How do you prove Osborn's rule?

Question

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.

Answer 1

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 ...

109. [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.

Answer 2

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,

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

and, by adding both sides,

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