2018年10月2日 · And hence every trigonometric identity can be easily transformed into a hyperbolic identity and vice versa. Once you prove that $\exp' = \exp$ , you can recover all the basic properties of $\exp$ and hence $\cosh,\sinh,\cos,\sin$ , including:

2020年10月7日 · Indeed, they are a symmetry group for the Minkowski product of hyperbolic $2$-vectors. So the difference between the circular and hyperbolic identities' geometric motivations can be thought of not as the difference between a Euclidean plane's circles and hyperbolae, but instead as the difference between the norms of Euclidean and hyperbolic spaces.

2021年1月25日 · Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts. The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$ .

2020年1月4日 · We know that $\cos(\theta)=\cosh(i\theta)$ and $\sin(\theta)=-i\sinh(i\theta)$.I have figured out and have been told on various other posts and websites that these relationships can be used to prove identities involving hyperbolic trigonometric functions simply by converting them to their circular forms, applying the corresponding circular identity, and then reconverting.

2017年4月18日 · I have been asked to convert a circular trigonometric identity into its corresponding hyperbolic version using Osborn’s rule. The identity is $\cos2x = 1-2\sin^2x$ I know that $\cos2x = \cos^2x-\sin^2x $ Then converting to hyperbolic version gives: $\cosh^2x-\sinh^2x = 1$ Therefore left hand side conversion is $\cos2x = 1$

I've been working on hyperbolic functions lately. All is well, however I seem to come across a couple difficulties here and there when it comes to actually proving hyperbolic identities, thus I'm

Here is the handout from a talk I gave on deriving the hyperbolic trig functions—this is actually a packet guiding a student through the derivation. More or less, it starts with the circular trig functions, shifts the definition to depend on area rather than arc length, constructs the comparable definition in terms of a unit hyperbola, and ...

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 ...

2019年2月12日 · I'm pretty sure the missing steps are just alegbra involving hyperbolic trig functions, which is why I'm asking here instead of on physics stack exchange. The derivation arrives at the expression $\frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} = e^{2 \eta} $. They next say, "Then, introducing the hyperbolic trigonometric functions,

2015年3月21日 · For 4n+2 if it only occurs with identities involving cosh(a+b) you don't divide because cos(ix) = cosh(x). for similar reasons with 4n+3 you divide by -i. This rule can be proven with individual cases but imo they require far too much unnecessary thinking to describe using some sort of intuition