-y^2=1$. Now identify the point on the hyperbola intercepted by $\rho$. The coordinates of this point will be $(\cosh 2\alpha, \sinh 2\alpha)$. Note that the $\alpha$ values you can get in this way are arbitrarily large, even though to bound an area $\rho$ can't pass (or hit) $\frac{\pi}{4}$ off the $x$-axis - using some calculus (and probably an integral table - the integral is ugly) you can see that the area between the diagonal line and the hyperbola, in the first quadrant, is infinite. If you want negative $\alpha$, just move your ray down and call the area negative.

This gives your second identity easily. The first one takes a little more work.

share|improve this answer
    
See also this answer of mine, which goes through the calculus involved. –  Blue Oct 30 at 0:28
    
Very cool, I have a feeling I'll enjoy a closer look at that in the not too distant future. I'm doing some old Thurston exercises & noticing some holes in my knowledge base! –  j0equ1nn Oct 30 at 0:59

Of course, they admit just the same interpretation in 1+1-dimensional pseudo-Euclidean space (with “dx2 − dv2” vector magnitude form) as trigonometric functions have in 2-dimensional Euclidean space. That pseudo-Euclidean space is also known as (the plane of) split-complex numbers and has its “trigonometric circle” with equation “x2 − v2 = 1” and hyperbolic angles. One minor difference: the “x > 0” inequality should be added to aforementioned equation to restrict trigonometric eh… hyperbola only to polar angles from −∞ to +∞.

Explanation with analytic functions on (standard) complex numbers is not mutually exclusive with metric signature argument presented above. If you consider ℂ2 = {(z, w)} where z = x + iy, w = u + iv, x, y, u, v ∈ ℝ, with quadratic form dz2 + dw2, then restriction to a totally real subspace of {(x, u)} with y = v = 0 gives Euclidean geometry with standard trigonometry, whereas restriction to the one of {(x, v)} with y = u = 0 gives aforementioned pseudo-Euclidean geometry with hyperbolic trigonometry. In other words, complexification of both 2-dimensional real spaces gives the same thing. Now you can write a parametric presentation of a line: $$ z = \frac{e^α + e^{-α}}2 t\\ w = i\frac{e^α - e^{-α}}2 t $$ and say that real α are hyperbolic angles (with corresponding lines lying in the pseudo-Euclidean plane for real t) and imaginary α are standard (circular) angles (likewise, in Euclidean plane). Note an identity $$(\frac{e^α + e^{-α}}2)^2 + (i\frac{e^α - e^{-α}}2)^2 = 1,$$ so t is (generally, complex) natural parameter: t2 = z2 + w2.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.