# How do you prove Osborn's rule?

Given a trigonometric identity written in terms of sine and cosine, it is possible to write down the corresponding hyperbolic identity using Osborn's rule:

• Replace $$\cos$$ with $$\cosh$$
• Replace $$\sin$$ with $$\sinh$$

However, negate the product of two $$\sinh$$ terms. For example, $$\cos^2x+\sin^2x \equiv 1 \implies \cosh^2x-\sinh^2x=1 \, .$$ I tried proving this rule using the following two identities \begin{align} \cos ix &\equiv \cosh x \\ \sin ix &\equiv i\sinh x \, . \end{align} Given an arbitrary trigonometric identity of the form $$f(\cos z,\sin z) \equiv g(\cos z, \sin z)$$ then letting $$z=ix$$ we see that \begin{align} f(\cos ix,\sin ix) &\equiv g(\cos ix, \sin ix) \\ f(\cosh x,i\sinh x) &\equiv g(\cosh x, i\sinh x) \\ \end{align} From here, it seems the rule makes sense because the product of two $$i\sinh x$$ terms gives us $$\color{red}{-}\sinh x$$. But I'm not quite sure how to formalise this argument. Moreover, an identity such as $$\sin 2z \equiv 2\sin z \cos z$$ would correspond to $$i\sinh x \equiv 2i\sinh x \cosh x \, .$$ Here, we could divide through by $$i$$ and the expected hyperbolic identity pops out, but I'm unsure if this 'divide by $$i$$' trick always works, since there may be other terms that don't contain an $$i$$ in them. To give a slightly contrived example, $$\sin x + \cos 2x \equiv \sin x + \cos^2 x - \sin^2x$$ corresponds to $$i\sinh x + \cosh 2x \equiv i\sinh x + \cosh^2 x + \sinh^2 x \, ,$$ and here, dividing by $$i$$ does not help. So I think I've understood the crux of how to prove Osborn's rule, but there are a few loose ends that I need to sort out.

• It is an interesting property, I didn't know this. Thank you :) Commented Feb 5, 2021 at 15:05
• Does this answer your question, though? Proof of Osborne's Rule Commented Feb 5, 2021 at 15:07
• @VIVID Thank you very much for providing me with the link, but I'm not quite satisfied with the answer given. I think Willie got to the stage that I got to in my attempted proof but he didn't give any more detail. Basically, I'm unsure how to deal with the '$i$' that results from the identity $\sin ix = i\sinh x$.
– user883638
Commented Feb 5, 2021 at 15:11

It's worth noting the origin of Osborn's Rule among a list of brief "Mathematical Notes" in The Mathematical Gazette, Vol 2 No. 34 (Jul 1902), pg 189 (JSTOR link). See this screenshot.

Transcribing in its entirety ...

1. [D.6.d.] Mnemonic for hyperbolic formulae.

Hyperbolic functions are now so constantly used, that a brief mnemonic for their somewhat confusing formulae may not be unwelcome.

In any Trigonometrical formula for $$\theta$$, $$2\theta$$, $$3\theta$$, or $$\theta$$ and $$\phi$$, after changing sin to sinh, cos to cosh, etc., change the sign of any term that contains (or implies) a product of sinhs, eg $$\tanh\theta \tanh\phi$$ implies a product of sinhs,

\begin{align} \therefore \tanh(\theta+\phi) &= \frac{\tanh\theta+\tanh\phi}{1+\tanh\theta\tanh\phi}\; ; \\[4pt] \sinh3\theta &= 3\sinh\theta + 4\sinh3\theta\; ; \\[4pt] \cosh\theta-\cosh\phi &= +2\sinh\frac{\theta+\phi}{2}\sinh\frac{\theta-\phi}{2}\; ; \end{align} and so on. This rule would likely fail for terms of the fourth degree, but it covers everything that is likely to be required, and is very convenient for teaching purposes.

G. Osborn

Despite the overly-ambitious ---one might say, hyperbolic--- claim of applicability to "any Trigonometrical formula", it seems clear that what has come to be known as "Osborn's Rule" was intended, and has ever only served, as something more like "Osborn's Rule of Thumb": it's a quick-and-easy way (especially for students) to remember when to change signs in familiar fundamental Trigonometrical formulas.

The Rule is a teaching tool, not an over-arching general principle.

• Thanks Blue. It does make more sense to think of it as a 'rule of thumb', especially because my attempts to formalise what a 'trigonometric identity' actually is probably weren't that successful. Do you know of a specific example, though, of when Osborn's rule might fail?
– user883638
Commented Feb 5, 2021 at 18:42
• @JoeLamond: Do you know of a specific example, though, of when Osborn's rule might fail? ... Your "slightly contrived example" already does that trick. ... Formally, the principle is just: replace "$\cos$" with "$\cosh$", and "$\sin$" with "$i\sinh$" (in accordance with the identities you noted). Thus, Osborn's Rule works when and only when ("whenn"?) sine products cause the "$i$"s combine to become "$-1$"s. That said, if you're sophisticated enough to introduce "$i$"s into formulas, then you don't need the Rule. Osborn likely meant the Rule for students not-yet-aware of complex numbers.
– Blue
Commented Feb 5, 2021 at 18:59
• +1 for the aptly-inserted pun. heh Commented Apr 29, 2022 at 4:35

You could proceed in the following way:$$\sin x + \cos 2x \equiv \sin x + \cos^2 x - \sin^2x$$ corresponds to $$i\sinh x + \cosh 2x \equiv i\sinh x + \cosh^2 x + \sinh^2 x \, ,$$

consequently, by equalling the imaginary parts and the real ones, we get that,

$$\begin{cases} \sinh x\equiv\sinh x\\ \cosh 2x\equiv\cosh^2 x+\sinh^2 x \end{cases}$$

$$\sinh x + \cosh 2x \equiv \sinh x + \cosh^2 x + \sinh^2 x\,.$$